What is the expected solubility, in grams per 100 mL, of magnesium hydroxide, Mg

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Explanation / Answer

1) First write the equation down: Mg(OH)2(s) ---> Mg2+(aq) + 2 OH-(aq) The concentrations at equilibrium will be: Mg2+(aq) = x OH-(aq) = (1x10^-4) + 2x Plug these values into the equilibrium expression: Ksp = [Mg2+][OH-]^2 (2.06x10^-13) = (x)(0.0001 + 2x)^2 Since the value of the Ksp is so small, we can assume that 0.0001 >> 2x, thus the equation reduces to: (2.06x10^-13) = (x)(0.0001)^2 x = 2.06x10^-5 2) Write the equilibrium expression just the same way as above: Ksp = [Mg2+][OH-]^2 (2.06x10^-13) = (x)(2x)^2 (2.06x10^-13) = 4x^3 x = [(2.06x10^-13)/4]^1/3 x = 3.72x10^-5 Thus, you can see that the solubility is greater in pure water than in a basic aqueous solution.

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