Calculate the solubility (in grams per 1.00*10^2mL of solution) of magnesium hyd

Question

Calculate the solubility (in grams per 1.00*10^2mL of solution) of magnesium hydroxide in a solution buffered at pH = 12.
S1= 1.2*10^-8 g/(1.00*10^2mL)

How does this compare to the solubility of MgOH2 in pure water?
S1/S= ?

Explanation / Answer

S1= 1.2*10^-8 g/(1.00*10^2mL)=2.059×10^-9M 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-] = 2.059×10^-9 / 10^(12-14) = 2.059×10-7M 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) = 2.059×10-7mol/L · 58.34 g/mol = 1.2×10-5g/L =1.2×10-5g/L ×1L/1000ml ×100ml =1.2×10^-6g/(1.00*10^2mL)

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