Calculate the pH at equivalence.

Question

A chemist titrates 160.0 mL of a 0.0730M nitrous acid (HNO₂) solution with 0.6587 M KOH solution at 25 °C. Calculate the pH at equivalence. The pK_a of nitrous acid is 3.35

Round your answer to 2 decimal places.

you may assume the total volume of the solution equals the initial volume plus the volume of KOH solution added.

pH =

Answer 1

The equivalence point is when mols of acid = mols of base.

HNO₂ + KOH = H2O + KNO₂ since HNO₂ is a weak acid, and KOH is a strong base, pH @ equivalence will be >7. And to calculate that pH, we need to look at the hydrolysis of KNO₂.

mols HNO₂ initially present = 160.0 ml x 1 L / 1000 ml x 0.0730 mol / L = 0.01168 mols

mols KOH needed for equivalence = 0.01168 mols

volume KOH needed = 0.01168 mols x 1 L / 0.6587 mols = 0.01773 L

mols KNO₂ formed = 0.01168 mols @ equivalence

final volume of solution = 0.1600 L + 0.01773 L = 0.1777 L

final [KNO₂] = 0.01168 mols / 0.1777 L = 0.0657 M

Hydrolysis of KNO₂ can be written as

NO₂^- + H2O ==> HNO₂ + OH^- (omitting spectator ions) Here NO₂ is acting as a base

Kb = [HNO2][OH-] / [NO2-] and to find Kb for NO2- we use KaKb = 1x10^-14 and we get Ka from pKa

Ka = 1x10^-3.35 = 4.47x10^-4

Kb = 1x10^-14 / 4.47x10^-4 = 2.24x10^-11

2.24x10^-11 = (x)(x) / 0.0657 - x (ignore x in denominator since Kb is so small x will be insignificant)

2.24x10^-11 = x² / 0.0657

x² = 1.47x10^-12

x = [OH-] = 1.21x10^-6 M

pOH = -log [OH-] = -log 1.21x10^-6

pOH = 5.92

pH = 14 - pOH

pH = 14 - 5.92

pH = 8.08 (basic, as predicted above)