Salt Hydrolysis — Buffer solutions, Henderson equation
Ionic Equilibrium
6
JEE Qs
8%
Hard
90
min
Thoroughly understand the derivation of hydrolysis and buffer pH formulas to avoid rote memorization, enabling you to adapt to varied problem types and troubleshoot common mistakes effectively.
🧮 Key Formulas
✅ Key Points for JEE
- 1Salt hydrolysis involves the reaction of the cation/anion of a salt with water, which is the conjugate of a weak acid/base, to produce H3O+ or OH- ions, thereby changing the pH of the solution.
- 2Buffer solutions contain a weak acid and its conjugate base, or a weak base and its conjugate acid, and resist significant pH changes upon the addition of small amounts of strong acid or base.
- 3The Henderson-Hasselbalch equation is a convenient way to calculate the pH of a buffer solution, but it is an approximation valid for dilute solutions where the concentrations of the weak acid/base and its salt are not extremely low.
- 4The maximum buffer capacity occurs when the concentrations of the weak acid/base and its conjugate salt are equal ([Acid] = [Salt] or [Base] = [Salt]), where pH = pKa or pOH = pKb respectively.
- 5A buffer is most effective within a pH range of pKa ± 1 (for acidic buffers) or pOH range of pKb ± 1 (for basic buffers).
⚠️ Common Mistakes
- ✕Confusing salt hydrolysis with simple dissociation; only the ion forming from a weak acid or weak base undergoes hydrolysis.
- ✕Incorrectly identifying the hydrolyzing ion (e.g., assuming Na+ from CH3COONa hydrolyzes) or the type of salt, leading to wrong formula application.
- ✕Applying the Henderson-Hasselbalch equation indiscriminately for solutions that are not buffers, or for buffer calculations after adding strong acids/bases that exceed the buffer's capacity.
- ✕Not accounting for changes in concentration due to dilution when mixing solutions or adding substances to a buffer, which can lead to errors in pH calculation.
📝 Practice Questions
See allQ74.If 1 mM solution of ethylamine produces pH = 9, then the ionization constant (Kb) of ethylamine is 10−x . The value of x is ______ (nearest integer). [The degree of ionization of ethylamine can be neglected with respect to unity.]
Q61. Ksp for Cr(OH)3 is 1.6 × 10−30 . What is the molar solubility of this salt in water? (1) 1.8×10−30 (2) 5√1.8 × 10−30 27 × 10−30 (3) 4√1.6×10−3027 (4) 2√1.6
Q58.The molar solubility(s) of zirconium phosphate with molecular formula (Zr4+)3(PO3−4 )4 is given by relation : (1) Ksp 13 (2) Ksp 17 ( 9612 ) ( 6912 ) (3) Ksp 17 (4) Ksp 16 ( 8435 ) ( 5348 )
Q62.Which of the following happens when NH4OH is added gradually to the solution containing 1 M A2+ and 1MB3+ ions? Given : Ksp [A(OH)2] = 9 × 10−10 and Ksp [B(OH)3] = 27 × 10−18 at 298 K . (1) Both A(OH)2 and B(OH)3 do not show (2) A(OH)2 will precipitate before B(OH)3 precipitation with NH4OH (3) B(OH)3 will precipitate before A(OH)2 (4) A(OH)2 and B(OH)3 will precipitate together excess HCHO alkali−
Q68.pH of water is 7 at 25∘C. If water is heated to 80∘C., it's pH will : (1) Decrease (2) H+ concentration increases, OH− concentration decreases (3) Remains the same (4) Increase 2025 (23 Jan Shift 2) JEE Main Previous Year Paper
Q58.A weak acid HA has degree of dissociation x . Which option gives the correct expression of ( pH pKa )? (1) 0 (2) log(1 + 2x) (3) log ( 1−xx ) (4) log ( 1−xx )
NCERT Chapters
- Class 11 Chemistry Ch 7: Equilibrium