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The molar solubility of Mg(OH)2 in pure water is found to be 1.0 x…
Question
The molar solubility of Mg(OH)2 in pure water is found to be 1.0 x 10^-4 mol L^-1 at 298 K. Calculate the solubility product Ksp of Mg(OH)2. Give the answer in scientific form.
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Step-by-step solution
Mg(OH)2 -> Mg2+ + 2OH-. If solubility = s, then [Mg2+] = s and [OH-] = 2s. Ksp = [Mg2+][OH-]^2 = s(2s)^2 = 4s^3. With s = 1.0e-4: Ksp = 4 * (1.0e-4)^3 = 4 * 1.0e-12 = 4.0e-12. Final answer: 4e-12
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