Calculate the solubility (in grams per 1.00 102 mL of solution) of magnesium hyd

Question

Calculate the solubility (in grams per 1.00 102 mL of solution) of magnesium hydroxide in a solution buffered at pH = 10. I got 0.0012 but it's wrong. Thanks in advance!

Explanation / Answer

Mg2+(aq) + OH(aq) ---> Mg(OH)2(s) => Ksp = [Mg2+] · [OH-] Hydroxide concentration is constant due to buffering: [OH-] = 10^(pH-14) Hence the maximum magnesium ion concentration is [Mg2+] = Ksp / [OH-] = 1.2×10–11 / 10^(10-14) = 1.2×10-7M The maximum soluble amount of magnesium hydroxide per liter is the same. Multiply by molar mass and you get the maximum solubility in g/L: c(Mg(OH)2) = [Mg(OH)2] · M(Mg(OH)2) = 1.2×10-7mol/L · 58.34 g/mol = 7.0×10-6g/L

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