Question

\(\mathrm{KI}(\mathrm{aq})\) is slowly added to a solution with \(\left[\mathrm{Pb}^{2+}\right]=\) \(\left[\mathrm{Ag}^{+}\right]=0.10 \mathrm{M} .\) For \(\mathrm{PbI}_{2}, K_{\mathrm{sp}}=7.1 \times 10^{-9} ;\) for \(\mathrm{AgI},\) \(K_{\mathrm{sp}}=8.5 \times 10^{-17}\). (a) Which precipitate should form first, \(\mathrm{PbI}_{2}\) or AgI? (b) What \(\left[\mathrm{I}^{-}\right]\) is required for the second cation to begin to precipitate? (c) What concentration of the first cation to precipitate remains in solution at the point at which the second cation begins to precipitate? (d) \(\operatorname{Can} \mathrm{Pb}^{2+}(\mathrm{aq})\) and \(\mathrm{Ag}^{+}(\) aq) be effectively separated by fractional precipitation of their iodides?

Short answer

The precipitate that forms first is PbI2 as it requires a lower iodide concentration to precipitate out than AgI. The calculated iodide concentration for the second precipitate to begin to form and the concentration of the first cation at the point when the second begins to precipitate can be found by using the \(K_{sp}\) equations and the ion concentrations. The effectiveness of the fractional precipitation depends on the significant difference in the solubilities of \(Pb^{2+}\) and \(Ag^+\) under the condition.

Step-by-step solution

Determine which precipitate forms first

Use the solubility product (\(K_{sp}\)) expressions for both salts, i.e., \(AgI \rightarrow Ag^+ + I^-\) and \(PbI2 \rightarrow Pb^{2+} + 2I^-\). Plug both ion concentrations into their respective \(K_{sp}\) expression: \(K_{sp}(AgI) = [Ag^+][I^-]\) and \(K_{sp}(PbI2) = [Pb^{2+}][I^-]^{2}\). Calculate the iodide concentration necessary for each salt to precipitate out of solution by using the given \(K_{sp}\) values. The substance needing the lower iodide concentration to precipitate out will be the one that forms first.

Calculate the iodide concentration for the second precipitate

Using the \(K_{sp}\) expression for the precipitate that does not form first, calculate what the iodide concentration must be for the second precipitate to form. This can be done by setting up the \(K_{sp}\) equation and solving for the iodide ion concentration.

Determine the concentration of the first cation when the second begins to precipitate

When the second cation begins to precipitate, the concentration of the first cation will be what's left in the solution. This can be determined by using the \(K_{sp}\) equation for the first precipitate, inserting the calculated iodide concentration from Step 2, and solving for the first cation concentration.

Evaluate the effectiveness of fractional precipitation

Fractional precipitation of iodides will be effective in separating \(Pb^{2+}\) and \(Ag^+\) if there is a significant difference in their solubilities under the condition. If the concentrations of the remaining cations in solution after precipitation of one of the cations are significantly different, the separation can be considered effective.

Key concepts

Solubility Product (Ksp)

The solubility product constant, often abbreviated as \(K_{sp}\), is a critical value used when assessing the solubility of ionic compounds in a solution. It's a measure of the maximum amount of compound that can dissolve in water to form a saturated solution. The concept of \(K_{sp}\) is grounded in the equilibrium state of the dissolution process.
For instance, consider an ionic compound such as silver iodide (\(AgI\)). The dissolution of \(AgI\) in water can be expressed as the following equilibrium reaction:

• \(AgI_{(s)} \rightleftharpoons Ag^+_{(aq)} + I^-_{(aq)}\)

The \(K_{sp}\) expression for \(AgI\) is given by \([Ag^+][I^-]\), indicating the ionic concentrations at equilibrium. A similar expression applies for lead(II) iodide (\(PbI_2\)):

• \(PbI_2_{(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2I^-_{(aq)}\)

The \(K_{sp}\) for \(PbI_2\) is \([Pb^{2+}][I^-]^2\).
When comparing the \(K_{sp}\) values between different compounds like \(AgI\) and \(PbI_2\), the one with the lower \(K_{sp}\) will generally precipitate out of solution first, assuming equivalent initial conditions.

Precipitation Reactions

Precipitation reactions occur when two aqueous solutions combine to form a solid, known as a precipitate. This process is governed by the concentrations of the reacting ions and the solubility product constant (\(K_{sp}\)) of the compound formed.
In the given problem, \(AgI\) and \(PbI_2\) are allowed to potentially precipitate through the addition of iodide ions (\(I^-\)). A systematic way to determine which compound will precipitate first involves examining their respective \(K_{sp}\) values. A lower \(K_{sp}\) means the compound requires fewer ions to exceed saturation and begin forming a precipitate.
For this specific case:

• \(AgI\) has a \(K_{sp}\) of \(8.5 \times 10^{-17}\).
• \(PbI_2\) has a \(K_{sp}\) of \(7.1 \times 10^{-9}\).

Because \(AgI\) has the lower \(K_{sp}\), it will precipitate first as the iodide ion concentration increases. Once \([I^-]\) is high enough to form \(PbI_2\), this new compound will also begin to precipitate, but this occurs only after \(AgI\) has largely been removed from the solution.

Iodide Concentration

Understanding the role of iodide concentration is essential in predicting when precipitation reactions will occur. The concentration of iodide ions \([I^-]\) must reach a certain threshold to fulfill the solubility product condition for a compound to precipitate.
In a situation where multiple ions are competing to form precipitates, as with \(AgI\) and \(PbI_2\), you can calculate required iodide concentration thresholds for each precipitate using their \(K_{sp}\) values.

• For \(AgI\), the critical concentration is determined by its \(K_{sp}\) expression: \([Ag^+][I^-] = 8.5 \times 10^{-17}\).
• For \(PbI_2\), it is calculated from \([Pb^{2+}][I^-]^2 = 7.1 \times 10^{-9}\).

The iodide concentration necessary for \(AgI\) to start precipitating is significantly smaller than that for \(PbI_2\). Thus, as you increase \([I^-]\) by adding \(KI_{(aq)}\) to the solution, \(AgI\) will begin to precipitate first, followed by \(PbI_2\). This difference in necessary ion concentration can be leveraged to achieve fractional precipitation, effectively separating the ion species.