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The solubility of Mn(OH)2 is 3.04 x 10^-4 gram per 100 ml of solution .
A) write the balanced chem equation for Mn(OH)2 in aqueous solution
B) calculate the molar solubility of Mn(OH)2 at 25 degrees celcius
C) calculate the value of the solubility product constant, Ksp, for Mn(OH)2 at 25 degrees celcius

  • Chem -

    mols Mn(OH)2 = grams/molar mass = approx 3E-6 but this is an estimate. You should recalculate this and all that follows.
    Then M = mols/L = about 3E-6/0.1L = about 3E-5.

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

    Ksp = (Mn^2+)(OH^-)^2
    (Mn^2+) = about 3E-5
    (OH^-) = about 2*3E-5 = about 6E-5

    Substitute this into Ksp expression and solve for Ksp.

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