Chemistry

Calculate the molar solubility of Cr(OH)3 in a solution with a PH of 11.6 knowing that the Ksp(Cr(OH)3)=6.7×10^-31.

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  1. pH+pOH=14

    14-pH=pOH

    pOH=14-11.6=2.4

    pOH=-log[OH^-]

    [OH^-]=10^-[2.4]

    [OH^-]=3.98 x 10^-3 M= initial concentration

    Cr(OH)3 -------> Cr^3+ + 3OH^-

    Ksp=[x][3.98 x 10^-3 M]^3

    Ksp/[3.98 x 10^-3 M]^3=x

    x=6.7×10^-31/[3.98 x 10^-3 M]^3

    Solve for x

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