Question

The \(K_{\mathrm{sp}}\) for lead iodide \(\left(\mathrm{Pb} \mathrm{I}_{2}\right.\) ) is \(1.4 \times 10^{-8}\). Calculate the solubility of lead iodide in each of the following. a. water b. \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) c. \(0.010 \mathrm{M}\) NaI

Short answer

The solubilities of PbI₂ in different solutions are: a. In water: \(1.52 \times 10^{-3}\) M b. In 0.10 M Pb(NO₃)₂: \(5.92 \times 10^{-5}\) M c. In 0.010 M NaI: \(1.4 \times 10^{-4}\) M

Step-by-step solution

Write the balanced equation for the dissolution of PbI₂ in water

The dissolution of lead iodide in water can be represented as: \[PbI_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2I^-(aq)\]

Set up an expression for Ksp and the equilibrium concentrations

We will use s as the solubility of PbI₂ in moles per liter. Based on the stoichiometric coefficients in the balanced equation: \[ [Pb^{2+}] = s \] \[ [I^-] = 2s \] The \(K_{sp}\) expression for this system will be: \[K_{sp} = [Pb^{2+}][I^-]^2\] Plug in the equilibrium concentrations: \[1.4 \times 10^{-8} = (s)(2s)^2\]

Solve for s (solubility in water)

Solve the above equation for s: \[1.4 \times 10^{-8} = 4s^3\] \[s^3 = \frac{1.4\times 10^{-8}}{4}\] \[s^3 = 3.5\times 10^{-9}\] \[s = \sqrt[3]{3.5\times 10^{-9}}\] \[s = 1.52\times 10^{-3}\, M\] The solubility of PbI₂ in water is \(1.52\times 10^{-3}\) M. #b. Solubility in 0.10 M Pb(NO₃)₂#

Consider the common ion effect on equilibrium

In the presence of 0.10 M Pb(NO₃)₂, more Pb²⁺ ions are available in solution, which will affect the solubility of PbI₂. It will shift the equilibrium to the left, so the solubility will be lower.

Write the equilibrium concentrations

Since the lead ion concentration is already 0.10 M, the equilibrium concentrations will be: \[ [Pb^{2+}] = 0.10 + s \] \[ [I^-] = 2s \]

Calculate the solubility in 0.10 M Pb(NO₃)₂

Since s is very small compared to 0.10 M, we can assume that the concentration of Pb²⁺ remains approximately constant at 0.10 M. \[K_{sp} = (0.10)(2s)^2\] \[1.4 \times 10^{-8} = (0.10)(4s^2)\] \[s^2 = \frac{1.4 \times 10^{-8}}{0.40}\] \[s^2 = 3.5 \times 10^{-8}\] \[s = \sqrt{3.5\times 10^{-8}}\] \[s = 5.92\times 10^{-5}\, M\] The solubility of PbI₂ in 0.10 M Pb(NO₃)₂ is \( 5.92\times 10^{-5}\) M. #c. Solubility in 0.010 M NaI#

Consider the common ion effect on I⁻ concentration

In the presence of 0.010 M NaI, more I⁻ ions are available in solution, which will affect the solubility of PbI₂. It will shift the equilibrium to the left, so the solubility will be lower.

Write the equilibrium concentrations

Since the iodide concentration is already 0.010 M: \[ [Pb^{2+}] = s \] \[ [I^-] = 0.010 + 2s \]

Calculate the solubility in 0.010 M NaI

Since s is very small compared to 0.010 M, we can assume that the concentration of I⁻ remains approximately constant at 0.010 M. \[K_{sp} = (s)(0.010 + 2s)^2\] \[1.4 \times 10^{-8} = (s)(0.010)^2\] \[s = \frac{1.4 \times 10^{-8}}{1 \times 10^{-4}}\] \[s = 1.4\times 10^{-4}\, M\] The solubility of PbI₂ in 0.010 M NaI is \(1.4\times 10^{-4}\) M.

Key concepts

Common Ion Effect

The concept of the common ion effect is crucial when studying solubility. It refers to the reduction in solubility of a salt when a solution already contains one of the ions present in the salt. For instance, when lead iodide (\(\mathrm{PbI}_2\)) dissolves, it releases \(\mathrm{Pb}^{2+}\) and \(\mathrm{I}^-\) ions into the solution.
However, if there's already a significant concentration of either ion in the solution, the system equilibrium shifts. This shift arises because according to Le Chatelier's principle, the system compensates for the additional ions by decreasing the dissolution of the salt.

• For example, when added to a solution containing \(0.10 \, \text{M} \; \mathrm{Pb(NO}_3)_2\), the extra \(\mathrm{Pb}^{2+}\) ions suppress further dissolution of \(\mathrm{PbI}_2\).
• Similarly, in a \(0.010 \, \text{M}\) solution of \(\text{NaI}\), excess \(\mathrm{I}^-\) ions in the solution lead to decreased solubility of \(\mathrm{PbI}_2\).

Understanding this effect helps in predicting how solubility changes with varying ion concentrations in solutions.

Equilibrium Concentrations

Equilibrium concentrations are vital in determining how much of a solute can dissolve in a solution at equilibrium. In a saturated solution, the rate at which the solute dissolves equals the rate at which it precipitates.
When calculating equilibrium concentrations for a dissolution reaction, like \(\mathrm{PbI}_2 \rightarrow \mathrm{Pb}^{2+} + 2\mathrm{I}^-\), the concentrations of the ions depend directly on the solubility of the initial compound. This is expressed in terms of "s," where:

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

Hence, the product of these concentrations gives us the solubility product constant (\(K_{sp}\)), defined as \(K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{I}^-]^2\). This balance keeps the solution stable and provides a pathway to calculate solubility changes under different conditions.

Solubility Calculation

Calculating solubility requires you to understand various equilibrium expressions. It involves using the \(K_{sp}\) to find out how much of a compound will dissolve under standard conditions. For lead iodide, we equate its solubility expressions to \(K_{sp}\). Starting from the relation:
\[K_{sp} = [Pb^{2+}][I^-]^2\]as derived from the balanced dissolution reaction. Substituting equilibrium concentrations into this expression allows for the calculation of solubility, denoted by "s."

• For water: Calculate by solving \([Pb^{2+}] = s\) and \([I^-] = 2s\).
• Adjust for ion presence in solutions like \(0.10 \, \text{M} \; \mathrm{Pb(NO}_3)_2\) or \(0.010 \, \text{M} \; \mathrm{NaI}\) by modifying the concentrations to \([Pb^{2+}] = 0.10 + s\) and \([I^-] \approx 0.010\) respectively.

This approach leads to solving equations that consider both the original ion contribution and additional ions from the compound, hence deriving the altered solubility level.

Dissolution Reaction

The process of a solid dissolving in a liquid involves a dissolution reaction, which transforms the solid into its constituent ions in solution. The expression for the dissolution of lead iodide, for instance, is:
\[\mathrm{PbI}_2(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{I}^-(aq)\] It clearly denotes the shift from solid \(\mathrm{PbI}_2\) to dissolved \(\mathrm{Pb}^{2+}\) and \(\mathrm{I}^-\) ions.
This reaction does not only describe dissolution, but also the dynamic equilibrium established in the system. At equilibrium, the rate of dissolution equals the rate of precipitation, meaning the system dynamically balances the forward and reverse reactions.

• The state ensures that the concentration of ions in the solution remains constant.
• In practical terms, it explains why dissolving stops when you reach a saturation point — additional solid remains undissolved.

This explanation provides foundational knowledge required for tackling problems related to chemical solubility and equilibrium in chemistry.