Answer to Question #168603 in General Chemistry for Mesfin

Question

Hydrogen cyanide, HCN, is a weak acid with a dissociation constant of 4.8 × 10–10.

Calculate the percent dissociation, the pH, and the OH– concentration of a 0.15 M

solution of HCN in water.

Answer 1

HCN $\leftrightarrow$ H⁺ + CN^-

0.15____--___--

0.15-x____x___x

$\frac{x^2}{0.15-x}$ = 4.8×10^-10

let suppose 0.15-x $\sim$ 0.15

$\frac{x^2}{0.15}$ = 4.8×10^-10

x² = 4.8×10^-10 ×0.15

x² = 0.72×10^-10 = 72×10^-12

x = 8.48 × 10^-6 M

pH = -log_oX

pH = 5.09

%dissociation =( 8.48×10^-6/0.15 )×100

= 56.5×10^-4 %