Question

The \(K_{\text {? }}\) for \(\mathrm{PbI}_{2}(s)\) is \(1.4 \times 10^{-8} .\) Calculate the solubility of \(\mathrm{PbI}_{2}(s)\) in \(0.048 M\) NaI.

Short answer

The solubility of \(\mathrm{PbI_2}(s)\) in a \(0.048\, \text{M}\) NaI solution is approximately \(6.1 \times 10^{-6}\, \text{M}\).

Step-by-step solution

Identify the solubility product constant

The solubility product constant, K\(_{\text {sp}}\), for \(\mathrm{PbI}_{2}(s)\) is given as \(1.4 \times 10^{-8}\).

Write the balanced dissolution reaction and the corresponding solubility product expression

Write the balanced dissolution reaction for \(\mathrm{PbI}_{2}(s)\) and the corresponding solubility product expression for K\(_{\text {sp}}\): \[\mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2 \mathrm{I^{-}}(aq)\] \[K_{\text{sp}} = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2\]

Set up the reaction table

Use a table to represent the initial concentration of ions before dissolution, change in concentration during dissolution, and the equilibrium concentration of ions at the end of dissolution. Before dissolution: \[ [\mathrm{Pb^{2+}}] = 0\] \[ [\mathrm{I^{-}}] = 0.048\,\text{M}\, \text{(from NaI)}\] Change during dissolution: \[ [\mathrm{Pb^{2+}}] \rightarrow +s\] \[ [\mathrm{I^{-}}] \rightarrow +2s\] At equilibrium: \[ [\mathrm{Pb^{2+}}] = s\] \[ [\mathrm{I^{-}}] = 0.048 \, \text{M} + 2s\]

Substitute the equilibrium concentrations into the K\(_{\text {sp}}\) expression

Substitute the equilibrium concentrations into the expression for K\(_{\text {sp}}\): \[K_{\text{sp}}= 1.4 \times 10^{-8} = s (0.048 + 2s)^2\]

Solve for solubility 's'

Since K\(_{\text {sp}}\) is a small value, the value of 's' will be also small in comparison to 0.048. Therefore, we can approximate the equation by neglecting the 2s term: \[1.4 \times 10^{-8} = s (0.048)^2\] Now, solve for 's': \[s = \frac{1.4 \times 10^{-8}}{(0.048)^2} = 6.1 \times 10^{-6}\, \text{M}\]

Conclusion

The solubility of \(\mathrm{PbI_2}(s)\) in a \(0.048\, \text{M}\) NaI solution is approximately \(6.1 \times 10^{-6}\, \text{M}\).

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state of balance in a chemical reaction where the rates of the forward and reverse reactions are equal. This means that the concentrations of reactants and products remain constant over time. For reactions involving solids and ions in solution, like the dissolution of a salt, equilibrium is crucial in determining how much solid will dissolve. In this context, when a salt like \( \mathrm{PbI}_2 \) is placed in water, it does not dissolve completely. Instead, it reaches a point where the rate of dissolution equals the rate of precipitation.

Dissolution Reaction

A dissolution reaction is the process through which a solid compound breaks down into its individual ions in a solution. For \( \mathrm{PbI}_2 \), the reaction can be written as:

• \( \mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb^{2+}}(aq) + 2 \mathrm{I^{-}}(aq) \)

Here, solid lead(II) iodide dissociates into lead and iodide ions in water. This dissociation is part of reaching equilibrium. The better we understand this reaction, the more accurately we can predict the solubility of the salt under different conditions.

Ksp Expression

The solubility product constant, \( K_{\text{sp}} \), is a measure of a compound's solubility when it is in equilibrium. It is a type of equilibrium constant specifically for salts. For \( \mathrm{PbI}_2 \), the \( K_{\text{sp}} \) expression is:

• \( K_{\text{sp}} = [\mathrm{Pb^{2+}}][\mathrm{I^{-}}]^2 \)

This expression shows the relationship between the concentration of ions in a saturated solution of \( \mathrm{PbI}_2 \). It gives us the mathematical framework to calculate solubility. In other words, knowing the \( K_{\text{sp}} \) helps predict how much of a salt can dissolve in a solution.

Solubility Calculation

Solubility calculation is the process of determining how much solute can dissolve in a solvent at equilibrium. Using the \( K_{\text{sp}} \) value, we set up an equation based on the dissolution reaction's equilibrium concentrations. For \( \mathrm{PbI}_2 \) in a NaI solution, we use:

• \( 1.4 \times 10^{-8} = s (0.048 + 2s)^2 \)

By simplifying, neglecting the \( 2s \) term due to its small size, we solve:

• \( s = \frac{1.4 \times 10^{-8}}{(0.048)^2} \approx 6.1 \times 10^{-6} \, \text{M} \)

This result indicates the molar solubility of \( \mathrm{PbI}_2 \) in the presence of \( 0.048 \, \text{M} \) NaI. Understanding this allows chemists to control conditions for desired solubility outcomes.