Question

Cyanide is used in several industrial applications, such as mining gold and electroplating silver. A danger of working with cyanide solutions is the potential hazard of releasing highly toxic hydrogen cyanide gas. What conditions are necessary to ensure that less than 1% of the cyanide in a particular solution is chemically present as HCN rather than CN-? (HCN pKa= 9.2)

(A) pH > 11.2

(B) pH > 10.2

(C) pH > 9.2

(D) pH > 8.2

(E) pH > 7.2

(F) pH < 11.2

(G) pH < 10.2

(H) pH < 9.2

(I) pH < 8.2

(J) pH < 7.2

-Please explain, Thank you.

Answer 1

From the question it is clear that [HCN]/[CN-] must be less than 1%

Since, less than 1% of the cyanide should be present as HCN rather than CN-. This implies [HCN] < 1% Ionization reaction of HCN is: HCN ---> H⁺ + CN^-   pK_a = 9.2 K_a = [H⁺][CN^-] / [HCN] For pK_a = pH , CN^- and HCN will be present in equal amounts i.e [CN^-] = [HCN] For pK_a < pH ,  [CN^-] > [HCN] For pK_a > pH ,  [CN^-] < [HCN]

Hence when pH > 9.2 then more conc of CN^- would be there than HCN.

Now , we want to calculate the conditions when [HCN]< 1%

So, [HCN]/[CN-] < 1% .........(1)

since ,  K_a = [H⁺][CN^-] / [HCN] , so  [HCN]/[CN-] = [H⁺]/K_a Putting this in equation (1)   [H⁺]/K_a = [HCN]/[CN-] < 1%    [H⁺]/K_a < 1/100 Taking log both sides : log [H⁺] - log(K_a) < log(10^-2) Multiplying -1 both sides: -log [H⁺] - (-log(K_a)) > 2 (inequality sign changes on multiplication by a negative number) pH - pK_a > 2     pH > 2 + pK_a        pH > 2 + 9.2 pH > 11.2

For pH>11.2, less than 1% of HCN is present in the solution.