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. CH₃COOH (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 1dos. 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 all know you to definitely weak acidic dissociates in order to a very short the amount as compared to one, an effective is so brief. formula (1) becomes,

This isn’t entirely dissociated for the an aqueous service so because of this the next harmony can be acquired

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 – ] = 3 x 10 step three M. [pH + pOH = 1cuatro] pH = fourteen – pOH pH = 14 – ( – journal [OH – ]) = fourteen + diary [OH – ] = fourteen + log (three times ten -3 ) = 14 + record step 3 + log ten -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 = – log (step 1.25 x 10 -dos ) = 2 – 0.0969 = step one.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 – log (10 -5 ) = – 0.3010 + 5 pH = cuatro.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 Salem escort reviews 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)