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Calculate the solubility of magnesium hydroxide (Mg(OH)2), ksp=1.1x10^-11?

i have done this problem so many times and keep getting it wrong, i know the answer is 1.4x10^-4 but how do i get there? i did an ice chart and tried doing the cube root but its wrong. help!

  • chem -

    Mg(OH)_2 ---------> Mg^2+ + 2OH^-

    I. X 0 0

    C. -X X 2X


    K_sp <=========> [Mg^2+] [OH^-]^2


    1.1*10^-11 = X (2X)^2

    X = 3_/ 2.75*10^-12

    X= 1.4*10^-4

    hope i helped you :)

    Cheers!

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