Calculate the pH of a 0.173 M solution of NaNO2. The ionization constant, 𝐾a, for the acid, HNO2, is 4.60×10−4. pH=

Answer 1

NaNO₂ is the salt of a strong base (NaOH) and a weak acid HNO₂. Therefore, the pH of the solution should be > 7 (alkaline).

We look at the hydrolysis of the NO₂^- anion as follows:

NO₂- + H2O ==> HNO₂ + OH^-

pKb for NO₂^- = Kw/Ka = 1x10^-14 / 4.60x10^-4 = 2.17x10^-11

Write the expression for Kb

Kb = [HNO₂][OH-] / [NO₂^-]

2.17x10^-11 = (x)(x) / 0.173 - x and assuming x is small relative to 0.173, ignore it in the denominator

2.17x10^-11 = x² / 0.173

x² = 3.75x10^-12

x = [OH-] = 1.94x10^-6

pOH = -log [OH-]

pOH = 5.71

pH = 14 - pOH

pH = 14 - 5.71

pH = 8.29 (alkaline, as predicted)