6.10 ACIDS, BASES AND SALTS a very high dielectric constant of 80. Thus, Acids, bases and salts find widespread when sodium chloride is dissolved in water, occurrence in nature. Hydrochloric acid the electrostatic interactions are reduced by present in the gastric juice is secreted by the a factor of 80 and this facilitates the ions to lining of our stomach in a significant amount move freely in the solution. Also, they are of 1.2-1.5 L/day and is essential for digestive well-separated due to hydration with water processes. Acetic acid is known to be the main molecules. constituent of vinegar. Lemon and orange juices contain citric and ascorbic acids, and tartaric acid is found in tamarind paste. As most of the acids taste sour, the word “acid” has been derived from a latin word “acidus” meaning sour. Acids are known to turn blue litmus paper into red and liberate dihydrogen on reacting with some metals. Similarly, bases are known to turn red litmus paper blue, taste bitter and feel soapy. A common example of a base is washing soda used for washing purposes. When acids and bases are mixed in the right proportion they react with each Fig.6.10 Dissolution of sodium chloride in water.other to give salts. Some commonly known Na+ and Cl– ions are stablised by their examples of salts are sodium chloride, barium hydration with polar water molecules. sulphate, sodium nitrate. Sodium chloride (common salt) is an important component of Comparing, the ionization of hydrochloric our diet and is formed by reaction between acid with that of acetic acid in water we find hydrochloric acid and sodium hydroxide. It that though both of them are polar covalent Faraday was born near London into a family of very limited means. At the age of 14 he was an apprentice to a kind bookbinder who allowed Faraday to read the books he was binding. Through a fortunate chance he became laboratory assistant to Davy, and during 1813-4, Faraday accompanied him to the Continent. During this trip he gained much from the experience of coming into contact with many of the leading scientists of the time. In 1825, he succeeded Davy as Director of the Royal Institution laboratories, and in 1833 he also became the first Fullerian Professor of Chemistry. Faraday’s first important work was on analytical chemistry. After 1821 Michael Faraday much of his work was on electricity and magnetism and different electromagnetic (1791–1867) phenomena. His ideas have led to the establishment of modern field theory. He discovered his two laws of electrolysis in 1834. Faraday was a very modest and kind hearted person. He declined all honours and avoided scientific controversies. He preferred to work alone and never had any assistant. He disseminated science in a variety of ways including his Friday evening discourses, which he founded at the Royal Institution. He has been very famous for his Christmas lecture on the ‘Chemical History of a Candle’. He published nearly 450 scientific papers. Reprint 2025-26 190 chemistry molecules, former is completely ionized into its constituent ions, while the latter is only Hydronium and Hydroxyl Ions partially ionized (< 5%). The extent to which Hydrogen ion by itself is a bare proton with very ionization occurs depends upon the strength small size (~10–15 m radius) and intense electric of the bond and the extent of solvation field, binds itself with the water molecule at of ions produced. The terms dissociation one of the two available lone pairs on it giving and ionization have earlier been used with H3O+. This species has been detected in many different meaning. Dissociation refers to the compounds (e.g., H3O+Cl–) in the solid state. In process of separation of ions in water already aqueous solution the hydronium ion is further existing as such in the solid state of the solute, hydrated to give species like H5O2+, H7O3 + and as in sodium chloride. On the other hand, H9O4+. Similarly the hydroxyl ion is hydrated to give several ionic species like H3O2–, H5O3–ionization corresponds to a process in which – and H7O4 etc.a neutral molecule splits into charged ions in the solution. Here, we shall not distinguish between the two and use the two terms interchangeably. 6.10.1 Arrhenius Concept of Acids and Bases According to Arrhenius theory, acids are H9O4+ substances that dissociates in water to give hydrogen ions H+(aq) and bases are 6.10.2 The Brönsted-Lowry Acids and substances that produce hydroxyl ions Bases OH –(aq). The ionization of an acid HX (aq) can The Danish chemist, Johannes Brönsted and be represented by the following equations: the English chemist, Thomas M. Lowry gave HX (aq) → H+(aq) + X– (aq) a more general definition of acids and bases. or According to Brönsted-Lowry theory, acid HX(aq) + H2O(l) → H3O+(aq) + X –(aq) is a substance that is capable of donating a hydrogen ion H+ and bases are substances A bare proton, H+ is very reactive and capable of accepting a hydrogen ion, H+. Incannot exist freely in aqueous solutions. short, acids are proton donors and bases areThus, it bonds to the oxygen atom of a solvent proton acceptors.water molecule to give trigonal pyramidal Consider the example of dissolution of NH3hydronium ion, H3O+ {[H (H2O)]+} (see box). In in H2O represented by the following equation:this chapter we shall use H+(aq) and H3O+(aq) interchangeably to mean the same i.e., a hydrated proton. Similarly, a base molecule like MOH ionizes in aqueous solution according to the equation: MOH(aq) → M+(aq) + OH–(aq) The hydroxyl ion also exists in the hydrated form in the aqueous solution. Arrhenius concept of acid and base, however, suffers The basic solution is formed due to the from the limitation of being applicable only to presence of hydroxyl ions. In this reaction, aqueous solutions and also, does not account water molecule acts as proton donor and for the basicity of substances like, ammonia ammonia molecule acts as proton acceptor which do not possess a hydroxyl group. and are thus, called Lowry-Brönsted acid and Reprint 2025-26 EQUILIBRIUM 191 Arrhenius was born near Uppsala, Sweden. He presented his thesis, on the conductivities of electrolyte solutions, to the University of Uppsala in 1884. For the next five