Question

What mass of Mg(OH)2 will dissolve in 1.0 L of 0.050 mol/L solution of MgSO4?
Ksp = 5.6 x 10^-12

Answer 1

Question #57946, Chemistry, Other

What mass of $\mathrm{Mg(OH)}_2$ will dissolve in 1.0 L of 0.050 mol/L solution of $\mathrm{MgSO}_4$?

$$K_{\mathrm{sp}} = 5.6 \times 10^{-12}$$

Answer:

$$K_{\mathrm{sp}} = [\mathrm{Mg}^{2+}] \cdot [\mathrm{OH}^-]$$

If we will plug these values in the $K_{\mathrm{sp}}$ expression: $5.6 \times 10^{-12} = (x + 0.05)^* (2x)^2$

If we suppose that $x \leq 0.050$, we have $0.20x^2 = 5.6 \times 10^{-12}$;

$x = 5.29 \times 10^{-6}$ moles/liter

Molar mass of $\mathrm{Mg(OH)}_2$ is 58.32 g/mol

So we have $58.32\mathrm{g/mol}^* 5.29\times 10^{-6}\mathrm{mol/liter} = 3.09\times 10^{-4}\mathrm{g/liter}$

$3.09 \times 10^{-4} \mathrm{~g}$ of $\mathrm{Mg(OH)}_2$ dissolve per liter of $0.05 \mathrm{M} \mathrm{Mg}^{2+}$ solution at $25^{\circ} \mathrm{C}$.

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