A solution of an unknown base has a concentration of 0.250M, and a pH of 12.72. Calculate Kb for this base.
The reaction is as followed:
B + H2O ------------> BH + OH^-
Kb=[BH][OH^-]/[B]
At equilibrium,
BH=OH^-
B=0.250M-OH^-
pH+pOH=14
14-pH=pOH
14-12.72=1.28
pOH=-log[OH^-]
OH^-=10^(-pOH)
OH^-=10^(-1.28)
OH^-=5.25 x 10^-2 M
BH=5.25 x 10 ^-2 M
B at Eq=B-OH^-=0.25M-5.25 x 10^-2=0.198M
Kb=[BH][OH^-]/[B]
Kb=[5.25 x 10^-2 M][5.25 x 10^-2 M]/[0.198M]
Kb=1.40 x 10^-2