Chemistry HW help

Question

Calculate the pH when 33.0 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

KOH + HBrO ==> KBrO + H2O .. balanced equation

moles KOH used = 33.0 ml x 1 L / 1000 ml x 0.150 mol / L = 4.95x10^-3 mols KOH

moles HBrO used = 20.0 ml x 1 L / 1000 ml x 0.300 mol / L = 6x10^-3 mols HBrO

Using an ICE table, we have...

KOH + HBrO ==> KBrO + H2O

0.00495 ..0.0060..........0..................Initial

-0.00495...-0.00495....+0.00495.......Change

0...............0.00105.......0.00495.......Equilibrium

So, @ equilibrium, we have a mixture of 0.00105 mols HBrO and 0.00495 mols KBrO in a final volume of 33 ml + 20 ml = 50 ml = 0.050 L. This creates a buffer composed of a weak acid (HBrO) and the conjugate base BrO^-.

Using the Henderson Hasselbalch equation, we can find the pH

pH = pKa + log [conj.base] / [acid]

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

pH = 8.60 + log [0.00495] / [0.00105] = 8.60 + log 4.71

pH = 8.60 + 0.67

pH = 9.27