Question 11. Answer: Common ion Effect: When a salt of a weak acid is added to the acid itself, the dissociation of the weak acid is suppressed further. Acetic acid is a weak acid. CH_{step three}COOH (aq) \(\rightleftharpoons\) H + (aq)+ CH₃COO – (aq)

However, the added salt, sodium acetate, completely dissociates to produce Na + and CH₃COO ion. CH₃COONa (aq) > Na + (aq) + CH₃COO (aq) Hence, the overall concentration ofCH3COO is increased, and the acid dissociation equilibrium is disturbed.

We know from Le chatelier’s principle that when stress is applied to a system at equilibrium, the system adjusts itself to nullify the effect produced by that stress. So, in order to maintain the equilibrium, the excess CH₃COO – ions combine with H ions to produce much more unionized CH₃COOH i.e.,

the equilibrium will shift towards the left. In other words, the dissociation of CH₃COOH is suppressed. Thus, the dissociation of a weak acid (CH₃COOH) is suppressed in the presence of a salt (CH₃COONa) containing an ion common to the weak electrolyte. It is called the common ion effect.

Question 12. Derive an expression for Ostwald’s dilution law. Answer: Ostwald’s dilution law: It relates the dissociation constant of the weak acid (K_a) with its degree of dissociation (?) and the concentration (c). Considering a weak acid, acetic acid. The dissociation of acetic acid can be represented as, CH₃COOH \(\rightleftharpoons\) CH₃COO – + H + The dissociation constant of acetic acid is,

We realize one weakened acid dissociates in order to a escort Kent highly quick the quantity compared to one to, an effective is so brief. equation (1) gets,

This isn’t completely dissociated inside the a keen aqueous services so because of this another balance can be found

Question 13. Define pH. Answer: pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution. pH = – log₁₀ [H₃O] (or) pH = – log₁₀ [H + ]

[OH – ] = three times 10 3 Yards. [pH + pOH = 1cuatro] pH = fourteen – pOH pH = 14 – ( – diary [OH – ]) = fourteen + diary [OH – ] = 14 + journal (3 times ten -3 ) = 14 + log step 3 + log 10 -step 3 = 11 + 0.4771 pH =

Question 15. 50 ml of 0.05 M HNO₃ is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution. Solution. Number of moles of HNO₃ = 0.05 x 50 x = 2.5 x 10 -3 Number of moles of KOH = 0.025 x 50 x 10 -3 = 1.25 x 10 -3 Number of moles of HNO₃ after mixing = 2.5 x 10 -3 – 1.5 x 10 -3 = 1.25 x 10 -3

Question17

pH = – record [H + ] pH = – diary (1.twenty five x 10 -2 ) = 2 – 0.0969 = 1.9031

Question 16. The K_a value for HCN is 10 -9 . What is the pH of 0.4 M HCN solution? Answer: K_a =10 -9 c = O.4M pH = – log [H + ]

? pH = – log(dos x 10 -5 ) = – log 2 – diary (ten -5 ) = – 0.3010 + 5 pH = 4.699

Calculate the extent of hydrolysis and the pH of 0.1 M ammonium acetate Given that. K_a = K_b = 1.8 x 10 -5 Solution.

Question 18. Derive an expression for the hydrolysis constant and degree of hydrolysis of salt of strong acid and weak base. Answer: Let us consider the reactions between a strong acid, HCl, and a weak base, NH₄OH, to produce a salt, NH₄Cl, and water. HCl (aq) + NH₄OH (aq) \(\rightleftharpoons\) NH₄Cl (aq) + H₂O (I) NH₄CI(aq) > NH₄ + + Cl – (aq)