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A soln has an initial molarity of 0.250 M and a final [OH] of 0.0000200M. What is Kb of the base?

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B(aq) H2O(l)<--->BH(aq) OH(aq)

Kb=[BH][OH]/[B]

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  1. [BH] = [OH-] = 0.0000200 M

    [B] = 0.250 - 0.0000200 =0.250 M

    Kb = ( 0.0000200)^2 / 0.250 =2 x 10^-9

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