What is the pH?

Question

Calculate the pH when 40.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)

Answer 1

Write the reaction:

KOH + HBrO ==> KBrO + H₂O

moles KOH present = 40.0 ml x 1 L / 1000 ml x 0.150 mol/L = 0.006 moles KOH

moles HBrO present = 20.0 ml x 1 L / 1000 ml x 0.300 mol/L = 0.006 moles HBrO

Since the moles are the same, this tells us that ALL of the KOH was converted to KBrO.

[BrO-] = 0.006 mols / 0.06 L = 0.1 M (0.06 L comes from adding 40 ml + 20 mls to get final volume)

To find the pH, we need to look at the hydrolysis of the salt.

KBrO + H2O ==> KOH + HBrO

BrO- + H2O ==> HBrO + OH- (without spectator ions)

Kb for BrO^- = 1x10^-14 / 2.5x10^-9 = 4x10^-6

Kb = 4x10^-6 = [HBrO][OH-] / [BrO-]

4x10^-6 = (x)(x) / 0.1

x² = 4x10^-7

x = 6.34x10^-4 M = [OH-]

pOH = -log 6.34x10^-4 = 3.2

pH = 14 - pOH

pH = 10.8