Question

Determine the concentration of a NaNO₂ solution that has a pH of 8.22. K_a for HNO₂ is 4.5 x 10^-4

Answer 1

Ka = 4.5 x 10^-4

Kb = (1.0 x 10^-14) / (4.5 x 10^-4) = 2.2 x 10^-11

pH = 8.22

So, pOH = 14 - pH = 14 - 8.22 = 5.78

- log [OH^-] = 5.78

[OH^-] = 10^-5.78

[OH^-] = 1.66 x 10^-6

So,

Now,

NO₂^- will react as a base:

NO₂^- + H₂O HNO₂ + OH^-
  

Kb = [HNO₂] [OH^-] / [NO₂^-]

2.2 x 10^-11 = (1.66 x 10^-6) (1.66 x 10^-6) / [NO₂^-]

2.2 x 10^-11 = (2.76 x 10^-12) / [NO₂^-]

  [NO₂^-] = (2.76 x 10^-12) / (2.2 x 10^-11)

  [NO₂^-] = 1.25 x 10^-1

So,   [NO₂^-] = 0.125 M