Hydrolysis of Salts - UNSOLVED PRACTICE SET

Q7. Define hydrolysis of salts. Why do salts of strong acid and strong base not undergo hydrolysis?

Q8. Write the hydrolysis reaction for NH₄Cl in water. Explain why the resulting solution is acidic.

Q9. For a salt of weak acid and strong base, derive the relationship between Kh, Kw, and Ka.

Q10. Predict whether the aqueous solutions of the following salts will be acidic, basic, or neutral:

(a) NaCl

(b) CH₃COONa

(c) NH₄Cl

(d) K₂SO₄

Q11. Explain why a solution of CH₃COONH₄ is approximately neutral, even though both CH₃COO⁻ and NH₄⁺ undergo hydrolysis.

Q12. The pH of 0.1 M CH₃COONa solution is 8.87. Explain how this pH value confirms that CH₃COO⁻ undergoes hydrolysis.

Q13. (a) Explain the phenomenon of salt hydrolysis. Classify salts into four categories based on the nature of the acid and base from which they are derived, and describe the hydrolysis behaviour of each category.

(b) Write the hydrolysis equation for CH₃COONa and derive the expression for:

(i) Hydrolysis constant (Kh)

(ii) Degree of hydrolysis (h)

(iii) pH of the solution

(c) Calculate the pH of 0.1 M CH₃COONa solution. (Ka of CH₃COOH = 1.8 × 10⁻⁵, Kw = 1.0 × 10⁻¹⁴)

Q14. (a) Derive the expression for the pH of a solution of a salt of weak base and strong acid (e.g., NH₄Cl).

(b) Calculate the pH of 0.1 M NH₄Cl solution. (Kb of NH₄OH = 1.8 × 10⁻⁵)

(c) Explain why the degree of hydrolysis increases with dilution for a salt of weak acid and strong base.

Q15. (a) Explain what happens when a salt of weak acid and weak base (e.g., CH₃COONH₄) is dissolved in water. Derive the expression for its pH.

(b) Show that for such a salt, pH = 7 + ½(pKa – pKb).

(c) Calculate the pH of 0.1 M CH₃COONH₄ solution. (Ka = Kb = 1.8 × 10⁻⁵)

(d) What would be the pH if Ka > Kb? Explain with reasoning.

Q16. Calculate the pH of the following salt solutions:

(a) 0.1 M NaCl

(b) 0.05 M CH₃COONa (Ka of CH₃COOH = 1.8 × 10⁻⁵)

(c) 0.1 M NH₄Cl (Kb of NH₄OH = 1.8 × 10⁻⁵)

(d) 0.1 M CH₃COONH₄ (Ka = Kb = 1.8 × 10⁻⁵)

For each case, also calculate the degree of hydrolysis.

Q17. A 0.05 M solution of sodium benzoate (C₆H₅COONa) has a pH of 8.6.

(a) Write the hydrolysis reaction of benzoate ion.

(b) Calculate the hydrolysis constant (Kh) of sodium benzoate.

(c) Calculate the Ka of benzoic acid.

(d) Calculate the degree of hydrolysis of sodium benzoate.

(Kw = 1.0 × 10⁻¹⁴)

Q18. In India, understanding salt hydrolysis has practical applications in daily life and industry:

(a) Baking soda (NaHCO₃) is used in cooking. Write the hydrolysis reaction of HCO₃⁻ and explain why a solution of NaHCO₃ is slightly basic. (Ka₁ of H₂CO₃ = 4.3 × 10⁻⁷, Ka₂ = 5.6 × 10⁻¹¹)

(b) Calculate the pH of 0.1 M NaHCO₃ solution.

(c) Alum (KAl(SO₄)₂·12H₂O) is used in water purification. Explain how the hydrolysis of Al³⁺ helps in coagulating impurities. Write the hydrolysis reaction of Al³⁺.

(d) A farmer tests the pH of soil treated with ammonium sulphate fertiliser and finds it to be acidic. Explain this observation using the concept of salt hydrolysis. Calculate the pH of 0.05 M (NH₄)₂SO₄ solution. (Kb of NH₄OH = 1.8 × 10⁻⁵)