What is the pH of a 0.770 M solution of Ca(NO₂)₂ (Ka of HNO₂ is 4.5 × 10⁻⁴)?

Answer 1

Just to add the the answer provided by @Jacques D.

Ca(NO₂)₂ is the salt of a strong base, Ca(OH)₂ and a weak acid HNO₂. So, the final pH of the salt must be basic (>7).

We look at the hydrolysis of the salt:

Ca(NO₂)₂ + 2H₂O ==> Ca(OH)₂ + 2HNO₂

Looking at just the net ionic we have..

NO₂^- + H₂O ==> HNO₂ + OH^-

Here NO₂^- acts as a base, so we find Kb for NO₂^-

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

Kb = 2.22x10^-11

Kb = [HNO₂][OH^-] / [NO₂^-]

2.22x10^-11 = (x)(x) / 1.54 (note: 1.54 is 2 x 0.77 since that will be the conc. of the NO₂^- ion)

x² = 3.42x10^-11

x = 5.85x10^-6 = [OH^-]

pOH = - log [OH^-]

pOH = 5.23

pH = 14 - [OH

pH = 8.77 (basic as predicted)

Answer 2

This is a weird question because the Ca ion affects the pH as well. Assuming that the NO₂^- is what determines the pH, we want to look at the equilibrium of

H⁺ + NO₂^- → HNO₂ which has a K = 1/K_A of HNO₂

If we let x be the H⁺ reacted to equilibrium we get that

1/K_A = [HNO₂]/([NO₂^-]{H⁺]) or x/((1.54)(10^-7-x))

The 1.54 came from the assumption that the calcium nitrite is completely soluble. 2*(.77 M) = 1.54 M NO₂^-

solve for x (I get 9.9971 x 10^-8 M H⁺

and the pH = -logx