Question

The solubility product of \(\mathrm{PbBr}_{2}\) is \(8.9 \times 10^{-6} .\) Determine the molar solubility in (a) pure water, (b) \(0.20 M \mathrm{KBr}\) solution, and (c) \(0.20 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

Short answer

Molar solubility is 0.0121 M in pure water, 2.225 x 10^-4 M in 0.20 M KBr, and 0.00333 M in 0.20 M Pb(NO3)2.

Step-by-step solution

Understanding the solubility product

For the dissociation of \( \text{PbBr}_2 \), we have the equation: \( \text{PbBr}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Br}^- (aq) \). The solubility product expression is given by \( K_{sp} = [\text{Pb}^{2+}][\text{Br}^-]^2 \). The solubility product \( K_{sp} \) can be used to find the molar solubility under different conditions.

Calculate molar solubility in pure water

Let \( s \) be the molar solubility of \( \text{PbBr}_2 \) in pure water. Then \([\text{Pb}^{2+}] = s\) and \([\text{Br}^-] = 2s\). Substituting these into the solubility product expression: \( K_{sp} = s(2s)^2 = 4s^3\). Solving for \( s \) gives \( s = \sqrt[3]{\frac in \frac{K_{sp}}{4}} = \sqrt[3]{\frac{8.9 \times 10^{-6}}{4}} \), which calculates to approximately \( \mathbf{0.0121} \) M.

Calculate molar solubility in 0.20 M KBr solution

In \( 0.20 \text{ M} \) \( \text{KBr} \), \([\text{Br}^-] = 0.20 + 2s \). Assuming \( 2s \) is negligible compared to 0.20 M, \([\text{Br}^-] \approx 0.20\). Substituting in the solubility product expression: \( K_{sp} = s(0.20)^2 \). Solving gives \( s = \frac{K_{sp}}{0.20^2} = \frac{8.9 \times 10^{-6}}{0.04} \), which calculates to approximately \( \mathbf{2.225 \times 10^{-4}} \) M.

Calculate molar solubility in 0.20 M Pb(NO3)2 solution

In \( 0.20 \text{ M} \) \( \text{Pb}\left(\text{NO}_3\right)_2 \), \([\text{Pb}^{2+}] = 0.20 + s\). Assuming \( s \) is negligible compared to 0.20 M, \([\text{Pb}^{2+}] \approx 0.20\). Substituting in the solubility product expression: \( K_{sp} = 0.20(2s)^2 \). Solving gives \( 4s^2 = \frac{K_{sp}}{0.20} = \frac{8.9 \times 10^{-6}}{0.20} \) resulting in \( 4s^2 = 4.45 \times 10^{-5} \) and \( s = \sqrt{\frac{4.45 \times 10^{-5}}{4}} \), which calculates to approximately \( \mathbf{0.00333} \) M.

Key concepts

Molar Solubility

Molar solubility is a measure of how much of a substance can dissolve in a solvent to form a saturated solution. It's expressed in moles per liter (M). When a solid dissolves, it dissociates into its constituent ions in the solution.
For instance, the molar solubility of lead(II) bromide, represented as \( \text{PbBr}_2 \), indicates how many moles of \( \text{PbBr}_2 \) can dissolve in 1 liter of water until the solution becomes saturated. In pure water, molar solubility helps us understand the maximum amount of solute that can be dissolved without any other ions interfering.
In the exercise example, the molar solubility in pure water was found to be approximately 0.0121 M. This means that in 1 L of water, about 0.0121 moles of \( \text{PbBr}_2 \) can dissolve before reaching the saturation point.

Ksp Expression

The solubility product constant (\( K_{sp} \)) is a crucial concept in predicting the solubility of compounds. It represents the equilibrium between the dissolving solid and its ions in saturated solution.
For lead(II) bromide, \( \text{PbBr}_2 \), upon dissolving, dissociates as follows:
- \( \text{PbBr}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Br}^- (aq) \).
Based on this dissociation, the \( K_{sp} \) expression is formulated as \( K_{sp} = [\text{Pb}^{2+}][\text{Br}^-]^2 \).
Here, \( [\text{Pb}^{2+}] \) denotes the concentration of lead ions and \( [\text{Br}^-] \) the concentration of bromide ions in the solution at equilibrium.
The \( K_{sp} \) value, like \( 8.9 \times 10^{-6} \) in this example, allows us to calculate molar solubility under various conditions, such as different solutions and under the influence of common ions.

Equilibrium Calculations

Equilibrium calculations for solubility can help determine how variables affect solubility. By calculating the concentrations of ions at equilibrium, we can understand and predict how solutes behave in a solution.
In these calculations, we establish a relationship between the solubility product constant and the molar solubility of the ion. For instance, in the case of \( \text{PbBr}_2 \), the equilibrium expression is \( 4s^3 = K_{sp} \) in pure water, where \( s \) stands for molar solubility.
This simplifies to solving for \( s \) and involves taking the cube root or solving algebraic equations. Likewise, equilibrium calculations apply when common ions exist in the solution, requiring adjustments for their existing concentrations and recognizing their influence on further solute dissociation.

Common Ion Effect

The common ion effect refers to the decrease in the solubility of an ionic compound when a solution already contains one of the ions in the compound. This is due to Le Chatelier's principle, which shifts equilibrium towards the undissolved solid to counteract the increase in ion concentration.
For example, when \( \text{PbBr}_2 \) is placed in a solution containing \( \text{KBr} \), the bromide ions (\( \text{Br}^- \)) already present will reduce the solubility of \( \text{PbBr}_2 \). Assuming that \( 2s \) is negligible in this context, the solubility product expression changes, simplifying calculations.
Similarly, when placed in a \( \text{Pb}(\text{NO}_3)_2 \) solution, the presence of additional lead ions \( (\text{Pb}^{2+}) \) also reduces its molar solubility. This illustrates how the presence of common ions can significantly alter dissolution processes and highlights the importance of calculating adjusted solubility.