Question

calculate the percent hydrolysis In A 0.06 M solution of KCN

Answer 1

HCN <---> H+ + CN- 
H2O <---> H+ + OH- 

K(HCN) = [H+][CN-] / [HCN] 
Kw = [H+][OH-] = 1x10^-14 

K(rxn) = [HCN] [OH-] / [CN-] = Kw / K(HCN) 
1x10^-14 / 6x10^-10 = x^2 / (0.06 - x) << round 
1.667x10^-5 = x^2 / 0.06 
1x10^-6 = x^2 
x = 0.001 M 

% dissociation = 0.001 / 0.06 X 100 = 1.66667 %

Answer 2

The percent hydrolysis in 0.66 M solution of KCN is 1.66667 %

Explanation:

Given Data

KCN solution = 0.06 M

K(HCN) = 6  × $6^(-10)$

Kw = 1 × $10^(-14)$

Kw ---> dissociation constant or ionization constant

The equation for KCN can be written as

 KCN --> K+ + CN-

Also, CN- + H2O <---> HCN + OH-  

The equation for HCN and H2O can be written as

HCN <-------> H+ + CN-

H2O <---> H+ + OH-

K(HCN)  = [H+][CN-] / [HCN]

Kw = [H+][OH-] = 1x10^-14

K(rxn) = [HCN] [OH-] / [CN-] = Kw / K(HCN)

            ====> x^2 / (0.06 ) = 1x10^-14 / 6x10^-10

            ====> x^2 / 0.06  = 1.667x10^-5

            ====> x^2  = 1x10^-6

Then x = 0.001 M

Percent dissociation = (0.001 / (0.06)) X 100 = 1.66667 %

Therefore 0.06 M solution of KCN contains 1.66667 % hydrolysis.

To learn more ...

1. https://brainly.in/question/3600407