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5 September, 11:24

Calculate the molar solubility of fe (oh) 2 in pure water. (the value of ksp for fe (oh) 2 is 4.87*10-17.)

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  1. 5 September, 12:51
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    When we have this balanced equation for a reaction:

    Fe (OH) 2 (s) ↔ Fe+2 + 2OH-

    when Fe (OH) 2 give 1 mole of Fe+2 & 2 mol of OH-

    so we can assume [Fe+2] = X and [OH-] = 2 X

    when Ksp = [Fe+2][OH-]^2

    and have Ksp = 4.87x10^-17

    [Fe+2] = X

    [OH-] = 2X

    so by substitution

    4.87x10^-17 = X * (2X) ^2

    ∴X^3 = 4.8x10^-17 / 4

    ∴the molar solubility X = 2.3x10^-6 M
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