Chemistry

Calculate the molar solubility of manganese II hydroxide in ph 12.85 solution.

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  1. I would convert pH to pOH first, then convert that to OH^-
    pH + pOH = pKw = 14
    You know pH and pKw, solve for pOH, then
    pOH = -log(OH^-)
    Solve for OH^- and substitute in equations below. I obtained about 0.07M

    ...........Mn(OH)2 ==> Mn^2+ + 2OH^-
    I..........solid........0.......0.07
    C..........solid........x.......2x
    E..........solid........x.......2x+0.07

    Ksp = (Mn^2+)(OH^-)^2
    Look up Ksp = (x)(2x+0.07)^2
    Solve for x.
    Be sure and run the numbers yourself, the above are estimates.
    x = solubility Mn = solubility Mn(OH)2.

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  2. The Ksp is 2.1 times 10 ^-13
    But I don't know how to solve for x the answer is 4.2 times 10^-11

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