Question

Calculate the pH of a 0.375 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 H⁺ + NO₂^-

Initial concentration: 0.375 M 0 0

equilibrium concentration: (0.375 -x)M x M xM (Let us assume x M is the equilibrium concentration of the conjugate base)

Ka = [H+]. [NO2-] / [ HNO2]

= x² / (0.375 -x)

4.5 x 10 ^-4 =  x² / (0.375 -x)

=  x² / 0.375 (as x is very small)

x² =    4.5 x 10 ^-4 x 0.375 = 1.6875 x 10 ^-4

   x = 1.299 x 10 ^-2

= 0.01299 M = 0.013 M

pH = - log [ H+] = - log [ 0.013] = 1.886 (answer)

Concentration of weak acid (HNO2) = (0.375 - x) = (0.375 - 0.013) M = 0.362 M (answer)

Concentration of conjugate base of weak acid (NO₂^-) = x = 0.013 M (answer)