6.11 IONIZATION OF ACIDS AND BASES and the equilibrium will shift in the direction of weaker acid. Say, if HA is a stronger acidArrhenius concept of acids and bases than H3O+, then HA will donate protons andbecomes useful in case of ionization of acids not H3O+, and the solution will mainly containand bases as mostly ionizations in chemical A– and H3O+ ions. The equilibrium moves inand biological systems occur in aqueous the direction of formation of weaker acid medium. Strong acids like perchloric acid Reprint 2025-26 EQUILIBRIUM 193 and weaker base because the stronger acid H2O(l) + H2O(l) H3O+(aq) + OH–(aq) donates a proton to the stronger base. acid base conjugate conjugate It follows that as a strong acid dissociates acid base completely in water, the resulting base formed The dissociation constant is represented by, would be very weak i.e., strong acids have K = [H3O+] [OH–] / [H2O] (6.26)very weak conjugate bases. Strong acids like perchloric acid (HClO4), hydrochloric acid The concentration of water is omitted from (HCl), hydrobromic acid (HBr), hydroiodic acid the denominator as water is a pure liquid and (HI), nitric acid (HNO3) and sulphuric acid its concentration remains constant. [H2O] is (H2SO4) will give conjugate base ions ClO4–, Cl, incorporated within the equilibrium constant Br–, I–, NO3– and HSO4– , which are much weaker to give a new constant, Kw, which is called the bases than H2O. Similarly a very strong base ionic product of water. would give a very weak conjugate acid. On the Kw = [H+][OH–] (6.27) other hand, a weak acid say HA is only partially The concentration of H+ has been founddissociated in aqueous medium and thus, the out experimentally as 1.0 × 10–7 M at 298 K.solution mainly contains undissociated HA And, as dissociation of water produces equalmolecules. Typical weak acids are nitrous number of H+ and OH– ions, the concentrationacid (HNO2), hydrofluoric acid (HF) and acetic of hydroxyl ions, [OH–] = [H+] = 1.0 × 10–7 M.acid (CH3COOH). It should be noted that the Thus, the value of Kw at 298K,weak acids have very strong conjugate bases. For example, NH2–, O 2– and H– are very good Kw = [H3O+][OH–] = (1 × 10–7)2 = 1 × 10–14 M2 proton acceptors and thus, much stronger (6.28) bases than H2O. The value of Kw is temperature dependent Certain water soluble organic compounds as it is an equilibrium constant. like phenolphthalein and bromothymol blue The density of pure water is 1000 g / Lbehave as weak acids and exhibit different and its molar mass is 18.0 g /mol. From thiscolours in their acid (HIn) and conjugate base the molarity of pure water can be given as,(In– ) forms. [H2O] = (1000 g /L)(1 mol/18.0 g) = 55.55 M.HIn(aq) + H2O(l) H3O+(aq) + In–(aq) Therefore, the ratio of dissociated water toacid conjugate conjugate that of undissociated water can be given as: indicator acid base 10–7 / (55.55) = 1.8 × 10–9 or ~ 2 in 10–9 (thus, colour A colour B equilibrium lies mainly towards undissociated Such compounds are useful as indicators water) in acid-base titrations, and finding out H+ ion We can distinguish acidic, neutral andconcentration. basic aqueous solutions by the relative values 6.11.1 The Ionization Constant of Water of the H3O+ and OH– concentrations: and its Ionic Product Acidic: [H3O+] > [OH– ]Some substances like water are unique in Neutral: [H3O+] = [OH– ]their ability of acting both as an acid and a base. We have seen this in case of water in Basic : [H3O+] < [OH–] section 6.10.2. In presence of an acid, HA it accepts a proton and acts as the base while 6.11.2 The pH Scale in the presence of a base, B– it acts as an Hydronium ion concentration in molarity is acid by donating a proton. In pure water, one more conveniently expressed on a logarithmic H2O molecule donates proton and acts as an scale known as the pH scale. The pH of a acid and another water molecules accepts a solution is defined as the negative logarithm proton and acts as a base at the same time. of hydrogen to base 10 of the activity  aH The following equilibrium exists: Reprint 2025-26 194 chemistry ion. In dilute solutions (< 0.01 M), activity when the hydrogen ion concentration, [H+] of hydrogen ion (H+) is equal in magnitude changes by a factor of 100, the value of pH to molarity represented by [H+]. It should changes by 2 units. Now you can realise why be noted that activity has no units and is the change in pH with temperature is often defined as: ignored. = [H+] / mol L–1 Measurement of pH of a solution is very essential as its value should be known From the definition of pH, the following when dealing with biological and cosmeticcan be written, applications. The pH of a solution can be pH = – log aH+ = – log {[H+] / mol L–1} found roughly with the help of pH paper that has different colour in solutions of different Thus, an acidic solution of HCl (10–2 M) will have a pH = 2. Similarly, a basic solution pH. Now-a-days pH paper is available with of NaOH having [OH–] =10–4 M and [H3O+] = four strips on it. The different strips have 10–10 M will have a pH = 10. At 25 °C, pure different colours (Fig. 6.11) at the same pH. water has a concentration of hydrogen ions, The pH in the range of 1-14 can be determined [H+] = 10–7 M. Hence, the pH of pure water is with an accuracy of ~0.5 using pH paper. given as: pH = –log(10–7) = 7 Acidic solutions possess a concentration of hydrogen ions, [H+] > 10–7 M, while basic solutions possess a concentration of hydrogen ions, [H+] < 10–7 M. thus, we can summarise that Fig.6.11 pH-paper with four strips that may haveAcidic solution has pH < 7 different colours at the same pH Basic solution has pH > 7 For greater accuracy pH meters are used.Neutral solution has pH = 7 pH meter is a device that measures the Now again, consider the equation (6.28) pH-dependent electrical potential of the testat 298 K solution within 0.001 precision. pH meters Kw = [H3O+] [OH–] = 10–14 of the size of a writing pen are now available Taking negative logarithm on both sides in the market. The pH of some very common of equation, we obtain substances are given in Table 6.5 (page 195). –log Kw = – log {[H3O+] [OH–]} Problem 6.16 = – log [H3O+] – log [OH–] –14 The concentration of hydrogen ion in a = – log 10 sample of soft drink is 3.8 × 10–3M. what pKw = pH + pOH = 14 (6.29) is its pH ? Note that although Kw may change with temperature the variations in pH with Solution temperature are so small that we often pH = – log[3.8 × 10–3] ignore it. = – {log[3.8] + log[10–3]} pKw is a very important quantity for = – {(0.58) + (– 3.0)} = – { – 2.42} = 2.42 aqueous solutions and controls the relative Therefore, the pH of the soft drink is 2.42 concentrations of hydrogen and hydroxyl and it can be inferred that it is acidic. ions as their product is a constant. It should Problem 6.17be noted that as the pH scale is logarithmic, Calculate pH of a 1.0 × 10 –8 M solution ofa change in pH by just one unit also means HCl.change in [H+] by a factor of 10. Similarly, Reprint 2025-26 EQUILIBRIUM 195 Table 6.5 The pH of Some Common Substances Name of the Fluid pH Name of the Fluid pH Saturated solution of NaOH ~15 Black Coffee 5.0 0.1 M NaOH solution 13 Tomato juice ~4.2 Lime water 10.5 Soft drinks and vinegar ~3.0 Milk of magnesia 10 Lemon juice ~2.2 Egg white, sea water 7.8 Gastric juice ~1.2 Human blood 7.4 1M HCl solution ~0 Milk 6.8 Concentrated HCl ~–1.0 Human Saliva 6.4 equilibrium constant for the above discussed Solution acid-dissociation equilibrium: 2H2O (l) H3O+ (aq) + OH–(aq) Ka = c2α2 / c(1-α) = cα2 / 1-α Kw = [OH–][H3O+] Ka is called the dissociation or ionization = 10–14 constant of acid HX. It can be represented Let, x = [OH–] = [H3O+] from H2O. The alternatively in terms of molar concentration H3O+ concentration is generated (i) from as follows, the ionization of HCl dissolved i.e., Ka = [H+][X–] / [HX] (6.30) HCl(aq) + H2O(l) H3O+ (aq) + Cl –(aq), At a given temperature T, Ka is a and (ii) from ionization of H2O. In these very measure of the strength of the acid HX i.e., dilute solutions, both sources of H3O+ must larger the value of Ka, the stronger is the be considered: acid. Ka is a dimensionless quantity with [H3O+] = 10–8 + x the understanding that the standard state Kw = (10–8 + x)(x) = 10–14 concentration of all species is 1M. or x2 + 10–8 x – 10–14 = 0 The values of the ionization constants [OH– ] = x = 9.5 × 10–8 of some selected weak acids are given in Table 6.6. So, pOH = 7.02 and pH = 6.98 Table 6.6 The Ionization Constants of Some 6.11.3 Ionization Constants of Weak Acids Selected Weak Acids (at 298K) Consider a weak acid HX that is partially Acid Ionization Constant, ionized in the aqueous solution. The Ka equilibrium can be expressed by: Hydrofluoric Acid (HF) 3.5 × 10–4 HX(aq) + H2O(l) H3O+(aq) + X–(aq) Nitrous Acid (HNO2) 4.5 × 10–4 Initial Formic Acid (HCOOH) 1.8 × 10–4 concentration (M) c 0 0 Niacin (C5H4NCOOH) 1.5 × 10–5 Let α be the extent of ionization Acetic Acid (CH3COOH) 1.74 × 10–5 Change (M) Benzoic Acid (C6H5COOH) 6.5 × 10–5 -cα +cα +cα Hypochlorous Acid (HCIO) 3.0 × 10–8 Equilibrium concentration (M) Hydrocyanic Acid (HCN) 4.9 × 10–10 c-cα cα cα Phenol (C6H5OH) 1.3 × 10–10 Here, c = initial concentration of the undissociated acid, HX at time, t = 0. α = The pH scale for the hydrogen ion extent up to which HX is ionized into ions. concentration has been so useful that besides Using these notations, we can derive the pKw, it has been extended to other species and Reprint 2025-26 196 chemistry quantities. Thus, we have: Solution pKa = –log (Ka) (6.31) The following proton transfer reactions are Knowing the ionization constant, Ka possible: of an acid and its initial concentration, c, 1) HF + H2O H3O+ + F–it is possible to calculate the equilibrium Ka = 3.2 × 10–4concentration of all species and also the 2) H2O + H2O H3O+ + OH–degree of ionization of the acid and the pH of the solution. Kw = 1.0 × 10–14 As Ka >> Kw, [1] is the principle reaction. A general step-wise approach can be adopted to evaluate the pH of the weak HF + H2O H3O+ + F– electrolyte as follows: Initial Step 1. The species present before concentration (M) dissociation are identified as Brönsted-Lowry 0.02 0 0 acids/bases. Change (M) Step 2. Balanced equations for all possible –0.02α +0.02α +0.02α reactions i.e., with a species acting both as Equilibriumacid as well as base are written. concentration (M)Step 3. The reaction with the higher Ka is identified as the primary reaction whilst the 0.02 – 0.02 α 0.02 α 0.02α other is a subsidiary reaction. Substituting equilibrium concentrations Step 4. Enlist in a tabular form the following in the equilibrium reaction for principal reaction gives:values for each of the species in the primary reaction Ka = (0.02α)2 / (0.02 – 0.02α) (a) Initial concentration, c. = 0.02 α2 / (1 –α) = 3.2 × 10–4 (b) Change in concentration on proceeding We obtain the following quadratic equation: to equilibrium in terms of α, degree of α2 + 1.6 × 10–2α – 1.6 × 10–2 = 0 ionization. The quadratic equation in α can be solved (c) Equilibrium concentration. and the two values of the roots are: α = + 0.12 and – 0.12Step 5. Substitute equilibrium concentrations into equilibrium constant equation for The negative root is not acceptable and hence,principal reaction and solve for α. Step 6. Calculate the concentration of species α = 0.12 in principal reaction. This means that the degree of ionization, α = 0.12, then equilibrium concentrationsStep 7. Calculate pH = – log[H3O+] – of other species viz., HF, F and H3O+ are The above mentioned methodology has given by: been elucidated in the following examples. [H3O+] = [F –] = cα = 0.02 × 0.12 = 2.4 × 10–3 M Problem 6.18 [HF] = c(1 – α) = 0.02 (1 – 0.12) The ionization constant of HF is = 17.6 × 10-3 M 3.2 × 10–4. Calculate the degree of dissociation of HF in its 0.02 M solution. pH = – log[H+] = –log(2.4 × 10–3) = 2.62 Calculate the concentration of all species Problem 6.19 present (H3O+, F – and HF) in the solution The pH of 0.1M monobasic acid is 4.50. and its pH. Calculate the concentration of species H+, A– Reprint 