Anonymous
Anonymous asked in Science & MathematicsChemistry · 5 months ago

The pH of a 0.15 mol/L solution of LiF(aq) is?

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  • 5 months ago

    F- acts as a base by the reaction:

    F- + H2O <--> HF + OH-

    Kb = [HF][OH-]/[F-]

    Ka for HF = 6.6X10^-4, so Kb for F- = 1X10^-14 / 6.6X10^-4 = 1.52X10^-11

    In the solution, leg [HF] = [OH-] = x and [F-] = 0.15 mol/L. Then,

    1.52X10^-11 = x^2 / 0.15

    x = [OH-] = 1.5X10^-6 M

    pOH = 5.82

    pH = 14.00 - pOH = 8.18

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