Question

Calculate the solubility (in grams per liter) of magnesium hydroxide in the following. (a) pure water (b) \(0.041 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) (c) \(0.0050 \mathrm{M} \mathrm{MgCl}_{2}\)

Short answer

Based on the calculations, the solubility of magnesium hydroxide (Mg(OH)\(_2\)) is approximately: (a) 7.00 g/L in pure water (b) 0.20 g/L in 0.041M Ba(OH)\(_2\) solution (c) 0.98 g/L in 0.0050M MgCl\(_2\) solution

Step-by-step solution

Write the Solubility Equilibrium and K\(_{sp}\) Expression for Mg(OH)\(_2\)

The solubility equilibrium for magnesium hydroxide is as follows: Mg(OH)\(_2\)(s) \(\leftrightharpoons\) Mg\(^{2+}\)(aq) + 2OH\(^-\) (aq) The solubility product constant (K\(_{sp}\)) expression corresponding to this reaction is: K\(_{sp}\) = [Mg\(^{2+}\)] [OH\(^-\)]\(^2\)

Find the K\(_{sp}\) Value for Mg(OH)\(_2\)

The K\(_{sp}\) value for magnesium hydroxide can be found in reference tables. For Mg(OH)\(_2\), the K\(_{sp}\) value is \(5.61 \times10^{-12}\).

Calculate Solubility in Pure Water

In pure water, the equilibrium concentrations for Mg\(^{2+}\) and OH\(^-\) are equal, let's say, x and 2x. We can set up the equation: \(5.61\times10^{-12}\) = (x)(2x)\(^2\) Solve for x: x = \(1.20 \times 10^{-4} \mathrm{M}\) (Mg\(^{2+}\)) Now convert the concentration of Mg\(^{2+}\) to grams per liter of Mg(OH)\(_2\): Solubility (Mg(OH)\(_2\) in pure water) = (\((1.20 \times 10^{-4}) mol/L) \times (58.32 \thinspace g/mol) \approx 7.00 \thinspace g/L\)

Calculate Solubility in 0.041M Ba(OH)\(_2\) Solution

In the presence of 0.041M Ba(OH)\(_2\), the initial concentration of OH\(^-\) ions is 0.082M due to the dissociation of Ba(OH)\(_2\). Now, let's say the increase in Mg\(^{2+}\) concentration is x and the increase in OH\(^-\) will be 2x: \(5.61\times10^{-12}\) = (x)(0.082 + 2x)\(^2\) Solve for x: x = 3.40 x \(10^{-6} \mathrm{M}\) (Mg\(^{2+}\)) Now convert the concentration of Mg\(^{2+}\) to grams per liter of Mg(OH)\(_2\): Solubility (Mg(OH)\(_2\) in Ba(OH)\(_2\) solution) = (\((3.40 \times 10^{-6}) mol/L) \times (58.32 \thinspace g/mol) \approx 0.20 \thinspace g/L\)

Calculate Solubility in 0.0050M MgCl\(_2\) Solution

In the presence of 0.0050M MgCl\(_2\), the initial concentration of Mg\(^{2+}\) ions is 0.0050M. Now, let's say the increase in OH\(^-\) concentration is 2x: \(5.61\times10^{-12}\) = (0.0050+x)(2x)\(^2\) Solve for x: x ≈ 1.68 × 10\(^{-5}\)M (OH\(^-\)) Now convert the concentration of OH\(^-\) to grams per liter of Mg(OH)\(_2\): Solubility (Mg(OH)\(_2\) in MgCl\(_2\) solution) = (\((1.68 \times 10^{-5}) mol/L) \times (58.32 \thinspace g/mol) \approx 0.98 \thinspace g/L\) So, the solubility of Mg(OH)\(_2\) is approximately: (a) 7.00 g/L in pure water (b) 0.20 g/L in 0.041M Ba(OH)\(_2\) solution (c) 0.98 g/L in 0.0050M MgCl\(_2\) solution