Question

Calculate the pH of a 0.383 M aqueous solution of nitrous acid (HNO₂, K_a = 4.5×10^-4) and the equilibrium concentrations of the weak acid and its conjugate base.

pH	=
[HNO2 ]equilibrium	=	M
[NO2- ]equilibrium	=	M

Answer 1

           HNO₂ + H₂O H₃O⁺ + NO₂^-

Initial concentration (M)                                       0.383                         0             0

Change in concentration (M)                                 -X                           X             X

Equilibrium concentration (M)                         (0.383 - X)                      X             X

Given, K_a of HNO₂ = 4.5 x 10^-4

Now,

K_a = [H₃O⁺][NO₂^-]/[HNO₂]

or, 4.5 x 10^-4 = (X)(X)/(0.383 - X)

or, 4.5 x 10^-4 = (X)(X)/0.383      [(0.383 - X) 0.383 as X << 0.383]

or, X² = 0.00017

or, X = 0.013

or, [H₃O⁺] = 0.013

or, -log[H₃O⁺] = -log(0.013)     (taking - log function on both sides of the equation)

or, pH = 1.9

Now, [NO₂^- ]_equilibrium = X = 0.013 M

[HNO₂ ]_equilibrium = (0.383 - X) = (0.383 - 0.013) M = 0.370 M