What is the pH of a 0.560 M solution of Ca(NO₂)₂ (Ka of HNO₂ is 4.5 × 10⁻⁴)?

Question

What is the pH of a 0.560 M solution of Ca(NO₂)₂ (Ka of HNO₂ is 4.5 × 10⁻⁴)?

Answer 1

As in your previous question, look at the hydrolysis of the salt. This is the salt formed from a weak acid (HNO₂) and a strong base Ca(OH)₂. The pH of the salt solution will be basic (>7)

NO₂^- + H₂O ==> HNO₂ + OH^-

Kb = [HNO₂][OH^-] / [NO₂^-]

Kb = 1x10^-14 / Ka = 1x10^-14 / 4.5x10^-4 = 2.22x10^-11

2.22x10^-11 = (x)(x) / 1.12 (use 1.12 because 0.56 M Ca(NO₂)₂ => 2x0.56 M NO₂^-

x² = 2.49x10^-11

x = 4.99x10^-6 = [OH^-]

pOH = 5.3

pH = 14 - 5.3

pH = 8.7