2025-26 EQUILIBRIUM 197 and HA at equilibrium. Also, determine the Percent dissociation value of Ka and pKa of the monobasic acid. = {[HOCl]dissociated / [HOCl]initial }× 100 Solution = 1.41 × 10–3 × 102/ 0.08 = 1.76 %. pH = –log(1.41 × 10–3) = 2.85. pH = – log [H+] Therefore, [H+] = 10 –pH = 10–4.50 6.11.4 Ionization of Weak Bases = 3.16 × 10–5 The ionization of base MOH can be represented by equation: [H+] = [A–] = 3.16 × 10–5 MOH(aq) M+(aq) + OH–(aq) Thus, Ka = [H+][A-] / [HA] In a weak base there is partial ionization [HA]eqlbm = 0.1 – (3.16 × 10-5)  0.1 of MOH into M+ and OH–, the case is similar to that of acid-dissociation equilibrium. The Ka = (3.16 × 10–5)2 / 0.1 = 1.0 × 10–8 equilibrium constant for base ionization pKa = – log(10–8) = 8 is called base ionization constant and is represented by Kb. It can be expressed in Alternatively, “Percent dissociation” is terms of concentration in molarity of various another useful method for measure of strength of a weak acid and is given as: species in equilibrium by the following equation: Percent dissociation Kb = [M+][OH–] / [MOH] (6.33) = [HA]dissociated/[HA]initial × 100% (6.32) Alternatively, if c = initial concentration Problem 6.20 of base and α = degree of ionization of base i.e. the extent to which the base ionizes. Calculate the pH of 0.08M solution of hypochlorous acid, HOCl. The ionization When equilibrium is reached, the equilibrium constant of the acid is 2.5 × 10–5. Determine constant can be written as: the percent dissociation of HOCl. Kb = (cα)2 / c (1-α) = cα2 / (1-α) Solution The values of the ionization constants of some selected weak bases, Kb are given in HOCl(aq) + H2O (l) H3O+(aq) + ClO–(aq) Table 6.7. Initial concentration (M) Table 6.7 The Values of the Ionization 0.08 0 0 Constant of Some Weak Bases at Change to reach 298 K equilibrium concentration Base Kb (M) Dimethylamine, (CH3)2NH 5.4 × 10–4 – x + x +x Triethylamine, (C2H5)3N 6.45 × 10–5 equilibrium concentartion (M) Ammonia, NH3 or NH4OH 1.77 × 10–5 0.08 – x x x Quinine, (A plant product) 1.10 × 10–6 Ka = {[H3O+][ClO–] / [HOCl]} Pyridine, C5H5N 1.77 × 10–9 = x2 / (0.08 –x) Aniline, C6H5NH2 4.27 × 10–10 As x << 0.08, therefore 0.08 – x  0.08 Urea, CO (NH2)2 1.3 × 10–14 x2 / 0.08 = 2.5 × 10–5 Many organic compounds like amines x2 = 2.0 × 10–6, thus, x = 1.41 × 10–3 are weak bases. Amines are derivatives of [H+] = 1.41 × 10–3 M. ammonia in which one or more hydrogen Therefore, atoms are replaced by another group. For example, methylamine, codeine, quinine and Reprint 2025-26 198 chemistry nicotine all behave as very weak bases due to Kb = 10–4.75 = 1.77 × 10–5 Mtheir very small Kb. Ammonia produces OH– in aqueous solution: NH3 + H2O NH4+ + OH– Initial concentration (M) NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) 0.10 0.20 0 The pH scale for the hydrogen ion Change to reachconcentration has been extended to get: pKb = –log (Kb) (6.34) equilibrium (M) –x +x +x At equilibrium (M) Problem 6.21 0.10 – x 0.20 + x x The pH of 0.004M hydrazine solution is 9.7. Kb = [NH4+][OH–] / [NH3] Calculate its ionization constant Kb and pKb. = (0.20 + x)(x) / (0.1 – x) = 1.77 × 10–5 Solution As Kb is small, we can neglect x in comparison NH2NH2 + H2O NH2NH3+ + OH– to 0.1M and 0.2M. Thus, From the pH we can calculate the hydrogen [OH–] = x = 0.88 × 10–5 ion concentration. Knowing hydrogen ion Therefore, [H+] = 1.12 × 10–9 concentration and the ionic product of pH = – log[H+] = 8.95. water we can calculate the concentration of hydroxyl ions. Thus we have: 6.11.5 Relation between Ka and Kb [H+] = antilog (–pH) As seen earlier in this chapter, Ka and Kb = antilog (–9.7) = 1.67 ×10–10 represent the strength of an acid and a base, [OH–] = Kw / [H+] = 1 × 10–14 / 1.67 × 10–10 respectively. In case of a conjugate acid-base pair, they are related in a simple manner so = 5.98 × 10–5 that if one is known, the other can be deduced. The concentration of the corresponding + Considering the example of NH4 and NH3 hydrazinium ion is also the same as that we see, of hydroxyl ion. The concentration of both these ions is very small so the concentration NH4+(aq) + H2O(l) H3O+(aq) + NH3(aq) of the undissociated base can be taken Ka = [H3O+][ NH3] / [NH4+] = 5.6 × 10–10 equal to 0.004M. NH3(aq) + H2O(l) NH4+(aq) + OH–(aq) Thus, Kb =[ NH4+][ OH–] / NH3 = 1.8 × 10–5 Kb = [NH2NH3+][OH–] / [NH2NH2] Net: 2 H2O(l) H3O+(aq) + OH–(aq) = (5.98 × 10–5)2 / 0.004 = 8.96 × 10–7 Kw = [H3O+][ OH– ] = 1.0 × 10–14 M pKb = –logKb = –log(8.96 × 10–7) = 6.04. + Where, Ka represents the strength of NH4 Problem 6.22 as an acid and Kb represents the strength of Calculate the pH of the solution in which NH3 as a base. 0.2M NH4Cl and 0.1M NH3 are present. The It can be seen from the net reaction that pKb of ammonia solution is 4.75. the equilibrium constant is equal to the product of equilibrium constants Ka and Kb Solution for the reactions added. Thus, NH3 + H2O NH4+ + OH– Ka × Kb = {[H3O+][ NH3] / [NH4+ ]} × {[NH4 +] The ionization constant of NH3, [OH–] / [NH3]} Kb = antilog (–pKb) i.e. = [H3O+][OH–] = Kw = (5.6 ×10–10) × (1.8 × 10–5) = 1.0 × 10–14 M Reprint 2025-26 EQUILIBRIUM 199 This can be extended to make a + NH3 + H2O NH4 + OH–generalisation. The equilibrium constant We use equation (6.33) to calculatefor a net reaction obtained after adding hydroxyl ion concentration,two (or more) reactions equals the product [OH–] = c α = 0.05 αof the equilibrium constants for individual reactions: Kb = 0.05 α2 / (1 – α) The value of α is small, therefore the KNET = K1 × K2 × …… (6.35) quadratic equation can be simplified by Similarly, in case of a conjugate acid-base neglecting α in comparison to 1 in the pair, denominator on right hand side of the Ka × Kb = Kw (6.36) equation, Knowing one, the other can be obtained. Thus, It should be noted that a strong acid will have Kb = c α2 or α = √ (1.77 × 10–5 / 0.05) a weak conjugate base and vice-versa. = 0.018. Alternatively, the above expression [OH–] = c α = 0.05 × 0.018 = 9.4 × 10–4M. Kw = Ka × Kb, can also be obtained by [H+] = Kw / [OH–] = 10–14 / (9.4 × 10–4)considering the base-dissociation equilibrium reaction: = 1.06 × 10–11 B(aq) + H2O(l) BH+(aq) + OH–(aq) pH = –log(1.06 × 10–11) = 10.97. Kb = [BH+][OH–] / [B] Now, using the relation for conjugate As the concentration of water remains acid-base pair, constant it has been omitted from the Ka × Kb = Kw denominator and incorporated within the using the value of Kb of NH3 fromdissociation constant. Then multiplying and Table 6.7.dividing the above expression by [H+], we get: We can determine the concentration of Kb = [BH+][OH–][H+] / [B][H+] + conjugate acid NH4 ={[ OH–][H+]}{[BH+] / [B][H+]} Ka = Kw / Kb = 10–14 / 1.77 × 10–5 = Kw / Ka = 5.64 × 10–10. or Ka × Kb = Kw It may be noted that if we take negative logarithm of both sides of the equation, then 6.11.6 Di- and Polybasic Acids and Di- pK values of the conjugate acid and base are and Polyacidic Bases related to each other by the equation: Some of the acids like oxalic acid, sulphuric pKa + pKb = pKw = 14 (at 298K) acid and phosphoric acids have more than one ionizable proton per molecule of the Problem 6.23 acid. Such acids are known as polybasic or polyprotic acids. Determine the degree of ionization and pH of The ionization reactions for example for a 0.05M of ammonia solution. The ionization a dibasic acid H2X are represented by the constant of ammonia can be taken from equations: Table 6.7. Also, calculate the ionization H2X(aq) H+(aq) + HX–(aq) constant of the conjugate acid of ammonia. HX–(aq) H+(aq) + X2–(aq) Solution And the corresponding equilibrium The ionization of NH3 in water is represented constants are given below: by equation: Ka1= {[H+][HX–]} / [H2X] and Reprint 2025-26 200 chemistry Ka2 = {[H+][X2-]} / [HX-] In general, when strength of H-A bond decreases, that is, the energy required to break Here, Ka1and Ka2are called the first and second the bond decreases, HA becomes a stronger ionization constants respectively of the acid H2 acid. Also, when the H-A bond becomes more X. Similarly, for tribasic acids like H3PO4 we polar i.e., the electronegativity difference have three ionization constants. The values between the atoms H and A increases and of the ionization constants for some common there is marked charge separation, cleavage polyprotic acids are given in Table 6.8. of the bond becomes easier thereby increasing Table 6.8 The Ionization Constants of Some the acidity. Common Polyprotic Acids (298K) But it should be noted that while comparing elements in the same group of the periodic table, H-A bond strength is a more important factor in determining acidity than its polar nature. As the size of A increases down the group, H-A bond strength decreases and so the acid strength increases. For example, Size increases HF << HCl << HBr << HI It can be seen that higher order ionization Acid strength increasesconstants (Ka2, Ka3) are smaller than the Similarly, H2S is stronger acid than H2O.lower order ionization constant (Ka1) of a polyprotic acid. The reason for this is that But, when we discuss elements in the same it is more difficult to remove a positively row of the periodic table, H-A bond polarity charged proton from a negative ion due to becomes the deciding factor for determining electrostatic forces. This can be seen in the the acid strength. As the electronegativity case of removing a proton from the uncharged of A increases, the strength of the acid also H2CO3 as compared from a negatively charged increases. For example, HCO3–. Similarly, it is more difficult to remove 2– Electronegativity of A increasesa proton from a doubly charged HPO4 anion as compared to H2PO4–. CH4 < NH3 < H2O < HF Polyprotic acid solutions contain a Acid strength increases mixture of acids like H2A, HA– and A2– in case 6.11.8 Common Ion Effect in theof a diprotic acid. H2A being a strong acid, the Ionization of Acids and Basesprimary reaction involves the dissociation of H2 A, and H3O+ in the solution comes mainly Consider an example of acetic acid dissociation from the first dissociation step. equilibrium represented as: CH3COOH(aq) H+(aq) + CH3COO– (aq)6.11.7 Factors Affecting Acid Strength or HAc(aq) H+ (aq) + Ac– (aq)Having discussed quantitatively the strengths of acids and bases, we come to a stage where Ka = [H+][Ac– ] / [HAc] we can calculate the pH of a given acid Addition of acetate ions to an aceticsolution. But, the curiosity rises about why acid solution results in decreasing theshould some acids be stronger than others? concentration of hydrogen ions, [H+]. Also,What factors are responsible for making if H+ ions are added from an external sourcethem stronger? The answer lies in its being a then the equilibrium moves in the directioncomplex phenomenon. But, broadly speaking of undissociated acetic acid i.e., in a directionwe can say that the extent of dissociation of of reducing the concentration of hydrogenan acid depends on the strength and polarity ions, [H+]. This phenomenon is an exampleof the H-A bond. Reprint 2025-26 EQUILIBRIUM 201 of common ion effect. It can be defined as Thus, x = 1.33 × 10–3 = [OH–] a shift in equilibrium on adding a substance that provides more of an ionic species already Therefore, [H+] = Kw / [OH–] = 10–14 / present in the dissociation equilibrium. (1.33 × 10–3) = 7.51 × 10–12 Thus, we can say that common ion effect is pH = –log (7.5 × 10–12) = 11.12a phenomenon based on the Le Chatelier’s principle discussed in section 6.8. On addition of 25 mL of 0.1M HCl solution (i.e., 2.5 mmol of HCl) to 50 In order to evaluate the pH of the solution mL of 0.1M ammonia solution (i.e., 5resulting on addition of 0.05M acetate ion to mmol of NH3), 2.5 mmol of ammonia0.05M acetic acid solution, we shall consider molecules are neutralized. The resultingthe acetic acid dissociation equilibrium once 75 mL solution contains the remaining again, unneutralized 2.5 mmol of NH3 molecules HAc(aq) H+(aq) + Ac–(aq) and 2.5 mmol of NH4+. Initial concentration (M) NH3 + HCl → NH4+ + Cl– 0.05 0 0.05 2.5 2.5 0 0 At equilibrium Let x be