When 20.0 mL of 0.10 M NaOH are added to 20.0 mL of 0.10 M hypobromous acid, HOBr, what is the hydrogen ion concentration (H^+)? Ka for HOBr is 2 x 10^-9.

Question

When 20.0 mL of 0.10 M NaOH are added to 20.0 mL of 0.10 M hypobromous acid, HOBr, what is the hydrogen ion concentration (H^+)?   Ka for HOBr is 2 x 10^-9.

I get 5 x 10^-4. Is that right?

DrBob222 · Answer

I don't think so.
HOBr + NaOH ==> NaOBr + H2O
20 x 0.1 HOBr = 2 mmoles HOBr.
20 x 0.1 NaOH = 2 mmoles NaOH

Those two exactly neutralize each other; therefore, the pH is determined by the salt, NaOBr.
(NaOBr) = 2 mmoles/40 mL = 0.05 M.
It hydrolyzes as a base,
BrO^- + HOH ==> HOBr + OH^-

Set up ICE chart.
Kb = (Kw/Ka) = (x)(x)/0.05
solve for x = (OH^-) and convert to H^+.