Question

What is the pH of a buffer solution containing 0.025M of HNO2 and 0.02 M of NaNO2? (Ka = 7.2 x 10–4) please put ICE table

Answer 1

HNO2 ---> NO2^- + H3O

Initial 0.025 ---> 0.02 + 0

Change -x ---> +x +x

Equilibrium 0.025-x ---> 0.02+x +x

$K_a = \dfrac{[H_3O][NO_2^-]}{[HNO_2]}$

$[H_3O] = \dfrac{[HNO_2]×K_a}{[NO_2^-]}$

H3O = 0.025×7.2×10^-4/0.02

H3O = 0.0009M = 9×10^-4

pH = -log[H+]

pH = -log[9×10^-4]

pH = 3.05