the extent of ionization of acetic acid. 0 0 2.5 2.5 Change in concentration (M) The resulting 75 mL of solution contains 2.5 mmol of NH4+ ions (i.e., 0.033 M) and –x +x +x 2.5 mmol (i.e., 0.033 M ) of uneutralised Equilibrium concentration (M) NH3 molecules. This NH3 exists in the 0.05-x x 0.05+x following equilibrium: Therefore, NH4OH   NH4+ + OH– 0.033M – y y yKa= [H+][Ac– ]/[H Ac] = {(0.05+x)(x)}/(0.05-x) where, y = [OH–] = [NH4+]As Ka is small for a very weak acid, x<<0.05. The final 75 mL solution afterHence, (0.05 + x) ≈ (0.05 – x) ≈ 0.05 neutralisation already contains Thus, + 2.5 m mol NH4 ions (i.e. 0.033M), thus 1.8 × 10–5 = (x) (0.05 + x) / (0.05 – x) total concentration of NH4+ ions is given as:= x(0.05) / (0.05) = x = [H+] = 1.8 × 10–5M [NH4+] = 0.033 + ypH = – log(1.8 × 10–5) = 4.74 As y is small, [NH4OH]  0.033 M and Problem 6.24 [NH4+]  0.033M. We know, Calculate the pH of a 0.10M ammonia solution. Calculate the pH after 50.0 mL Kb = [NH4+][OH–] / [NH4OH] of this solution is treated with 25.0 mL of = y (0.033)/(0.033) = 1.77 × 10–5 M 0.10M HCl. The dissociation constant of Thus, y = 1.77 × 10–5 = [OH–] ammonia, Kb = 1.77 × 10–5 [H+] = 10–14 / 1.77 × 10–5 = 0.56 × 10–9 Solution + Hence, pH = 9.24 NH3 + H2O → NH4 + OH– Kb = [NH4 +][OH–] / [NH3] = 1.77 × 10–5 6.11.9 Hydrolysis of Salts and the pH of Before neutralization, their Solutions [NH4 +] = [OH–] = x Salts formed by the reactions between acids [NH3] = 0.10 – x  0.10 and bases in definite proportions, undergo x2 / 0.10 = 1.77 × 10–5 ionization in water. The cations/anions Reprint 2025-26 202 chemistry formed on ionization of salts either exist as increased of H+ ion concentration in solution hydrated ions in aqueous solutions or interact making the solution acidic. Thus, the pH of with water to reform corresponding acids/ NH4Cl solution in water is less than 7. bases depending upon the nature of salts. Consider the hydrolysis of CH3COONH4The later process of interaction between salt formed from weak acid and weak base. water and cations/anions or both of salts The ions formed undergo hydrolysis as follow: is called hydrolysis. The pH of the solution + gets affected by this interaction. The cations CH3COO– + NH4 + H2O CH3COOH + (e.g., Na+, K+, Ca2+, Ba2+, etc.) of strong bases NH4OH and anions (e.g., Cl–, Br–, NO3–, ClO4– etc.) of CH3COOH and NH4OH, also remain into strong acids simply get hydrated but do not partially dissociated form: hydrolyse, and therefore the solutions of CH3COOH CH3COO– + H+ salts formed from strong acids and bases are + NH4OH NH4 + OH–neutral i.e., their pH is 7. However, the other category of salts do undergo hydrolysis. H2O H+ + OH– We now consider the hydrolysis of the Without going into detailed calculation, salts of the following types : it can be said that degree of hydrolysis is (i) salts of weak acid and strong base e.g., independent of concentration of solution, and CH3COONa. pH of such solutions is determined by their pK values:(ii) salts of strong acid and weak base e.g., NH4Cl, and pH = 7 + ½ (pKa – pKb) (6.38) (iii) salts of weak acid and weak base, e.g., The pH of solution can be greater than 7, CH3COONH4. if the difference is positive and it will be less In the first case, CH3COONa being a salt of than 7, if the difference is negative. weak acid, CH3COOH and strong base, NaOH Problem 6.25gets completely ionised in aqueous solution. The pKa of acetic acid and pKb of ammoniumCH3COONa(aq) → CH3COO– (aq)+ Na+(aq) hydroxide are 4.76 and 4.75 respectively. Acetate ion thus formed undergoes Calculate the pH of ammonium acetate hydrolysis in water to give acetic acid and solution. OH– ions Solution CH3COO–(aq)+H2O(l) CH3COOH(aq)+OH–(aq) pH = 7 + ½ [pKa – pKb] Acetic acid being a weak acid = 7 + ½ [4.76 – 4.75] (Ka = 1.8 × 10–5) remains mainly unionised in = 7 + ½ [0.01] = 7 + 0.005 = 7.005solution. This results in increase of OH– ion concentration in solution making it alkaline. The pH of such a solution is more than 7. 6.12 BUFFER SOLUTIONS Many body fluids e.g., blood or urine have Similarly, NH4Cl formed from weak definite pH and any deviation in their pHbase, NH4OH and strong acid, HCl, in water indicates malfunctioning of the body. Thedissociates completely. + control of pH is also very important in NH4Cl(aq) → NH 4(aq) +Cl– (aq) many chemical and biochemical processes. Ammonium ions undergo hydrolysis with Many medical and cosmetic formulations water to form NH4OH and H+ ions require that these be kept and administered NH +4 (aq) + H2O (1) NH4OH(aq) + H+(aq) at a particular pH. The solutions which Ammonium hydroxide is a weak base resist change in pH on dilution or with (Kb = 1.77 × 10–5) and therefore remains the addition of small amounts of acid or almost unionised in solution. This results in alkali are called Buffer Solutions. Buffer Reprint 2025-26 EQUILIBRIUM 203 solutions of known pH can be prepared from acid present in the mixture. Since acid is a the knowledge of pKa of the acid or pKb of base weak acid, it ionises to a very little extent and and by controlling the ratio of the salt and acid concentration of [HA] is negligibly different or salt and base. A mixture of acetic acid and from concentration of acid taken to form sodium acetate acts as buffer solution around buffer. Also, most of the conjugate base, [A—], pH 4.75 and a mixture of ammonium chloride comes from the ionisation of salt of the acid. and ammonium hydroxide acts as a buffer Therefore, the concentration of conjugate around pH 9.25. You will learn more about base will be negligibly different from the buffer solutions in higher classes. concentration of salt. Thus, equation (6.40) takes the form:6.12.1 Designing Buffer Solution [Salt]Knowledge of pKa, pKb and equilibrium pH=pKa + log constant help us to prepare the buffer solution [Acid] of known pH. Let us see how we can do this. In the equation (6.39), if the concentration Preparation of Acidic Buffer of [A—] is equal to the concentration of [HA], then pH = pKa because value of log 1 is zero.To prepare a buffer of acidic pH we use weak Thus if we take molar concentration of acidacid and its salt formed with strong base. and salt (conjugate base) same, the pH of theWe develop the equation relating the pH, the buffer solution will be equal to the pKa of theequilibrium constant, Ka of weak acid and acid. So for preparing the buffer solution ofratio of concentration of weak acid and its the required pH we select that acid whose pKaconjugate base. For the general case where is close to the required pH. For acetic acidthe weak acid HA ionises in water, pKa value is 4.76, therefore pH of the buffer HA + H2O  H3O+ + A– solution formed by acetic acid and sodium For which we can write the expression acetate taken in equal molar concentration will be around 4.76. A similar analysis of a buffer made with a weak base and its conjugate acid leads to Rearranging the expression we have, the result, [Conjugate acid,BH+] pOH= p K b +log [Base,B] Taking logarithm on both the sides and (6.41) rearranging the terms we get — pH of the buffer solution can be calculated by using the equation pH + pOH =14. We know that pH + pOH = pKw and Or pKa + pKb = pKw. On putting these values in equation (6.41) it takes the form as follows:  (6.39) [Conjugate acid,BH ] p K w - pH= p K w  p Ka  log [Base,B] or + [Conjugate acid,BH ] pH= p Ka + log (6.40) [Base,B] (6.42) The expression (6.40) is known as If molar concentration of base and its Henderson–Hasselbalch equation. The conjugate acid (cation) is same then pH of the buffer solution will be same as pKa for the quantity is the ratio of concentration base. pKa value for ammonia is s9.25; therefore a buffer of pH close to 9.25 can be obtained of conjugate base (anion) of the acid and the by taking ammonia solution and ammonium Reprint 2025-26 204 chemistry chloride solution of same molar concentration. We shall now consider the equilibrium For a buffer solution formed by ammonium between the sparingly soluble ionic salt and chloride and ammonium hydroxide, equation its saturated aqueous solution. (6.42) becomes: 6.13.1 Solubility Product Constant + [Conjugate acid,BH ] pH= 9 .25 + log Let us now have a solid like barium sulphate [Base,B] in contact with its saturated aqueous solution. pH of the buffer solution is not affected by The equilibrium between the undisolved solid dilution because ratio under the logarithmic and the ions in a saturated solution can be term remains unchanged. represented by the equation: 6.13 SOLUBILITY EQUILIBRIA OF BaSO4(s)  Ba2+(aq) + SO42–(aq), SPARINGLY SOLUBLE SALTS We have already known that the solubility of The equilibrium constant is given by the ionic solids in water varies a great deal. Some of equation: these (like calcium chloride) are so soluble that K = {[Ba2+][SO42–]} / [BaSO4] they are hygroscopic in nature and even absorb For a pure solid substance thewater vapour from atmosphere. Others (such concentration remains constant and we canas lithium fluoride) have so little solubility writethat they are commonly termed as insoluble. The solubility depends on a number of factors Ksp = K[BaSO4] = [Ba2+][SO42–] (6.43) important amongst which are the lattice We call Ksp the solubility product constant enthalpy of the salt and the solvation enthalpy or simply solubility product. The experimental of the ions in a solution. For a salt to dissolve value of Ksp in above equation at 298K is in a solvent the strong forces of attraction 1.1 × 10–10. This means that for solid barium between its ions (lattice enthalpy) must be sulphate in equilibrium with its saturated overcome by the ion-solvent interactions. The solution, the product of the concentrations solvation enthalpy of ions is referred to in of barium and sulphate ions is equal terms of solvation which is always negative i.e. to its solubility product constant. The energy is released in the process of solvation. concentrations of the two ions will be equal to The amount of solvation enthalpy depends on the molar solubility of the barium sulphate. the nature of the solvent. In case of a non- If molar solubility is S, then polar (covalent) solvent, solvation enthalpy is 1.1 × 10–10 = (S)(S) = S2small and hence, not sufficient to overcome or S = 1.05 × 10–5.lattice enthalpy of the salt. Consequently, the salt does not dissolve in non-polar solvent. As Thus, molar solubility of barium sulphate a general rule, for a salt to be able to dissolve will be equal to 1.05 × 10–5 mol L–1. in a particular solvent its solvation enthalpy A salt may give on dissociation two or must be greater than its lattice enthalpy so more than two anions and cations carrying that the latter may be overcome by former. different charges. For example, consider a salt Each salt has its characteristic solubility which like zirconium phosphate of molecular formula depends on temperature. We classify salts on (Zr4+)3(PO43–)4. It dissociates into 3 zirconium the basis of their solubility in the following cations of charge +4 and 4 phosphate anions of charge –3. If the molar solubility ofthree categories. zirconium phosphate is S, then it can be seen Category I Soluble Solubility > 0.1M from the stoichiometry of the compound that Category II Slightly 0.01M<Solubility< 0.1M [Zr4+] = 3S and [PO43–] = 4S Soluble Category III Sparingly Solubility < 0.01M and Ksp = (3S)3 (4S)4 = 6912 (S)7 Soluble or S = {Ksp / (33 × 44)}1/7 = (Ksp / 6912)1/7 Reprint 2025-26 EQUILIBRIUM 205 A solid salt of the general formula M px  X qy  Table 6.9 The Solubility Product Constants, with molar solubility S in equilibrium with Ksp of Some Common Ionic Salts at its saturated solution may be represented by 298K. the equation: MxXy(s) xMp+(aq) + yXq– (aq) (where x × p+ = y × q–) And its solubility product constant is given by: Ksp = [Mp+]x[Xq– ]y = (xS)x(yS)y (6.44) = xx . yy . S(x + y) S(x + y) = Ksp / xx . yy S = (Ksp / xx . yy)1 / x + y (6.45) The term Ksp in equation is given by Qsp (section 6.6.2) when the concentration of one or more species is not the concentration under equilibrium. Obviously under equilibrium