pH = 10.3 = -log(H^+)
(H^+) = 5.01E-11 and (OH^-) = 2E-4
.........B + HOH ==> BH^+ + OH^-
I........x............0.......0
C........-y...........y.......y
E........x-y..........y........y
y = 2E-4
Ka = 9.00E-7 = (BH^+)(OH^-)/(B)
(BH^+) = (OH^-) = 2E-4
(B) = x-2E-4. Solve for x-y
A certain weak base has a Kb of 9.00x10^-7. What concentration of this base will produce a pH of 10.30?
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