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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asked by Tay
  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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