conditions Ksp = Qsp but otherwise it gives the direction of the processes of precipitation or dissolution. The solubility product constants of a number of common salts at 298K are given in Table 6.9. Problem 6.26 Calculate the solubility of A2X3 in pure water, assuming that neither kind of ion reacts with water. The solubility product of A2X3, Ksp = 1.1 × 10–23. Solution A2X3 → 2A3+ + 3X2– Ksp = [A3+]2 [X2–]3 = 1.1 × 10–23 If S = solubility of A2X3, then [A3+] = 2S; [X2–] = 3S therefore, Ksp = (2S)2(3S)3 = 108S5 = 1.1 × 10–23 thus, S5 = 1 × 10–25 S = 1.0 × 10–5 mol/L. Problem 6.27 The values of Ksp of two sparingly soluble salts Ni(OH)2 and AgCN are 2.0 × 10–15 and 6 × 0–17 respectively. Which salt is more soluble? Explain. Solution AgCN Ag+ + CN– Reprint 2025-26 206 chemistry Ksp = [Ag+][CN–] = 6 × 10–17 Dissolution of S mol/L of Ni(OH)2 provides S mol/L of Ni2+ and 2S mol/L of OH–, but Ni(OH)2 Ni2+ + 2OH– the total concentration of OH– = (0.10 + 2S) Ksp = [Ni2+][OH–]2 = 2 × 10–15 mol/L because the solution already contains Let [Ag+] = S1, then [CN-] = S1 0.10 mol/L of OH– from NaOH. Let [Ni2+] = S2, then [OH–] = 2S2 2 Ksp = 2.0 × 10–15 = [Ni2+] [OH–]2 S1 = 6 × 10–17 , S1 = 7.8 × 10–9 = (S) (0.10 + 2S)2 (S2)(2S2)2 = 2 × 10–15, S2 = 0.58 × 10–4 As Ksp is small, 2S << 0.10, Ni(OH)2 is more soluble than AgCN. thus, (0.10 + 2S) ≈ 0.10 6.13.2 Common Ion Effect on Solubility Hence, of Ionic Salts 2.0 × 10–15 = S (0.10)2 It is expected from Le Chatelier’s principle S = 2.0 × 10–13 M = [Ni2+] that if we increase the concentration of any one of the ions, it should combine with the ion of its opposite charge and some of the The solubility of salts of weak acids like salt will be precipitated till once again Ksp = phosphates increases at lower pH. This is Qsp. Similarly, if the concentration of one of because at lower pH the concentration of the the ions is decreased, more salt will dissolve anion decreases due to its protonation. This to increase the concentration of both the ions in turn increase the solubility of the salt so till once again Ksp = Qsp. This is applicable that Ksp = Qsp. We have to satisfy two equilibria even to soluble salts like sodium chloride simultaneously i.e., except that due to higher concentrations of the ions, we use their activities instead Ksp = [M+] [X–], of their molarities in the expression for Qsp. Thus if we take a saturated solution of sodium chloride and pass HCl gas through it, then sodium chloride is precipitated due to increased concentration (activity) of chloride [X–] / [HX] = Ka/[H+]ion available from the dissociation of HCl. Sodium chloride thus obtained is of very Taking inverse of both side and adding 1 high purity and we can get rid of impurities we get Hlike sodium and magnesium sulphates. The  HX   common ion effect is also used for almost 1   1   acomplete precipitation of a particular ion  X  K as its sparingly soluble salt, with very low    HX  H   H  K avalue of solubility product for gravimetric   aestimation. Thus we can precipitate silver ion  X  K as silver chloride, ferric ion as its hydroxide Now, again taking inverse, we get(or hydrated ferric oxide) and barium ion as its sulphate for quantitative estimations. [X–] / {[X–] + [HX]} = f = Ka/(Ka + [H+]) and it can be seen that ‘f’ decreases as pH decreases. Problem 6.28 If S is the solubility of the salt at a given Calculate the molar solubility of Ni(OH)2 in pH then 0.10 M NaOH. The ionic product of Ni(OH)2 is 2.0 × 10–15. Ksp = [S] [f S] = S2 {Ka/(Ka + [H+])} and S = {Ksp ([H+] + Ka)/Ka}1/2 (6.46) Solution Thus solubility S increases with increase Let the solubility of Ni(OH)2 be equal to S. in [H+] or decrease in pH. Reprint 2025-26 EQUILIBRIUM 207 SUMMARY When the number of molecules leaving the liquid to vapour equals the number of molecules returning to the liquid from vapour, equilibrium is said to be attained and is dynamic in nature. Equilibrium can be established for both physical and chemical processes and at this stage rate of forward and reverse reactions are equal. Equilibrium constant, Kc is expressed as the concentration of products divided by reactants, each term raised to the stoichiometric coefficient. For reaction, a A + b B c C +d D Kc = [C]c[D]d/[A]a[B]b Equilibrium constant has constant value at a fixed temperature and at this stage all the macroscopic properties such as concentration, pressure, etc. become constant. For a gaseous reaction equilibrium constant is expressed as Kp and is written by replacing concentration terms by partial pressures in Kc expression. The direction of reaction can be predicted by reaction quotient Qc which is equal to Kc at equilibrium. Le Chatelier’s principle states that the change in any factor such as temperature, pressure, concentration, etc. will cause the equilibrium to shift in such a direction so as to reduce or counteract the effect of the change. It can be used to study the effect of various factors such as temperature, concentration, pressure, catalyst and inert gases on the direction of equilibrium and to control the yield of products by controlling these factors. Catalyst does not effect the equilibrium composition of a reaction mixture but increases the rate of chemical reaction by making available a new lower energy pathway for conversion of reactants to products and vice-versa. All substances that conduct electricity in aqueous solutions are called electrolytes. Acids, bases and salts are electrolytes and the conduction of electricity by their aqueous solutions is due to anions and cations produced by the dissociation or ionization of electrolytes in aqueous solution. The strong electrolytes are completely dissociated. In weak electrolytes there is equilibrium between the ions and the unionized electrolyte molecules. According to Arrhenius, acids give hydrogen ions while bases produce hydroxyl ions in their aqueous solutions. Brönsted-Lowry on the other hand, defined an acid as a proton donor and a base as a proton acceptor. When a Brönsted-Lowry acid reacts with a base, it produces its conjugate base and a conjugate acid corresponding to the base with which it reacts. Thus a conjugate pair of acid-base differs only by one proton. Lewis further generalised the definition of an acid as an electron pair acceptor and a base as an electron pair donor. The expressions for ionization (equilibrium) constants of weak acids (Ka) and weak bases (Kb) are developed using Arrhenius definition. The degree of ionization and its dependence on concentration and common ion are discussed. The pH scale (pH = –log[H+]) for the hydrogen ion concentration (activity) has been introduced and extended to other quantities (pOH = – log[OH–]); pKa = –log[Ka]; pKb = –log[Kb]; and pKw = –log[Kw] etc.). The ionization of water has been considered and we note that the equation: pH + pOH = pKw is always satisfied. The salts of strong acid and weak base, weak acid and strong base, and weak acid and weak base undergo hydrolysis in aqueous solution. The definition of buffer solutions, and their importance are discussed briefly. The solubility equilibrium of sparingly soluble salts is discussed and the equilibrium constant is introduced as solubility product constant (Ksp). Its relationship with solubility of the salt is established. The conditions of precipitation of the salt from their solutions or their dissolution in water are worked out. The role of common ion and the solubility of sparingly soluble salts is also discussed. Reprint 2025-26 208 chemistry SUGGESTED ACTIVITIES FOR STUDENTS REGARDING THIS UNIT (a) The student may use pH paper in determining the pH of fresh juices of various vegetables and fruits, soft drinks, body fluids and also that of water samples available. (b) The pH paper may also be used to determine the pH of different salt solutions and from that he/she may determine if these are formed from strong/weak acids and bases. (c) They may prepare some buffer solutions by mixing the solutions of sodium acetate and acetic acid and determine their pH using pH paper. (d) They may be provided with different indicators to observe their colours in solutions of varying pH. (e) They may perform some acid-base titrations using indicators. (f) They may observe common ion effect on the solubility of sparingly soluble salts. (g) If pH meter is available in their school, they may measure the pH with it and compare the results obtained with that of the pH paper. EXERCISES 6.1 A liquid is in equilibrium with its vapour in a sealed container at a fixed temperature. The volume of the container is suddenly increased. a) What is the initial effect of the change on vapour pressure? b) How do rates of evaporation and condensation change initially? c) What happens when equilibrium is restored finally and what will be the final vapour pressure? 6.2 What is Kc for the following equilibrium when the equilibrium concentration of each substance is: [SO2]= 0.60M, [O2] = 0.82M and [SO3] = 1.90M ? 2SO2(g) + O2(g) 2SO3(g) 6.3 At a certain temperature and total pressure of 105Pa, iodine vapour contains 40% by volume of I atoms I2 (g) 2I (g) Calculate Kp for the equilibrium. 6.4 Write the expression for the equilibrium constant, Kc for each of the following reactions: (i) 2NOCl (g) 2NO (g) + Cl2 (g) (ii) 2Cu(NO3)2 (s) 2CuO (s) + 4NO2 (g) + O2 (g) (iii) CH3COOC2H5(aq) + H2O(l) CH3COOH (aq) + C2H5OH (aq) (iv) Fe3+ (aq) + 3OH– (aq) Fe(OH)3 (s) (v) I2 (s) + 5F2 2IF5 6.5 Find out the value of Kc for each of the following equilibria from the value of Kp: (i) 2NOCl (g) 2NO (g) + Cl2 (g); Kp= 1.8 × 10–2 at 500 K (ii) CaCO3 (s) CaO(s) + CO2(g); Kp= 167 at 1073 K Reprint 2025-26 EQUILIBRIUM 209 6.6 For the following equilibrium, Kc= 6.3 × 1014 at 1000 K NO (g) + O3 (g) NO2 (g) + O2 (g) Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is Kc, for the reverse reaction? 6.7 Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression? 6.8 Reaction between N2 and O2– takes place as follows: 2N2 (g) + O2 (g) 2N2O (g) If a mixture of 0.482 mol N2 and 0.933 mol of O2 is placed in a 10 L reaction vessel and allowed to form N2O at a temperature for which Kc= 2.0 × 10–37, determine the composition of equilibrium mixture. 6.9 Nitric oxide reacts with Br2 and gives nitrosyl bromide as per reaction given below: 2NO (g) + Br2 (g) 2NOBr (g) When 0.087 mol of NO and 0.0437 mol of Br2 are mixed in a closed container at constant temperature, 0.0518 mol of NOBr is obtained at equilibrium. Calculate equilibrium amount of NO and Br2 . 6.10 At 450K, Kp= 2.0 × 1010/bar for the given reaction at equilibrium. 2SO2(g) + O2(g) 2SO3 (g) What is Kc at this temperature ? 6.11 A sample of HI(g) is placed in flask at a pressure of 0.2 atm. At equilibrium the partial pressure of HI(g) is 0.04 atm. What is Kp for the given equilibrium ? 2HI (g) H2 (g) + I2 (g) 6.12 A mixture of 1.57 mol of N2, 1.92 mol of H2 and 8.13 mol of NH3 is introduced into a 20 L reaction vessel at 500 K. At this temperature, the equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) is 1.7 × 102. Is the reaction mixture at equilibrium? If not, what is the direction of the net reaction? 6.13 The equilibrium constant expression for a gas reaction is, NH 3 4 O 2 5 Kc  4  NO  H 2 O 6 Write the balanced chemical equation corresponding to this expression. 6.14 One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water (by mass) reacts with CO according to the equation, H2O (g) + CO (g) H2 (g) + CO2 (g) Calculate the equilibrium constant for the reaction. 6.15 At 700 K, equilibrium constant for the reaction: H2 (g) + I2 (g) 2HI (g) is 54.8. If 0.5 mol L–1 of HI(g) is present at equilibrium at 700 K, what are the concentration of H2(g) and I2(g) assuming that we initially started with HI(g) and allowed it to reach equilibrium at 700K? Reprint 2025-26 210 chemistry 6.16 What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M ? 