If you have a solution that is 0.4000 M LiNO2, 0.8000 M HNO2, and 0.0800 M KOH, what is the resulting pH?

Question

If you have a solution that is 0.4000 M LiNO₂, 0.8000 M HNO₂, and 0.0800 M KOH, what is the resulting pH? (HNO₂ Ka = 4.5 x 10^-4)

Answer 1

The combination of HNO₂ (weak acid) and KOH (strong base) will form a buffer consisting of the weak acid + the salt (KNO₂) of that weak acid. All of the compounds present will dissociate completely and so we will have the following situation:

HNO₂ + KOH + LiNO₂==> KNO₂ + LiNO₂ + H2O (K⁺ and Li⁺ are spectators)

0.800....0.0800....0.400.......0.............0.400............Initial

-0.08....-0.08.......................+0.08...........................Change

0.72........0...........................0.08.........0.40............Equilibrium

Final concentrations:

[HNO₂] = 0.7200 M

[NO₂^-] = 0.08 + 0.40 = 0.4800 M

Henderson Hasselbalch equation:

pH = pKa + log [NO₂^-] /[HNO₂]

pH = 3.35 + log 0.48 / 0.72

pH = 3.35 - 0.18

pH = 3.18