# Chemistry

A solution of an unknown base has a concentration of 0.250M, and a pH of 12.72. Calculate Kb for this base.

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1. 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

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posted by Devron

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