Calculate the molar concentration of OH- ions in a 0.724 M solution of hypobromite ion (BrO-: Kp = 4.0x10-6). What is the pH of this solution?

Answer 1

It isn't Kp = 4.0x10^-6, but rather is should be Kb = 4.0x10^-6

When you put BrO^- in H2O, you have the following hydrolysis reaction:

BrO^- + H2O ==> HBrO + OH^- where BrO^- is acting as a base and H2O as an acid

Kb = [HBrO][OH^-] / [BrO^-]

4.0x10^-6 = (x)(x) / 0.724 - x (assume x is small relative to 0.724 and ignore it in denomiator)

4.0x10^-6 = x² / 0.724

x² = 2.896x10^-6

x = 1.70x10^-3 (this is only ~0.2% of 0.724 so above assumption was valid)

[OH^-] = 1.70x10^-3

pH + pOH = 14

pOH = -log [OH^-] = -log 1.70x10-3 = 2.77

pH = 14 - 2.77

pH = 11.23