Question

References Use the References to access important values if needed for this question. Write a balanced net ionic equation to show why the solubility of Zn(CN)2 (s) increases in the presence of a strong acid and calculate the equilibrium constant for the reaction of this sparingly soluble salt with acid. Use the pull-down boxes to specify states such as (aq) or (s). Zn(CN)2 (s) |2н,о* (aq) Zn (aq) 21H2O 2HCN (aq) + K- 8x10 16 Submit Answer

please check the answers this is my last attempt.

Answer 1

Ans :-

Partial dissociation of Zn(CN)₂ in aqueous solution is :

Zn(CN)₂ (s) <----------------> Zn²⁺ (aq) + 2 CN^- (aq) , K_sp = 8.0 x 10^-12 M² ...............(1)

With the addition of strong acid say H⁺ :

2 H⁺ (aq) + 2 CN^- (aq) --------------> 2 HCN (aq) , K_f = (1/K_a of HCN)² ..............................(2)

Expression of formation constant (K_f) is :

K_f = (1/K_a of HCN)²

Here, K_a = Acid dissociation constant of HCN = 4.0 x 10^-10

So,

K_f = (1/4.0 x 10^-10 )²

K_f = 6.25 x 10¹⁸  

Adding equations (1) and (2) in order to obtain equation of sparingly soluble salt with acid

Zn(CN)₂ (s) + 2 H⁺ (aq) -----------------> Zn²⁺ (aq) + 2 HCN (aq) , K = K_sp x K_f

So,

K = K_sp x K_f

K = 8.0 x 10^-12 x 6.25 x 10¹⁸  

K = 5.0 x 10⁷

Hence, Equilibrium constant for sparingly soluble salt with acid = K = 5.0 x 10⁷