Calculate the pH when 29.0 mL of 0.150 M KOH is mixed with 20.0 mL of 0.300 M HBrO (Ka = 2.5 × 10⁻⁹)

Question

I ended up with a pH of 12.5 but it still was not correct...

Answer 1

Not sure how you arrived at pH = 12.5 since you didn't show your work, but here is my solution.

When adding a base to a weak acid, you will form the salt of that weak acid, and thus form a buffer.

HBrO + KOH ==> KBrO + H2O

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

moles KOH = 29.0 ml x 1 L / 1000 ml x 0.150 mol/L = 0.00435 mols KOH

Setting up an ICE table, we have...

HBrO + KOH ==> KBrO + H2O

0.006.........0.00435.....0.................Initial

-0.0043.....-0.00435....+0.00435....Change

0.00165........0............0.00435.......Equilibrium

Using the Henderson Hasselbalch equation

pH = pKa + log [salt] / [acid]

pKa = -log Ka = -log 2.5x10-9 = 8.60

pH = 8.60 + log [0.00435] / [0.00165]

pH = 8.60 + log 2.636

pH = 8.60 + 0.42

pH = 9.02