Question

Calculate the pH of a 0.0152 M aqueous solution of nitrous acid (HNO₂, K_a = 4.6×10^-4) and the equilibrium concentrations of the weak acid and its conjugate base.

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

Answer 1

HNO2 dissociates as:

HNO2 -----> H+ + NO2-

1.52*10^-2 0 0 (INITIAL)

1.52*10^-2-x x x (AT EQUILIBRIUM)

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

Ka = x*x/(c-x)

4.6*10^-4 = x*x / (1.52*10^-2 - x)

6.99*10^-6 - 4.6*10^-4 * x = x^2

x^2 + 4.6*10^-4 * x - 6.99*10^-6 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1.0

b = 4.6*10^-4

c = -6.99*10^-6

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 2.82*10^-5

roots are :

x = 2.42*10^-3 M and x = -2.88*10^-3 M

since x can't be negative, the possible value of x is

x = 2.42*10^-3 M

SO,

[H+] = 2.42*10^-3 M

use:

pH = -log [H+]

= -log ( 2.42*10^-3 )

= 2.62

So,

pH is 2.62

[HNO2] = 1.52*10^-2-x = 1.52*10^-2 - 2.42*10^-3 = 1.28*10^-2 M

[NO2-] = x = 2.42*10^-3 M