Question

Calculate the ph of an aqueous solution containing .01M HCL, .01M H2SO4, and .01M HCN. For HCN, ka=6.2*10^-10

Answer 1

We know that,

HCl and H2SO4 are strong acids. Hence they will dissociate completely in aqueous solution to produce H+

[H+] from HCl = 0.01 M

[H+] from H2SO4 = 2 x 0.01 = 0.02 M

Combined = 0.01 + 0.02 = 0.03 M

HCN is a weak acid. Hence, it will dissociate partially

HCN ----> H+   + CN-

0.01 - X...0.03+X....X

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

=> 6.2 x 10^-10 = (0.03 + X) (X) / (0.01 - X)

Neglecting X with respect to 0.03 and 0.01, we get

6.2 x 10^-10 = 0.03 * X / 0.01

=> X = 2.067 x 10^-10 M

[H+] = 0.03 + X = 0.03 (approx)

pH = - log [H+] = - log (0.03) = 1.523