Question

Calculate the pH when 10.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 x 10-9)

Answer 1

Given:

M(HBrO) = 0.3 M

V(HBrO) = 20 mL

M(KOH) = 0.15 M

V(KOH) = 10 mL

mol(HBrO) = M(HBrO) * V(HBrO)

mol(HBrO) = 0.3 M * 20 mL = 6 mmol

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.15 M * 10 mL = 1.5 mmol

We have:

mol(HBrO) = 6 mmol

mol(KOH) = 1.5 mmol

1.5 mmol of both will react

excess HBrO remaining = 4.5 mmol

Volume of Solution = 20 + 10 = 30 mL

[HBrO] = 4.5 mmol/30 mL = 0.15M

[BrO-] = 1.5/30 = 0.05M

They form acidic buffer

acid is HBrO

conjugate base is BrO-

Ka = 2.5*10^-9

pKa = - log (Ka)

= - log(2.5*10^-9)

= 8.602

use:

pH = pKa + log {[conjugate base]/[acid]}

= 8.602+ log {5*10^-2/0.15}

= 8.125

Answer: 8.12