2ICl (g) I2 (g) + Cl2 (g); Kc = 0.14 6.17 Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium? C2H6 (g) C2H4 (g) + H2 (g) 6.18 Ethyl acetate is formed by the reaction between ethanol and acetic acid and the equilibrium is represented as: CH3COOH (l) + C2H5OH (l) CH3COOC2H5 (l) + H2O (l) (i) Write the concentration ratio (reaction quotient), Qc, for this reaction (note: water is not in excess and is not a solvent in this reaction) (ii) At 293 K, if one starts with 1.00 mol of acetic acid and 0.18 mol of ethanol, there is 0.171 mol of ethyl acetate in the final equilibrium mixture. Calculate the equilibrium constant. (iii) Starting with 0.5 mol of ethanol and 1.0 mol of acetic acid and maintaining it at 293 K, 0.214 mol of ethyl acetate is found after sometime. Has equilibrium been reached? 6.19 A sample of pure PCl5 was introduced into an evacuated vessel at 473 K. After equilibrium was attained, concentration of PCl5 was found to be 0.5 × 10–1 mol L–1. If value of Kc is 8.3 × 10–3, what are the concentrations of PCl3 and Cl2 at equilibrium? PCl5 (g) PCl3 (g) + Cl2(g) 6.20 One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and CO2. FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K What are the equilibrium partial pressures of CO and CO2 at 1050 K if the initial partial pressures are: pCO= 1.4 atm and = 0.80 atm? 6.21 Equilibrium constant, Kc for the reaction N2 (g) + 3H2 (g) 2NH3 (g) at 500 K is 0.061 At a particular time, the analysis shows that composition of the reaction mixture is 3.0 mol L–1 N2, 2.0 mol L–1 H2 and 0.5 mol L–1 NH3. Is the reaction at equilibrium? If not in which direction does the reaction tend to proceed to reach equilibrium? 6.22 Bromine monochloride, BrCl decomposes into bromine and chlorine and reaches the equilibrium: 2BrCl (g) Br2 (g) + Cl2 (g) for which Kc= 32 at 500 K. If initially pure BrCl is present at a concentration of 3.3 × 10–3 mol L–1, what is its molar concentration in the mixture at equilibrium? 6.23 At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with soild carbon has 90.55% CO by mass C (s) + CO2 (g) 2CO (g) Calculate Kc for this reaction at the above temperature. Reprint 2025-26 EQUILIBRIUM 211 6.24 Calculate a) ∆G and b) the equilibrium constant for the formation of NO2 from NO and O2 at 298K NO (g) + ½ O2 (g) NO2 (g) where ∆fG (NO2) = 52.0 kJ/mol ∆fG (NO) = 87.0 kJ/mol ∆fG (O2) = 0 kJ/mol 6.25 Does the number of moles of reaction products increase, decrease or remain same when each of the following equilibria is subjected to a decrease in pressure by increasing the volume? (a) PCl5 (g) PCl3 (g) + Cl2 (g) (b) CaO (s) + CO2 (g) CaCO3 (s) (c) 3Fe (s) + 4H2O (g) Fe3O4 (s) + 4H2 (g) 6.26 Which of the following reactions will get affected by increasing the pressure? Also, mention whether change will cause the reaction to go into forward or backward direction. (i) COCl2 (g) CO (g) + Cl2 (g) (ii) CH4 (g) + 2S2 (g) CS2 (g) + 2H2S (g) (iii) CO2 (g) + C (s) 2CO (g) (iv) 2H2 (g) + CO (g) CH3OH (g) (v) CaCO3 (s) CaO (s) + CO2 (g) (vi) 4 NH3 (g) + 5O2 (g) 4NO (g) + 6H2O(g) 6.27 The equilibrium constant for the following reaction is 1.6 ×105 at 1024K H2(g) + Br2(g) 2HBr(g) Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a sealed container at 1024K. 6.28 Dihydrogen gas is obtained from natural gas by partial oxidation with steam as per following endothermic reaction: CH4 (g) + H2O (g) CO (g) + 3H2 (g) (a) Write as expression for Kp for the above reaction. (b) How will the values of Kp and composition of equilibrium mixture be affected by (i) increasing the pressure (ii) increasing the temperature (iii) using a catalyst? 6.29 Describe the effect of: a) addition of H2 b) addition of CH3OH c) removal of CO d) removal of CH3OH on the equilibrium of the reaction: 2H2(g) + CO (g) CH3OH (g) 6.30 At 473 K, equilibrium constant Kc for decomposition of phosphorus pentachloride, PCl5 is 8.3 ×10-3. If decomposition is depicted as, Reprint 2025-26 212 chemistry PCl5 (g) PCl3 (g) + Cl2 (g) ∆rH = 124.0 kJ mol–1 a) write an expression for Kc for the reaction. b) what is the value of Kc for the reverse reaction at the same temperature? c) what would be the effect on Kc if (i) more PCl5 is added (ii) pressure is increased (iii) the temperature is increased ? 6.31 Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction, CO (g) + H2O (g) CO2 (g) + H2 (g) If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pco = pH2O = 4.0 bar, what will be the partial pressure of H2 at equilibrium? Kp= 10.1 at 400°C 6.32 Predict which of the following reaction will have appreciable concentration of reactants and products: a) Cl2 (g) 2Cl (g) Kc = 5 ×10–39 b) Cl2 (g) + 2NO (g) 2NOCl (g) Kc = 3.7 × 108 c) Cl2 (g) + 2NO2 (g) 2NO2Cl (g) Kc = 1.8 6.33 The value of Kc for the reaction 3O2 (g) 2O3 (g) is 2.0 ×10–50 at 25°C. If the equilibrium concentration of O2 in air at 25°C is 1.6 ×10–2, what is the concentration of O3? 6.34 The reaction, CO(g) + 3H2(g) CH4(g) + H2O(g) is at equilibrium at 1300 K in a 1L flask. It also contain 0.30 mol of CO, 0.10 mol of H2 and 0.02 mol of H2O and an unknown amount of CH4 in the flask. Determine the concentration of CH4 in the mixture. The equilibrium constant, Kc for the reaction at the given temperature is 3.90. 6.35 What is meant by the conjugate acid-base pair? Find the conjugate acid/base for the following species: HNO2, CN–, HClO4, F –, OH–, CO , and S2– 6.36 Which of the followings are Lewis acids? H2O, BF3, H+, and NH4+ 6.37 What will be the conjugate bases for the Brönsted acids: HF, H2SO4 and HCO–3? 6.38 Write the conjugate acids for the following Brönsted bases: NH2–, NH3 and HCOO–. 6.39 The species: H2O, HCO3–, HSO4– and NH3 can act both as Brönsted acids and bases. For each case give the corresponding conjugate acid and base. 6.40 Classify the following species into Lewis acids and Lewis bases and show how these act as Lewis acid/base: (a) OH– (b) F– (c) H+ (d) BCl3 . 6.41 The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10–3 M. What is its pH? 6.42 The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen ion in it. 6.43 The ionization constant of HF, HCOOH and HCN at 298K are 6.8 × 10–4, 1.8 × 10–4 and 4.8 × 10–9 respectively. Calculate the ionization constants of the corresponding conjugate base. Reprint 2025-26 EQUILIBRIUM 213 6.44 The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate? 6.45 The first ionization constant of H2S is 9.1 × 10–8. Calculate the concentration of HS– ion in its 0.1M solution. How will this concentration be affected if the solution is 0.1M in HCl also? If the second dissociation constant of H2S is 1.2 × 10–13, calculate the concentration of S2– under both conditions. 6.46 The ionization constant of acetic acid is 1.74 × 10–5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH. 6.47 It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pKa. 6.48 Assuming complete dissociation, calculate the pH of the following solutions: (a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH 6.49 Calculate the pH of the following solutions: a) 2 g of TlOH dissolved in water to give 2 litre of solution. b) 0.3 g of Ca(OH)2 dissolved in water to give 500 mL of solution. c) 0.3 g of NaOH dissolved in water to give 200 mL of solution. d) 1mL of 13.6 M HCl is diluted with water to give 1 litre of solution. 6.50 The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate the pH of the solution and the pKa of bromoacetic acid. 6.51 The pH of 0.005M codeine (C18H21NO3) solution is 9.95. Calculate its ionization constant and pKb. 6.52 What is the pH of 0.001M aniline solution? The ionization constant of aniline can be taken from Table 6.7. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline. 6.53 Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M (b) 0.1M in HCl ? 6.54 The ionization constant of dimethylamine is 5.4 × 10–4. Calculate its degree of ionization in its 0.02M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in NaOH? 6.55 Calculate the hydrogen ion concentration in the following biological fluids whose pH are given below: (a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2 (c) Human blood, 7.38 (d) Human saliva, 6.4. 6.56 The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8, 5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion concentration in each. 6.57 If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH? 6.58 The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution. Reprint 2025-26 214 chemistry 6.59 The ionization constant of propanoic acid is 1.32 × 10–5. Calculate the degree of ionization of the acid in its 0.05M solution and also its pH. What will be its degree of ionization if the solution is 0.01M in HCl also? 6.60 The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization constant of the acid and its degree of ionization in the solution. 6.61 The ionization constant of nitrous acid is 4.5 × 10–4. Calculate the pH of 0.04 M sodium nitrite solution and also its degree of hydrolysis. 6.62 A 0.02M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine. 6.63 Predict if the solutions of the following salts are neutral, acidic or basic: NaCl, KBr, NaCN, NH4NO3, NaNO2 and KF 6.64 The ionization constant of chloroacetic acid is 1.35 × 10–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution? 6.65 Ionic product of water at 310 K is 2.7 × 10–14. What is the pH of neutral water at this temperature? 6.66 Calculate the pH of the resultant mixtures: a) 10 mL of 0.2M Ca(OH)2 + 25 mL of 0.1M HCl b) 10 mL of 0.01M H2SO4 + 10 mL of 0.01M Ca(OH)2 c) 10 mL of 0.1M H2SO4 + 10 mL of 0.1M KOH 6.67 Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 6.9. Determine also the molarities of individual ions. 6.68 The solubility product constant of Ag2CrO4 and AgBr are 1.1 × 10–12 and 5.0 × 10–13 respectively. Calculate the ratio of the molarities of their saturated solutions. 6.69 Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate Ksp = 7.4 × 10–8 ). 6.70 The ionization constant of benzoic acid is 6.46 × 10–5 and Ksp for silver benzoate is 2.5 × 10–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water? 6.71 What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10–18). 6.72 What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, Ksp is 9.1 × 10–6). 6.73 The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2. in which of these solutions precipitation will take place? Reprint 2025-26

4.16 Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with (i) iron(II) ions (ii) SO2 and (iii) oxalic acid? Write the ionic equations for the reactions. 4.17 For M2+/M and M3+/M 2+ systems the E o values for some metals are as follows: Cr2+/Cr -0.9V Cr3/Cr2+ -0.4 V Mn 2+/Mn -1.2V Mn3+/Mn2+ +1.5 V Fe2+/Fe -0.4V Fe3+/Fe2+ +0.8 V Use this data to comment upon: (i) the stability of Fe3+ in acid solution as compared to that of Cr3+ or Mn3+ and (ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal. 4.18 Predict which of the following will be coloured in aqueous solution? Ti 3+, V3+, Cu+, Sc3+, Mn 2+, Fe3+ and Co 2+. Give reasons for each. 4.19 Compare the stability of +2 oxidation state for the elements of the first transition series. 4.20 Compare the chemistry of actinoids with that of the lanthanoids with special reference to: (i) electronic configuration (iii) oxidation state (ii) atomic and ionic sizes and (iv) chemical reactivity. 4.21 How would you account for the following: (i) Of the d4 species, Cr2+ is strongly reducing while manganese(III) is strongly oxidising. (ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents it is easily oxidised. (iii) The d1 configuration is very unstable in ions. 4.22 What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution. 4.23 Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why? 4.24 Calculate the number of unpaired electrons in the following gaseous ions: Mn3+, Cr3+, V3+ and Ti3+. Which one of these is the most stable in aqueous solution? 4.25 Give examples and suggest reasons for the following features of the transition metal chemistry: (i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic. (ii) A transition metal exhibits highest oxidation state in oxides and fluorides. (iii) The highest oxidation state is exhibited in oxoanions of a metal. 