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Chapter 16: Problem 29

Calculate the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{-}, K_{\mathrm{sp}}=8.9 \times 10^{-12}\).

Short Answer

Expert verified
The molar solubility of \(\mathrm{Mg(OH)_2}\) is \(1.34 \times 10^{-4}\,\text{mol/L}\).

Step by step solution

01

Write the dissolution equation

In this step, we will write the balanced chemical equation for the dissolution of \(\mathrm{Mg(OH)_2}\) in water: \[\mathrm{Mg(OH)_2 (s) \rightleftharpoons Mg^{2+} (aq) + 2OH^{-} (aq)}\] #step 2: Write \(K_{\mathrm{sp}}\) expression#
02

Write the expression for \(K_{\mathrm{sp}}\)

Based on the dissolution equation, we can write the \(K_{\mathrm{sp}}\) expression as: \[K_{\mathrm{sp}} = [\mathrm{Mg^{2+}}][\mathrm{OH^{-}}]^2\] #step 3: Define the molar solubility and equilibrium concentrations#
03

Define the molar solubility and equilibrium concentrations

Let's assume that the molar solubility of \(\mathrm{Mg(OH)_2}\) is \(s\) moles per litre. Then, at equilibrium, the concentrations of the ions will be: \([\mathrm{Mg^{2+}}] = s\) and \([\mathrm{OH^{-}}] = 2s\) #step 4: Substitute the equilibrium concentrations into the \(K_{\mathrm{sp}}\) expression#
04

Substitute the equilibrium concentrations into the \(K_{\mathrm{sp}}\) expression

Using the equilibrium concentrations, \([\mathrm{Mg^{2+}}] = s\) and \([\mathrm{OH^{-}}] = 2s\), we can rewrite the \(K_{\mathrm{sp}}\) expression as: \[K_{\mathrm{sp}} = s(2s)^2\] #step 5: Solve for the molar solubility, s#
05

Solve for the molar solubility (s)

To find the molar solubility (s), we first plug in the given \(K_{\mathrm{sp}}\) value, \(8.9 \times 10^{-12}\), and solve for \(s\): \[8.9 \times 10^{-12} = s(2s)^2\] This simplifies to: \[8.9 \times 10^{-12} = 4s^3\] Now solving for \(s\): \[s = \sqrt[3]{\frac{8.9 \times 10^{-12}}{4}} = 1.34 \times 10^{-4}\,\text{mol/L}\] So, the molar solubility of \(\mathrm{Mg(OH)_2}\) is \(1.34 \times 10^{-4}\,\text{mol/L}\).

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Most popular questions from this chapter

For which salt in each of the following groups will the solubility depend on pH? a. \(\mathrm{AgF}, \mathrm{AgCl}, \mathrm{AgBr}\) c. \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Sr}\left(\mathrm{NO}_{2}\right)_{2}\) b. \(\mathrm{Pb}(\mathrm{OH})_{2}, \mathrm{PbCl}_{2}\) d. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}, \mathrm{Ni}(\mathrm{CN})_{2}\)

Aluminum ions react with the hydroxide ion to form the precipitate \(\mathrm{Al}(\mathrm{OH})_{3}(s)\), but can also react to form the soluble complex ion \(\mathrm{Al}(\mathrm{OH})_{4}^{-} .\) In terms of solubility, \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) will be more soluble in very acidic solutions as well as more soluble in very basic solutions. a. Write equations for the reactions that occur to increase the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in very acidic solutions and in very basic solutions. b. Let's study the \(\mathrm{pH}\) dependence of the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}(s)\) in more detail. Show that the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\), as a function of \(\left[\mathrm{H}^{+}\right]\), obeys the equation $$S=\left[\mathrm{H}^{+}\right]^{3} K_{\mathrm{sp}} / K_{\mathrm{w}}^{3}+K K_{\mathrm{w}} /\left[\mathrm{H}^{+}\right]$$ where \(S=\) solubility \(=\left[\mathrm{Al}^{3+}\right]+\left[\mathrm{Al}(\mathrm{OH})_{4}^{-}\right]\) and \(K\) is the equilibrium constant for $$\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{Al}(\mathrm{OH})_{4}^{-}(a q)$$ c. The value of \(K\) is \(40.0\) and \(K_{\mathrm{se}}\) for \(\mathrm{Al}(\mathrm{OH})_{3}\) is \(2 \times 10^{-32}\). Plot the solubility of \(\mathrm{Al}(\mathrm{OH})_{3}\) in the \(\mathrm{pH}\) range \(4-12\).

The common ion effect for ionic solids (salts) is to significantly decrease the solubility of the ionic compound in water. Explain the common ion effect.

What happens to the \(K_{\mathrm{xg}}\) value of a solid as the temperature of the solution changes? Consider both increasing and decreasing temperatures, and explain your answer.

A solution is prepared by mixing \(75.0 \mathrm{~mL}\) of \(0.020 \mathrm{M} \mathrm{BaCl}_{2}\) and \(125 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{K}_{2} \mathrm{SO}_{4}\). What are the concentrations of barium and sulfate ions in this solution? Assume only \(\mathrm{SO}_{4}^{2-}\) ions (no \(\left.\mathrm{HSO}_{4}^{-}\right)\) are present.
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