The solubility product constant for ${Mg(OH)}_{2}$ $"Mg(OH)"_2$ is $1.8 \times 10^{-11}$ $1.8 × 10^"-11"$ . What would be the solubility of ${Mg(OH)}_{2}$ $"Mg(OH)"_2$ in 0.345 M $NaOH$ $"NaOH"$ ?

Question

The solubility product constant for ${Mg(OH)}_{2}$ $"Mg(OH)"_2$ is $1.8 \times 10^{-11}$ $1.8 × 10^"-11"$ . What would be the solubility of ${Mg(OH)}_{2}$ $"Mg(OH)"_2$ in 0.345 M $NaOH$ $"NaOH"$ ?

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Answer 1 · 2016-06-03T02:05:18

The solubility would be $1.5 \times 10^{-10} l mol/L$ $1.5 × 10^"-10"color(white)(l) "mol/L"$ .

Explanation:

Set up the chemical equation for the equilibrium:

$m m m m m m {Mg(OH)}_{2} (s) ⇌ {Mg}^{2 +} (aq) + 2 {OH}^{-} (aq)$ $color(white)(mmmmmm)"Mg(OH)"_2"(s)" ⇌ "Mg"^(2+)"(aq)" + 2"OH"^"-""(aq)"$
${E:/mol\cdotL}^{-1} m m m m m m m m m m x m m m m 0.345 + 2 x$ $"E:/mol·L"^"-1" color(white)(mmmmmmmmmm)x color(white)(mmmm)0.345 + 2x$

Set up the solubility product expression:

$K_{sp} = [{Mg}^{2 +}] {[{OH}^{-}]}^{2} = x {(0.345 + 2 x)}^{2} = 1.8 \times 10^{-11}$ $K_"sp" = ["Mg"^(2+)]["OH"^"-"]^2 = x(0.345 +2x)^2 = 1.8 × 10^"-11"$

Check that $2 x l << 0.345$ $2xcolor(white)(l) "<< 0.345"$

$\frac{0.345}{1.8 \times 10^{-11}} = 1.9 \times 10^{10} > 400$ $0.345/(1.8 × 10^"-11") = 1.9 × 10^10 > 400$ . ∴ $2 x l << 0.345$ $2xcolor(white)(l) "<< 0.345"$ ,

Solve the $K_{sp}$ $K_"sp"$ expression:

$x {(0.345)}^{2} = 1.8 \times 10^{-11}$ $x(0.345)^2 = 1.8 × 10^"-11"$

$0.119 x = 1.8 \times 10^{-11}$ $0.119x = 1.8 × 10^"-11"$

$x = \frac{1.8 \times 10^{-11}}{0.119} = 1.5 \times 10^{-10}$ $x = (1.8 × 10^"-11")/0.119 = 1.5 × 10^"-10"$

∴ Solubility of ${Mg(OH)}_{2} = 1.5 \times 10^{-10} l mol/L$ $"Mg(OH)"_2 = 1.5 × 10^"-10" color(white)(l) "mol/L"$

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The solubility product constant for ${Mg(OH)}_{2}$ $"Mg(OH)"_2$ is $1.8 \times 10^{-11}$ $1.8 × 10^"-11"$ . What would be the solubility of ${Mg(OH)}_{2}$ $"Mg(OH)"_2$ in 0.345 M $NaOH$ $"NaOH"$ ?