4.26 Indicate the steps in the preparation of: (i) K2Cr2O7 from chromite ore. (ii) KMnO4 from pyrolusite ore. 4.27 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses. 4.28 What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104. 4.29 The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements. 4.30 Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element. Chemistry 116 Reprint 2025-26 4.31 Use Hund’s rule to derive the electronic configuration of Ce3+ ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula. 4.32 Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements. 4.33 Compare the chemistry of the actinoids with that of lanthanoids with reference to: (i) electronic configuration (ii) oxidation states and (iii) chemical reactivity. 4.34 Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109. 4.35 Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points: (i) electronic configurations (ii) oxidation states (iii) ionisation enthalpies and (iv) atomic sizes. 4.36 Write down the number of 3d electrons in each of the following ions: Ti 2+, V 2+, Cr3+, Mn 2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral). 4.37 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements. 4.38 What can be inferred from the magnetic moment values of the following complex species ? Example Magnetic Moment (BM) K4[Mn(CN)6) 2.2 [Fe(H2O)6]2+ 5.3 K2[MnCl4] 5.9 Answers to Some Intext Questions 4.1 Silver (Z = 47) can exhibit +2 oxidation state wherein it will have incompletely filled d-orbitals (4d), hence a transition element. 4.2 In the formation of metallic bonds, no eletrons from 3d-orbitals are involved in case of zinc, while in all other metals of the 3d series, electrons from the d-orbitals are always involved in the formation of metallic bonds. 4.3 Manganese (Z = 25), as its atom has the maximum number of unpaired electrons. 4.5 Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d-configurations (e.g., d 0, d 5, d 10 are exceptionally stable). 4.6 Because of small size and high electronegativity oxygen or fluorine can oxidise the metal to its highest oxidation state. 4.7 Cr 2+ is stronger reducing agent than Fe 2+ Reason: d 4  d 3 occurs in case of Cr 2+ to Cr 3+ But d 6  d 5 occurs in case of Fe2+ to Fe 3+ In a medium (like water) d 3 is more stable as compared to d 5 (see CFSE) 4.9 Cu + in aqueous solution underoes disproportionation, i.e., 2Cu +(aq) ® Cu 2+(aq) + Cu(s) The E0 value for this is favourable.

8.20 Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol. Why? Answers to Some Intext Questions 8.1 (i) (iv) (ii) (v) (iii) (vi) 257 Aldehydes, Ketones and Carboxylic Acids Reprint 2025-26 8.2 (i) (ii) (iii) (iv) 8.3 CH3CH2CH3 < CH3OCH3 < CH3CHO < CH3CH2OH 8.4 (i) Butanone < Propanone < Propanal < Ethanal (ii) Acetophenone < p-Tolualdehyde , Benzaldehyde < p-Nitrobenzaldehyde. 8.5 (i) (ii) (iii) (iv) 8.6 (i) 3-Phenylpropanoic acid (ii) 3-Methylbut-2-enoic acid (iii) 2-Methylcyclopentanecarboxylic acid. (iv) 2,4,6-Trinitrobenzoic acid 8.7 (i) (ii) (iii) (iv) 8.8 Chemistry 258 Reprint 2025-26 UnitUnitUnitUnit Unit9 AminesAminesmines9minesminesminesminesminesminesminesObjectives After studying this Unit, you will be able to “The chief commercial use of amines is as intermediates in the · describe amines as derivatives of synthesis of medicines and fibres” . ammonia having a pyramidal structure; Amines constitute an important class of organic · classify amines as primary, compounds derived by replacing one or more hydrogen secondary and tertiary; atoms of ammonia molecule by alkyl/aryl group(s). In· name amines by common names nature, they occur among proteins, vitamins, alkaloids and IUPAC system; and hormones. Synthetic examples include polymers,· describe some of the important dye stuffs and drugs. Two biologically active methods of preparation of amines; compounds, namely adrenaline and ephedrine, both· explain the properties of amines; containing secondary amino group, are used to increase· distinguish between primary, blood pressure. Novocain, a synthetic amino compound, secondary and tertiary amines; is used as an anaesthetic in dentistry. Benadryl, a well· describe the method of prepara- known antihistaminic drug also contains tertiary amino tion of diazonium salts and their importance in the synthesis of a group. Quaternary ammonium salts are used as series of aromatic compounds surfactants. Diazonium salts are intermediates in the including azo dyes. preparation of a variety of aromatic compounds including dyes. In this Unit, you will learn about amines and diazonium salts. I. AMINES Amines can be considered as derivatives of ammonia, obtained by replacement of one, two or all the three hydrogen atoms by alkyl and/or aryl groups. For example: 9.19.19.19.19.1 StructureStructureStructureStructureStructure ofofofofof AminesAminesAminesAminesAmines Like ammonia, nitrogen atom of amines is trivalent and carries an unshared pair of electrons. Nitrogen orbitals in amines are therefore, sp3 hybridised and the geometry of amines is pyramidal. Each of the three sp3 hybridised orbitals of nitrogen overlap with orbitals of hydrogen or carbon depending upon the composition of the amines. The fourth orbital of nitrogen in all amines contains an unshared pair of electrons. Due to the presence of unshared pair of electrons, the angle C–N–E, (where E is Reprint 2025-26 C or H) is less than 109.5°; for instance, it is 108o in case of trimethylamine as shown in Fig. 9.1. Fig. 9.1 Pyramidal shape of trimethylamine 9.29.29.29.29.2 ClassificationClassificationClassificationClassificationClassification Amines are classified as primary (1o), secondary (2o) and tertiary (3o) depending upon the number of hydrogen atoms replaced by alkyl or aryl groups in ammonia molecule. If one hydrogen atom of ammonia is replaced by R or Ar , we get RNH2 or ArNH2, a primary amine (1o). If two hydrogen atoms of ammonia or one hydrogen atom of R-NH2 are replaced by another alkyl/aryl(R’) group, what would you get? You get R-NHR’, secondary amine. The second alkyl/aryl group may be same or different. Replacement of another hydrogen atom by alkyl/aryl group leads to the formation of tertiary amine. Amines are said to be ‘simple’ when all the alkyl or aryl groups are the same, and ‘mixed’ when they are different. 9.39.39.39.39.3 NomenclatureNomenclatureNomenclatureNomenclatureNomenclature In common system, an aliphatic amine is named by prefixing alkyl group to amine, i.e., alkylamine as one word (e.g., methylamine). In secondary and tertiary amines, when two or more groups are the same, the prefix di or tri is appended before the name of alkyl group. In IUPAC system, primary amines are named as alkanamines. The name is derived by replacement of ‘e’ of alkane by the word amine. For example, CH3NH2 is named as methanamine. In case, more than one amino group is present at different positions in the parent chain, their positions are specified by giving numbers to the carbon atoms bearing –NH2 groups and suitable prefix such as di, tri, etc. is attached to the amine. The letter ‘e’ of the suffix of the hydrocarbon part is retained. For example, H2N–CH2–CH2–NH2 is named as ethane-1, 2-diamine. To name secondary and tertiary amines, we use locant N to designate substituent attached to a nitrogen atom. For example, CH3 NHCH2CH3 is Chemistry 260 Reprint 2025-26 named as N-methylethanamine and (CH3CH2)3N is named as N, N- diethylethanamine. More examples are given in Table 9.1. In arylamines, –NH2 group is directly attached to the benzene ring. C6H5NH2 is the simplest example of arylamine. In common system, it is known as aniline. It is also an accepted IUPAC name. While naming arylamines according to IUPAC system, suffix ‘e’ of arene is replaced by ‘amine’. Thus in IUPAC system, C6H5–NH2 is named as benzenamine. Common and IUPAC names of some alkylamines and arylamines are given in Table 9.1. Table 9.1: Nomenclature of Some Alkylamines and Arylamines Amine Common name IUPAC name CH3-–CH2–NH2 Ethylamine Ethanamine CH3–CH2–CH2–NH2 n-Propylamine Propan-1-amine Isopropylamine Propan-2-amine Ethylmethylamine N-Methylethanamine Trimethylamine N,N-Dimethylmethanamine N,N-Diethylbutylamine N,N-Diethylbutan-1-amine Allylamine Prop-2-en-1-amine Hexamethylenediamine Hexane-1,6-diamine Aniline Aniline or Benzenamine o-Toluidine 2-Methylaniline p-Bromoaniline 4-Bromobenzenamine or 4-Bromoaniline N,N-Dimethylaniline N,N-Dimethylbenzenamine 261 Amines Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 9.1 Classify the following amines as primary, secondary or tertiary: 9.2 (i) Write structures of different isomeric amines corresponding to the molecular formula, C4H11N. (ii) Write IUPAC names of all the isomers. (iii) What type of isomerism is exhibited by different pairs of amines? 9.49.49.49.49.4 PreparationPreparationPreparationPreparationPreparation Amines are prepared by the following methods: ofofofofof AminesAminesAminesAminesAmines 1. Reduction of nitro compounds Nitro compounds are reduced to amines by passing hydrogen gas in the presence of finely divided nickel, palladium or platinum and also by reduction with metals in acidic medium. Nitroalkanes can also be similarly reduced to the corresponding alkanamines. Reduction with iron scrap and hydrochloric acid is preferred because FeCl2 formed gets hydrolysed to release hydrochloric acid during the reaction. Thus, only a small amount of hydrochloric acid is required to initiate the reaction. 2. Ammonolysis of alkyl halides You have read (Unit 6, Class XII) that the carbon - halogen bond in alkyl or benzyl halides can be easily cleaved by a nucleophile. Hence, an alkyl or benzyl halide on reaction with an ethanolic solution of ammonia undergoes nucleophilic substitution reaction in which the halogen atom is replaced by an amino (–NH2) group. This process of cleavage of the C–X bond by ammonia molecule is known as ammonolysis. The reaction is carried out in a sealed tube at 373 K. The primary amine thus obtained behaves as a nucleophile and can further react with alkyl halide to form secondary and tertiary amines, and finally quaternary ammonium salt. Chemistry 262 Reprint 2025-26 The free amine can be obtained from the ammonium salt by treatment with a strong base: Ammonolysis has the disadvantage of yielding a mixture of primary, secondary and tertiary amines and also a quaternary ammonium salt. However, primary amine is obtained as a major product by taking large excess of ammonia. The order of reactivity of halides with amines is RI > RBr >RCl. Write chemical equations for the following reactions: ExampleExampleExampleExampleExample 9.19.19.19.19.1 (i) Reaction of ethanolic NH3 with C2H5Cl. (ii) Ammonolysis of benzyl chloride and reaction of amine so formed with two moles of CH3Cl. SolutionSolutionSolutionSolutionSolution 3. Reduction of nitriles Nitriles on reduction with lithium aluminium hydride (LiAlH4) or catalytic hydrogenation produce primary amines. This reaction is used for ascent of amine series, i.e., for preparation of amines containing one carbon atom more than the starting amine. 4. Reduction of amides The amides on reduction with lithium aluminium hydride yield amines. 263 Amines Reprint 2025-26 5. Gabriel phthalimide synthesis Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine. Aromatic primary amines cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. 6. Hoffmann bromamide degradation reaction Hoffmann developed a method for preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide. ExampleExampleExampleExampleExample 9.29.29.29.29.2 Write chemical equations for the following conversions: (i) CH3–CH2–Cl into CH3–CH2–CH2–NH2 (ii) C6H5–CH2–Cl into C6H5–CH2–CH2–NH2 SolutionSolutionSolutionSolutionSolution Chemistry 264 Reprint 2025-26 Write structures and IUPAC names of ExampleExampleExampleExampleExample 9.39.39.39.39.3 (i) the amide which gives propanamine by Hoffmann bromamide reaction. (ii) the amine produced by the Hoffmann degradation of benzamide. SolutionSolutionSolutionSolutionSolution (i) Propanamine contains three carbons. Hence, the amide molecule must contain four carbon atoms. Structure and IUPAC name of the starting amide with four carbon atoms are given below: Butanamide (ii) Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is aromatic primary amine containing six carbon atoms. Aniline or benzenamine IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 9.3 How will you convert (i) Benzene into aniline (ii) Benzene into N, N-dimethylaniline (iii) Cl–(CH2)4–Cl into hexan-1,6-diamine? 