This is a buffer problem. ICE table should be used
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
0.50 mol in 1.00 L = 0.50 M HOBr 0.30 mol in 1.00 L = 0.30 M KOBr (or OBr-) HOBr + H2O <-> H3O+ + OBr- 0.50 0 0.30 -x +x +x (0.50-x) x (0.30+x) Assume x is small comparing to 0.30 and 0.50 2.0 x 10^-9 = x (0.30) / 0.50) [H3O+] = x = 3.33 x 10^-9
So : pH = 8.48
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