The solubility product constant for #"Mg(OH)"_2# is #1.8 × 10^"-11"#. What would be the solubility of #"Mg(OH)"_2# in 0.345 M #"NaOH"#?

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Answer 1 · 2016-06-03T02:05:18

The solubility would be #1.5 × 10^"-10"color(white)(l) "mol/L"#.

Set up the chemical equation for the equilibrium:

#color(white)(mmmmmm)"Mg(OH)"_2"(s)" ⇌ "Mg"^(2+)"(aq)" + 2"OH"^"-""(aq)"#
#"E:/mol·L"^"-1" color(white)(mmmmmmmmmm)x color(white)(mmmm)0.345 + 2x#

Set up the solubility product expression:

#K_"sp" = ["Mg"^(2+)]["OH"^"-"]^2 = x(0.345 +2x)^2 = 1.8 × 10^"-11"#

Check that #2xcolor(white)(l) "<< 0.345"#

#0.345/(1.8 × 10^"-11") = 1.9 × 10^10 > 400#. ∴ #2xcolor(white)(l) "<< 0.345"#,

Solve the #K_"sp"# expression:

#x(0.345)^2 = 1.8 × 10^"-11"#

#0.119x = 1.8 × 10^"-11"#

#x = (1.8 × 10^"-11")/0.119 = 1.5 × 10^"-10"#

∴ Solubility of #"Mg(OH)"_2 = 1.5 × 10^"-10" color(white)(l) "mol/L"#