Question

Determine the pH of 0.35M HCN. The Ka of HCN is 6.2e-10 M.

Answer 1

HCN ---> H+ + CN-

using ICE table

initial conc of HCN , H+ , CN- are 0.35 , 0 , 0

change in conc of HCN , H+ , CN- are -y, +y , +y

equilibrium conc of HCN , H+ , CN- are 0.35-y , +y, +y

now

Ka = [H+] [CN-] / [HCN]

given

Ka = 6.2 x 10-10

so

6.2 x 10-10 = [y] [y] / [0.35-y]

6.2 x 10-10 = [y]^2 / [0.35-y]

as Ka value is very low , y will have a low value

so

0.35 - y ----> 0.35

so

6.2 x 10-10 = [y]^2 / 0.35

[y]^2 = 6.2 x 10-10 x 0.35

y = 1.4731 x 10-5

now

[H+] = y = 1.4731 x 10-5

now

we know that

pH= -log [H+]

pH = -log 1.4731 x 10-5

pH = 4.8318

so

pH of 0.35M HCN solution is 4.8318