9.59.59.59.59.5 PhysicalPhysicalPhysicalPhysicalPhysical The lower aliphatic amines are gases with fishy odour. Primary amines with three or more carbon atoms are liquid and still higher ones are PropertiesPropertiesPropertiesPropertiesProperties solid. Aniline and other arylamines are usually colourless but get coloured on storage due to atmospheric oxidation. Lower aliphatic amines are soluble in water because they can form hydrogen bonds with water molecules. However, solubility decreases with increase in molar mass of amines due to increase in size of the hydrophobic alkyl part. Higher amines are essentially insoluble in water. Considering the electronegativity of nitrogen of amine and oxygen of alcohol as 3.0 and 3.5 respectively, you can predict the pattern of solubility of amines and alcohols in water. Out of butan-1-ol and butan-1-amine, which will be more soluble in water and why? Amines are soluble in organic solvents like alcohol, ether and benzene. You may remember that alcohols are more polar than amines and form stronger intermolecular hydrogen bonds than amines. Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. This intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it. Tertiary amines do not have intermolecular association due to the absence of hydrogen atom available for hydrogen bond formation. Therefore, the order of boiling points of isomeric amines is as follows: 265 Amines Reprint 2025-26 Primary > Secondary > Tertiary Intermolecular hydrogen bonding in primary amines is shown in Fig. 9.2. Fig. 9.2 Intermolecular hydrogen bonding in primary amines Boiling points of amines, alcohols and alkanes of almost the same molar mass are shown in Table 9.2. Table 9.2: Comparison of Boiling Points of Amines, Alcohols and Alkanes of Similar Molecular Masses Sl. No. Compound Molar mass b.p./K 1. n-C4H9NH2 73 350.8 2. (C2H5)2NH 73 329.3 3. C2H5N(CH3)2 73 310.5 4. C2H5CH(CH3)2 72 300.8 5. n-C4H9OH 74 390.3 9.69.69.69.69.6 ChemicalChemicalChemicalChemicalChemical Difference in electronegativity between nitrogen and hydrogen atoms and the presence of unshared pair of electrons over the nitrogen atom makes ReactionsReactionsReactionsReactionsReactions amines reactive. The number of hydrogen atoms attached to nitrogen atom also decides the course of reaction of amines; that is why primary (–NH2), secondary N H and tertiary amines N differ in many reactions. Moreover, amines behave as nucleophiles due to the presence of unshared electron pair. Some of the reactions of amines are described below: 1. Basic character of amines Amines, being basic in nature, react with acids to form salts. Chemistry 266 Reprint 2025-26 Amine salts on treatment with a base like NaOH, regenerate the parent amine. Amine salts are soluble in water but insoluble in organic solvents like ether. This reaction is the basis for the separation of amines from the non basic organic compounds insoluble in water. The reaction of amines with mineral acids to form ammonium salts shows that these are basic in nature. Amines have an unshared pair of electrons on nitrogen atom due to which they behave as Lewis base. Basic character of amines can be better understood in terms of their Kb and pKb values as explained below:       R  NH 3   OH  K =  R  NH2  H2 O        R  N H3   O H  or K [H 2 O] =  R  NH 2        R  N H 3   O H  or K b =  R  NH2  pKb = –log Kb Larger the value of Kb or smaller the value of pKb, stronger is the base. The pKb values of few amines are given in Table 9.3. pKb value of ammonia is 4.75. Aliphatic amines are stronger bases than ammonia due to +I effect of alkyl groups leading to high electron density on the nitrogen atom. Their pKb values lie in the range of 3 to 4.22. On the other hand, aromatic amines are weaker bases than ammonia due to the electron withdrawing nature of the aryl group. Table 9.3: pKb Values of Amines in Aqueous Phase Name of amine pKb Methanamine 3.38 N-Methylmethanamine 3.27 N,N-Dimethylmethanamine 4.22 Ethanamine 3.29 N-Ethylethanamine 3.00 N,N-Diethylethanamine 3.25 Benzenamine 9.38 Phenylmethanamine 4.70 N-Methylaniline 9.30 N,N-Dimethylaniline 8.92 267 Amines Reprint 2025-26 You may find some discrepancies while trying to interpret the Kb values of amines on the basis of +I or –I effect of the substituents present in amines. Besides inductive effect, there are other effects like solvation effect, steric hinderance, etc., which affect the basic strength of amines. Just ponder over. You may get the answer in the following paragraphs. Structure-basicity relationship of amines Basicity of amines is related to their structure. Basic character of an amine depends upon the ease of formation of the cation by accepting a proton from the acid. The more stable the cation is relative to the amine, more basic is the amine. (a) Alkanamines versus ammonia Let us consider the reaction of an alkanamine and ammonia with a proton to compare their basicity. Due to the electron releasing nature of alkyl group, it pushes electrons towards nitrogen and thus makes the unshared electron pair more available for sharing with the proton of the acid. Moreover, the substituted ammonium ion formed from the amine gets stabilised due to dispersal of the positive charge by the +I effect of the alkyl group. Hence, alkylamines are stronger bases than ammonia. Thus, the basic nature of aliphatic amines should increase with increase in the number of alkyl groups. This trend is followed in the gaseous phase. The order of basicity of amines in the gaseous phase follows the expected order: tertiary amine > secondary amine > primary amine > NH3. The trend is not regular in the aqueous state as evident by their pKb values given in Table 9.3. In the aqueous phase, the substituted ammonium cations get stabilised not only by electron releasing effect of the alkyl group (+I) but also by solvation with water molecules. The greater the size of the ion, lesser will be the solvation and the less stabilised is the ion. The order of stability of ions are as follows: Decreasing order of extent of H-bonding in water and order of stability of ions by solvation. Chemistry 268 Reprint 2025-26 Greater is the stability of the substituted ammonium cation, stronger should be the corresponding amine as a base. Thus, the order of basicity of aliphatic amines should be: primary > secondary > tertiary, which is opposite to the inductive effect based order. Secondly, when the alkyl group is small, like –CH3 group, there is no steric hindrance to H-bonding. In case the alkyl group is bigger than CH3 group, there will be steric hinderance to H-bonding. Therefore, the change of nature of the alkyl group, e.g., from –CH3 to –C2H5 results in change of the order of basic strength. Thus, there is a subtle interplay of the inductive effect, solvation effect and steric hinderance of the alkyl group which decides the basic strength of alkyl amines in the aqueous state. The order of basic strength in case of methyl substituted amines and ethyl substituted amines in aqueous solution is as follows: (C2H5)2NH > (C2H5)3N > C2H5NH2 > NH3 (CH3)2NH > CH3NH2 > (CH3)3N > NH3 (b) Arylamines versus ammonia pKb value of aniline is quite high. Why is it so? It is because in aniline or other arylamines, the -NH2 group is attached directly to the benzene ring. It results in the unshared electron pair on nitrogen atom to be in conjugation with the benzene ring and thus making it less available for protonation. If you write different resonating structures of aniline, you will find that aniline is a resonance hybrid of the following five structures. On the other hand, anilinium ion obtained by accepting a proton can have only two resonating structures (kekule). We know that greater the number of resonating structures, greater is the stability. Thus you can infer that aniline (five resonating structures) is more stable than anilinium ion. Hence, the proton acceptability or the basic nature of aniline or other aromatic amines would be less than that of ammonia. In case of substituted aniline, it is observed that electron releasing groups like –OCH3, –CH3 increase basic strength whereas electron withdrawing groups like –NO2, –SO3H, –COOH, –X decrease it. 269 Amines Reprint 2025-26 ExampleExampleExampleExampleExample 9.49.49.49.49.4 Arrange the following in decreasing order of their basic strength: C6H5NH2, C2H5NH2, (C2H5)2NH, NH3 SolutionSolutionSolutionSolutionSolution The decreasing order of basic strength of the above amines and ammonia follows the following order: (C2H5)2NH > C2H5NH2 > NH3 > C6H5NH2 2. Alkylation Amines undergo alkylation on reaction with alkyl halides (refer Unit 6, Class XII). 3. Acylation Aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution reaction. This reaction is known as acylation. You can consider this reaction as the replacement of hydrogen atom of –NH2 or >N–H group by the acyl group. The products obtained by acylation reaction are known as amides. The reaction is carried out in the presence of a base stronger than the amine, like pyridine, which removes HCl so formed and shifts the equilibrium to the right hand side. Amines also react with benzoyl chloride (C6H5COCl). This reaction is known as benzoylation. CH 3 NH 2  C 6 H 5 CO Cl  CH 3 NH CO C 6 H 5  H Cl Methanamine Benzoyl chloride N  Methylbenzamide What do you think is the product of the reaction of amines with carboxylic acids ? They form salts with amines at room temperature. Chemistry 270 Reprint 2025-26 4. Carbylamine reaction Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines. 5. Reaction with nitrous acid Three classes of amines react differently with nitrous acid which is prepared in situ from a mineral acid and sodium nitrite. (a) Primary aliphatic amines react with nitrous acid to form aliphatic diazonium salts which being unstable, liberate nitrogen gas quantitatively and alcohols. Quantitative evolution of nitrogen is used in estimation of amino acids and proteins. (b) Aromatic amines react with nitrous acid at low temperatures (273-278 K) to form diazonium salts, a very important class of compounds used for synthesis of a variety of aromatic compounds discussed in Section 9.7. Secondary and tertiary amines react with nitrous acid in a different manner. 6. Reaction with arylsulphonyl chloride Benzenesulphonyl chloride (C6H5SO2Cl), which is also known as Hinsberg’s reagent, reacts with primary and secondary amines to form sulphonamides. (a) The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide. The hydrogen attached to nitrogen in sulphonamide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali. (b) In the reaction with secondary amine, N,N-diethyl- benzenesulphonamide is formed. O O S Cl + H N C 2H 5 S N C 2H 5 + HCl O C 2H 5 O C 2H 5 N,N-Diethylbenzenesulphonamide 271 Amines Reprint 2025-26 Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali. (c) Tertiary amines do not react with benzenesulphonyl chloride. This property of amines reacting with benzenesulphonyl chloride in a different manner is used for the distinction of primary, secondary and tertiary amines and also for the separation of a mixture of amines. However, these days benzenesulphonyl chloride is replaced by p-toluenesulphonyl chloride. 