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chemistry

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A 4.50e-2 M solution of unknown monoprotic weak base has a pH of 10.200 what is the value of kb for the base

  • chemistry - ,

    BOH ==> B^+ + OH^-

    Kb = (B^+)(OH^-)/(BOH)
    pH = 10.2; therefore, pOH = 3.8
    pOH = -log(OH^-). Solve for OH^- and substitute into Kb expression. For (BOH) substitute 0.045-(OH^-).
    Post your work if you get stuck.

  • chemistry - ,

    For pOH=-log(OH-) I solved to get OH=1.58e-4
    then I did kb=(4.50e-2)(1.58e-4)/.045-1.58e-4 to get 1.59e-4 but that's not the right answer

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