Answer :
To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
HOBr + H2O <-> H+ + OBr-
I 0.50 - 0 0.30
C -x x x
---------------------------------------------
E(0.50-x) x (0.30+x)
Assuming that the value of x is small as compared to 0.30 and 0.50
Ka = 2.0 x 10^-9 = x (0.30) / 0.50)
x = 3.33 x 10^-9 = H+
pH = 8.48
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
HOBr + H2O <-> H+ + OBr-
I 0.50 - 0 0.30
C -x x x
---------------------------------------------
E(0.50-x) x (0.30+x)
Assuming that the value of x is small as compared to 0.30 and 0.50
Ka = 2.0 x 10^-9 = x (0.30) / 0.50)
x = 3.33 x 10^-9 = H+
pH = 8.48