years he travelled extensively and visited a number of research centers in Europe. In 1895 he was appointed professor of physics at the newly formed University of Stockholm, serving its rector from 1897 to 1902. From 1905 until his death he was Director of physical chemistry at the Nobel Institute in Stockholm. He continued to work for many years on electrolytic solutions. In 1899 he discussed the temperature dependence of reaction rates on the basis of an equation, now usually known as Arrhenius equation. He worked in a variety of fields, and made important contributions to Svante Arrhenius immunochemistry, cosmology, the origin of life, and the causes of ice age. He was (1859-1927) the first to discuss the ‘green house effect’ calling by that name. He received Nobel Prize in Chemistry in 1903 for his theory of electrolytic dissociation and its use in the development of chemistry. base, respectively. In the reverse reaction, in case of ammonia it acts as an acid by H+ is transferred from NH4+ to OH–. In this donating a proton. case, NH4+ acts as a Bronsted acid while Problem 6.12OH– acted as a Brönsted base. The acid-base What will be the conjugate bases for thepair that differs only by one proton is called following Brönsted acids: HF, H2SO4 anda conjugate acid-base pair. Therefore, OH– – HCO3 ?is called the conjugate base of an acid H2O and NH4+ is called conjugate acid of the base Solution NH3. If Brönsted acid is a strong acid then The conjugate bases should have one its conjugate base is a weak base and vice- proton less in each case and therefore the versa. It may be noted that conjugate acid corresponding conjugate bases are: F –, has one extra proton and each conjugate base HSO4– and CO32– respectively. has one less proton. Problem 6.13 Consider the example of ionization of Write the conjugate acids for the following hydrochloric acid in water. HCl(aq) acts as Brönsted bases: NH2–, NH3 and HCOO–. an acid by donating a proton to H2O molecule which acts as a base. Solution The conjugate acid should have one extra proton in each case and therefore the corresponding conjugate acids are: NH3, NH4+ and HCOOH respectively. Problem 6.14 The species: H2O, HCO3–, HSO4– and NH3 can act both as Bronsted acids and bases. For each case give the corresponding conjugate It can be seen in the above equation, that acid and conjugate base. water acts as a base because it accepts the Solution proton. The species H3O+ is produced when The answer is given in the following Table:water accepts a proton from HCl. Therefore, Cl– is a conjugate base of HCl and HCl is the Species Conjugate Conjugate conjugate acid of base Cl –. Similarly, H2O is acid base a conjugate base of an acid H3O+ and H3O+ is H2O H3O+ OH–a conjugate acid of base H2O. – 2– HCO3 H2CO3 CO3 It is interesting to observe the dual role – 2– HSO4 H2SO4 SO4of water as an acid and a base. In case of + – reaction with HCl water acts as a base while NH3 NH4 NH2 Reprint 2025-26 192 chemistry 6.10.3 Lewis Acids and Bases (HClO4), hydrochloric acid (HCl), hydrobromic G.N. Lewis in 1923 defined an acid as a acid (HBr), hyrdoiodic acid (HI), nitric acid species which accepts electron pair and base (HNO3) and sulphuric acid (H2SO4) are termed which donates an electron pair. As far as bases strong because they are almost completely are concerned, there is not much difference dissociated into their constituent ions in an between Brönsted-Lowry and Lewis concepts, aqueous medium, thereby acting as proton as the base provides a lone pair in both the (H+) donors. Similarly, strong bases like cases. However, in Lewis concept many lithium hydroxide (LiOH), sodium hydroxide acids do not have proton. A typical example (NaOH), potassium hydroxide (KOH), caesium is reaction of electron deficient species BF3 hydroxide (CsOH) and barium hydroxide with NH3. Ba(OH)2 are almost completely dissociated into ions in an aqueous medium giving BF3 does not have a proton but still acts hydroxyl ions, OH–. According to Arrhenius as an acid and reacts with NH3 by accepting concept they are strong acids and bases asits lone pair of electrons. The reaction can be they are able to completely dissociate andrepresented by, produce H3O+ and OH– ions respectively in BF3 + :NH3 → BF3:NH3 the medium. Alternatively, the strength of an Electron deficient species like AlCl3, Co3+, acid or base may also be gauged in terms of Mg2+, etc. can act as Lewis acids while species Brönsted-Lowry concept of acids and bases, like H2O, NH3, OH– etc. which can donate a wherein a strong acid means a good proton pair of electrons, can act as Lewis bases. donor and a strong base implies a good proton acceptor. Consider, the acid-base dissociation Problem 6.15 equilibrium of a weak acid HA, Classify the following species into Lewis HA(aq) + H2O(l)   H3O+(aq) + A–(aq) acids and Lewis bases and show how these conjugate conjugate act as such: acid base acid base (a) HO– (b) F – (c) H+ (d) BCl3 In section 6.10.2 we saw that acid (or Solution base) dissociation equilibrium is dynamic involving a transfer of proton in forward and (a) Hydroxyl ion is a Lewis base as it can reverse directions. Now, the question arises donate an electron lone pair (:OH– ). that if the equilibrium is dynamic then with (b) Flouride ion acts as a Lewis base passage of time which direction is favoured? as it can donate any one of its four What is the driving force behind it? In order electron lone pairs. to answer these questions we shall deal (c) A proton is a Lewis acid as it can into the issue of comparing the strengths accept a lone pair of electrons from of the two acids (or bases) involved in the bases like hydroxyl ion and fluoride dissociation equilibrium. Consider the two ion. acids HA and H3O+ present in the above mentioned acid-dissociation equilibrium. (d) BCl3 acts as a Lewis acid as it can We have to see which amongst them is a accept a lone pair of electrons from stronger proton donor. Whichever exceeds species like ammonia or amine in its tendency of donating a proton over the molecules. other shall be termed as the stronger acid

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.

4.14 Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?