7. Electrophilic substitution You have read earlier that aniline is a resonance hybrid of five structures. Where do you find the maximum electron density in these structures? Ortho- and para-positions to the –NH2 group become centres of high electron density. Thus –NH2 group is ortho and para directing and a powerful activating group. (a) Bromination: Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline. The main problem encountered during electrophilic substitution reactions of aromatic amines is that of their very high reactivity. Substitution tends to occur at ortho- and para-positions. If we have to prepare monosubstituted aniline derivative, how can the activating effect of –NH2 group be controlled ? This can be done by protecting the -NH2 group by acetylation with acetic anhydride, then carrying out the desired substitution followed by hydrolysis of the substituted amide to the substituted amine. The lone pair of electrons on nitrogen of acetanilide interacts with oxygen atom due to resonance as shown below: Chemistry 272 Reprint 2025-26 Hence, the lone pair of electrons on nitrogen is less available for donation to benzene ring by resonance. Therefore, activating effect of –NHCOCH3 group is less than that of amino group. (b) Nitration: Direct nitration of aniline yields tarry oxidation products in addition to the nitro derivatives. Moreover, in the strongly acidic medium, aniline is protonated to form the anilinium ion which is meta directing. That is why besides the ortho and para derivatives, significant amount of meta derivative is also formed. However, by protecting the –NH2 group by acetylation reaction with acetic anhydride, the nitration reaction can be controlled and the p-nitro derivative can be obtained as the major product. (c) Sulphonation: Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate which on heating with sulphuric acid at 453-473K produces p-aminobenzene sulphonic acid, commonly known as sulphanilic acid, as the major product. Aniline does not undergo Friedel-Crafts reaction (alkylation and acetylation) due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. Due to this, nitrogen of aniline acquires positive charge and hence acts as a strong deactivating group for further reaction. 273 Amines Reprint 2025-26 IntextIntextIntextIntextIntext QuestionsQuestionsQuestionsQuestionsQuestions 9.4 Arrange the following in increasing order of their basic strength: (i) C2H5NH2, C6H5NH2, NH3, C6H5CH2NH2 and (C2H5)2NH (ii) C2H5NH2, (C2H5)2NH, (C2H5)3N, C6H5NH2 (iii) CH3NH2, (CH3)2NH, (CH3)3N, C6H5NH2, C6H5CH2NH2. 9.5 Complete the following acid-base reactions and name the products: (i) CH3CH2CH2NH2 + HCl ® (ii) (C2H5)3N + HCl ® 9.6 Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. 9.7 Write chemical reaction of aniline with benzoyl chloride and write the name of the product obtained. 9.8 Write structures of different isomers corresponding to the molecular formula, C3H9N. Write IUPAC names of the isomers which will liberate nitrogen gas on treatment with nitrous acid. II. DIAZONIUM SALTS  – The diazonium salts have the general formula R N 2 X where R stands – – – for an aryl group and X ion may be Cl Br, HSO,4 BF,4 etc. They are named by suffixing diazonium to the name of the parent hydrocarbon from which they are formed, followed by the name of anion such as  chloride, hydrogensulphate, etc. The N 2 group is called diazonium  – group. For example, C 6 H 5 N 2 Cl is named as benzenediazonium chloride and C6H5N2+HSO4– is known as benzenediazonium hydrogensulphate. Primary aliphatic amines form highly unstable alkyldiazonium salts (refer to Section 9.6). Primary aromatic amines form arenediazonium salts which are stable for a short time in solution at low temperatures (273-278 K). The stability of arenediazonium ion is explained on the basis of resonance. 9.79.79.79.79.7 MethodMethodMethodMethodMethod ofofofofof Benzenediazonium chloride is prepared by the reaction of aniline with PreparationPreparationPreparationPreparationPreparation nitrous acid at 273-278K. Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with hydrochloric acid. The ofofofofof DiazoniunDiazoniunDiazoniunDiazoniunDiazoniun conversion of primary aromatic amines into diazonium salts is known SaltsSaltsSaltsSaltsSalts as diazotisation. Due to its instability, the diazonium salt is not generally stored and is used immediately after its preparation.  – C 6 H 5 N H 2  NaNO 2  2H Cl 273  278K C 6 H 5 N 2 Cl  Na Cl  2H 2 O Chemistry 274 Reprint 2025-26 9.89.89.89.89.8 PhysicalPhysicalPhysicalPhysicalPhysical Benzenediazonium chloride is a colourless crystalline solid. It is readily soluble in water and is stable in cold but reacts with water when PropertiesPropertiesPropertiesPropertiesProperties warmed. It decomposes easily in the dry state. Benzenediazonium fluoroborate is water insoluble and stable at room temperature. 9.99.99.99.99.9 ChemicalChemicalChemicalChemicalChemical The reactions of diazonium salts can be broadly divided into two ReactionsReactionsReactionsReactionsReactions categories, namely (A) reactions involving displacement of nitrogen and (B) reactions involving retention of diazo group. A. Reactions involving displacement of nitrogen Diazonium group being a very good leaving group, is substituted by other groups such as Cl–, Br –, I – , CN – and OH– which displace nitrogen from the aromatic ring. The nitrogen formed escapes from the reaction mixture as a gas. 1. Replacement by halide or cyanide ion: The Cl –, Br– and CN – nucleophiles can easily be introduced in the benzene ring in the presence of Cu(I) ion. This reaction is called Sandmeyer reaction. 2 2 2 2 Alternatively, chlorine or bromine can also be introduced in the benzene ring by treating the diazonium salt solution with corresponding halogen acid in the presence of copper powder. This is referred as Gatterman reaction. The yield in Sandmeyer reaction is found to be better than Gattermann reaction. 2. Replacement by iodide ion: Iodine is not easily introduced into the benzene ring directly, but, when the diazonium salt solution is treated with potassium iodide, iodobenzene is formed. 3. Replacement by fluoride ion: When arenediazonium chloride is treated with fluoroboric acid, arene diazonium fluoroborate is precipitated which on heating decomposes to yield aryl fluoride. 4. Replacement by H: Certain mild reducing agents like hypophosphorous acid (phosphinic acid) or ethanol reduce diazonium salts to arenes and themselves get oxidised to phosphorous acid and ethanal, respectively. 275 Amines Reprint 2025-26 5. Replacement by hydroxyl group: If the temperature of the diazonium salt solution is allowed to rise upto 283 K, the salt gets hydrolysed to phenol. 6. Replacement by –NO2 group: When diazonium fluoroborate is heated with aqueous sodium nitrite solution in the presence of copper, the diazonium group is replaced by –NO2 group. B. Reactions involving retention of diazo group coupling reactions The azo products obtained have an extended conjugate system having both the aromatic rings joined through the –N=N– bond. These compounds are often coloured and are used as dyes. Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para position is coupled with the diazonium salt to form p-hydroxyazobenzene. This type of reaction is known as coupling reaction. Similarly the reaction of diazonium salt with aniline yields p-aminoazobenzene. This is an example of electrophilic substitution reaction. 9.10.10.10.10.10 ImportanceImportanceImportanceImportanceImportance From the above reactions, it is clear that the diazonium salts are very ofofofofof DiazoniumDiazoniumDiazoniumDiazoniumDiazonium good intermediates for the introduction of –F, –Cl, –Br, –I, –CN, –OH, –NO2 groups into the aromatic ring. SaltsSaltsSaltsSaltsSalts ininininin Aryl fluorides and iodides cannot be prepared by direct halogenation. SynthesisSynthesisSynthesisSynthesisSynthesis The cyano group cannot be introduced by nucleophilic substitution of ofofofofof AromaticAromaticAromaticAromaticAromatic chlorine in chlorobenzene but cyanobenzene can be easily obtained from diazonium salt. CompoundsCompoundsCompoundsCompoundsCompounds Thus, the replacement of diazo group by other groups is helpful in Chemistry 276 Reprint 2025-26 preparing those substituted aromatic compounds which cannot be prepared by direct substitution in benzene or substituted benzene. How will you convert 4-nitrotoluene to 2-bromobenzoic acid ? ExampleExampleExampleExampleExample 9.59.59.59.59.5 SolutionSolutionSolutionSolutionSolution IntextIntextIntextIntextIntext QuestionQuestionQuestionQuestionQuestion 9.9 Convert (i) 3-Methylaniline into 3-nitrotoluene. (ii) Aniline into 1,3,5 - tribromobenzene. SummarySummarySummarySummarySummary Amines can be considered as derivatives of ammonia obtained by replacement of hydrogen atoms with alkyl or aryl groups. Replacement of one hydrogen atom of ammonia gives rise to structure of the type R-NH2, known as primary amine. Secondary amines are characterised by the structure R2NH or R-NHR¢ and tertiary amines by R3N, RNR¢R¢¢¢¢¢¢¢¢¢¢ or R2NR¢.¢.¢.¢.¢. Secondary and tertiary amines are known as simple amines if the alkyl or aryl groups are the same and mixed amines if the groups are different. Like ammonia, all the three types of amines have one unshared electron pair on nitrogen atom due to which they behave as Lewis bases. Amines are usually formed from nitro compounds, halides, amides, imides, etc. They exhibit hydrogen bonding which influence their physical properties. In alkylamines, a combination of electron releasing, steric and H-bonding factors influence the stability of the substituted ammonium cations in protic polar solvents and thus affect the basic nature of amines. Alkyl amines are found to be stronger bases than ammonia. In aromatic amines, electron releasing and withdrawing groups, respectively increase and decrease their basic character. Aniline is a weaker base 277 Amines Reprint 2025-26 than ammonia. Reactions of amines are governed by availability of the unshared pair of electrons on nitrogen. Influence of the number of hydrogen atoms at nitrogen atom on the type of reactions and nature of products is responsible for identification and distinction between primary, secondary and tertiary amines. p-Toluenesulphonyl chloride is used for the identification of primary, secondary and tertiary amines. Presence of amino group in aromatic ring enhances reactivity of the aromatic amines. Reactivity of aromatic amines can be controlled by acylation process, i.e., by treating with acetyl chloride or acetic anhydride. Tertiary amines like trimethylamine are used as insect attractants. Aryldiazonium salts, usually obtained from arylamines, undergo replacement of the diazonium group with a variety of nucleophiles to provide advantageous methods for producing aryl halides, cyanides, phenols and arenes by reductive removal of the diazo group. Coupling reaction of aryldiazonium salts with phenols or arylamines give rise to the formation of azo dyes. Exercises 9.1 Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines. (i) (CH3)2CHNH2 (ii) CH3(CH2)2NH2 (iii) CH3NHCH(CH3)2 (iv) (CH3)3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2)2NCH3 (vii) m–BrC6H4NH2 9.2 Give one chemical test to distinguish between the following pairs of compounds. (i) Methylamine and dimethylamine (ii) Secondary and tertiary amines (iii) Ethylamine and aniline (iv) Aniline and benzylamine (v) Aniline and N-methylaniline. 9.3 Account for the following: (i) pKb of aniline is more than that of methylamine. (ii) Ethylamine is soluble in water whereas aniline is not. (iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. (iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. (v) Aniline does not undergo Friedel-Crafts reaction. (vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines. 9.4 Arrange the following: (i) In decreasing order of the pKb values: C2H5NH2, C6H5NHCH3, (C2H5)2NH and C6H5NH2 (ii) In increasing order of basic strength: C6H5NH2, C6H5N(CH3)2, (C2H5)2NH and CH3NH2 (iii) In increasing order of basic strength: (a) Aniline, p-nitroaniline and p-toluidine Chemistry 278 Reprint 2025-26 (b) C6H5NH2, C6H5NHCH3, C6H5CH2NH2. (iv) In decreasing order of basic strength in gas phase: C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3 (v) In increasing order of boiling point: C2H5OH, (CH3)2NH, C2H5NH2 (vi) In increasing order of solubility in water: C6H5NH2, (C2H5)2NH, C2H5NH2.