Chapter 17: Problem 81
What is the molar solubility of \(\mathrm{Mg}(\mathrm{OH})_{2}\) in a solution containing \(1.0 \times 10^{-1} M \mathrm{NaOH}\) ?
Short Answer
Expert verified
The molar solubility of \( \mathrm{Mg(OH)}_2 \) is \( 1.8 \times 10^{-9} \) M.
Step by step solution
01
Understand the Chemical Equation
The dissociation of \( \mathrm{Mg(OH)}_2 \) in water can be represented as \( \mathrm{Mg(OH)}_2 (s) \rightleftharpoons \mathrm{Mg}^{2+} (aq) + 2 \mathrm{OH}^{-} (aq) \). Since we are dealing with a saturated solution of \( \mathrm{Mg(OH)}_2 \), we are interested in the solubility product (\( K_{sp} \)).
02
Write the Expression for Ksp
The \( K_{sp} \) expression for \( \mathrm{Mg(OH)}_2 \) is \( K_{sp} = [\mathrm{Mg}^{2+}][\mathrm{OH}^{-}]^2 \). The provided \( K_{sp} \) value for \( \mathrm{Mg(OH)}_2 \) is typically \( 1.8 \times 10^{-11} \).
03
Determine OH- Contribution from NaOH
The solution contains \( 1.0 \times 10^{-1} M \) \( \mathrm{NaOH} \), which fully dissociates to provide \( 1.0 \times 10^{-1} M \) \( \mathrm{OH}^- \) ions. This initial \( \mathrm{OH}^- \) concentration shifts the equilibrium.
04
Set Up the Equilibrium Expression
Let \( s \) be the molar solubility of \( \mathrm{Mg(OH)}_2 \). At equilibrium, \([\mathrm{Mg}^{2+}] = s\) and \([\mathrm{OH}^{-}] = 2s + 1.0 \times 10^{-1}\). Substitute these into the \( K_{sp} \) expression.
05
Solve for Molar Solubility s
Substitute into the \( K_{sp} \) expression: \( 1.8 \times 10^{-11} = s(2s + 1.0 \times 10^{-1})^2 \). Assume \( 2s \) is negligible compared to \( 1.0 \times 10^{-1} \), simplifying to \( s = \frac{1.8 \times 10^{-11}}{1.0 \times 10^{-2}} = 1.8 \times 10^{-9} \).
06
Conclusion
The assumption that \( 2s \) is negligible is valid because \( 1.8 \times 10^{-9} \) is very small compared to \( 0.1 \). Therefore, the molar solubility of \( \mathrm{Mg(OH)}_2 \) in \( 1.0 \times 10^{-1} M \) \( \mathrm{NaOH} \) is \( 1.8 \times 10^{-9} \) M.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Solubility Product Constant (Ksp)
The Solubility Product Constant, commonly referred to as Ksp, is a crucial concept in understanding the solubility of ionic compounds. It defines the extent to which a compound dissociates into its constituent ions in a saturated solution. For a given ionic solid such as \[ \text{Mg(OH)}_2 \rightarrow \text{Mg}^{2+} + 2\text{OH}^- \]the Ksp expression is \[ K_{sp} = [\text{Mg}^{2+}][\text{OH}^-]^2. \]The Ksp value is a product of the molar concentrations of the ions, each raised to the power of its coefficient in the dissociation equation. In our example, it tells us how much \( \text{Mg(OH)}_2 \)dissolves to reach equilibrium in the presence of its ions. Some key points to remember:
- A higher Ksp indicates a more soluble compound.
- It is specific for each compound and depends on the conditions like temperature.
- Ksp values are typically determined experimentally.
Chemical Equilibrium
Chemical Equilibrium is the state of a reversible reaction where the rate of the forward reaction equals the rate of the backward reaction. In the case of dissolved solids like \( \text{Mg(OH)}_2 \)it establishes a balance where salts dissolve and ions re-associate to form solids simultaneously. Initially, when \( \text{Mg(OH)}_2 \)dissolves, it forms \( \text{Mg}^{2+} \)and\( \text{OH}^- \)ions in solution. As more dissolves, these ions start to recombine to form the solid again. At equilibrium:
- The concentration of dissolved ions remains constant.
- The solid’s dissolution and formation rates are equal.
- The system’s macroscopic properties (e.g., concentration) don’t change.
Ionic Dissociation
When ionic compounds like \( \text{Mg(OH)}_2 \)enter water, they undergo ionic dissociation, a process where the compound separates into its ions. This is depicted by the equation\[ \text{Mg(OH)}_2 (s) \rightleftharpoons \text{Mg}^{2+} (aq) + 2\text{OH}^- (aq). \]Once dissociated, these ions are free to move in the solution, conducting electricity.It's important to note:
- Dissociation is affected by the solution conditions, such as pH and ion concentrations.
- Complete dissociation is more likely in diluted solutions, while in saturated solutions, equilibrium can limit dissociation.
- Spectator ions like \( \text{Na}^+ \)from \( \text{NaOH} \)do not participate in the dissolution of \( \text{Mg(OH)}_2 \)but they affect the overall ion concentration.