Question

A solution contains \(2.0 \times 10^{-4} \mathrm{M} \mathrm{Ag}^{+}\) and \(1.5 \times 10^{-3} \mathrm{M} \mathrm{Pb}^{2+}\). If NaI is added, will AgI \(\left(K_{s p}=8.3 \times 10^{-17}\right)\) or \(\mathrm{PbI}_{2}\left(K_{s p}=7.9 \times 10^{-9}\right)\) precipitate first? Specify the concentration of \(\mathrm{I}^{-}\) needed to begin precipitation.

Short answer

AgI will precipitate first, and the minimum concentration of \(I^{-}\) needed for precipitation is \(4.15 \times 10^{-13} M\).

Step-by-step solution

Write the solubility product expressions for AgI and PbI₂

Write the balanced equations for the dissociation of AgI and PbI₂ in solution and their corresponding solubility product expressions. The balanced equation for AgI dissociation: \(AgI_{(s)} \rightleftharpoons Ag_{(aq)}^{+} + I_{(aq)}^{-}\) The solubility product expression for AgI: \( K_{sp, AgI} = [Ag^{+}][I^{-}]\) The balanced equation for PbI₂ dissociation: \(PbI_{2(s)} \rightleftharpoons Pb_{(aq)}^{2+} + 2I_{(aq)}^{-}\) The solubility product expression for PbI₂: \( K_{sp, PbI_2} = [Pb^{2+}][I^{-}]^2\)

Calculate the reaction quotient (Q) for AgI and PbI₂

Calculate the reaction quotient (Q) for AgI and PbI₂ using the given concentrations of Ag⁺ and Pb²⁺. We'll assume iodide-to-be-added has a molar concentration of "\([I^{-}]\)" for simplicity. For AgI: \(Q_{AgI} = [Ag^{+}][I^{-}] = (2.0 \times 10^{-4})([I^{-}])\) For PbI₂: \(Q_{PbI_2} = [Pb^{2+}][I^{-}]^2 = (1.5 \times 10^{-3})([I^{-}])^2\)

Compare Q to Ksp for both precipitates

To determine which compound precipitates first, compare the reaction quotient (Q) of each compound to the relative solubility product (Ksp). The compound that precipitates first is the one where the value of Q_for_added_iodide is greater than or equal to Ksp For AgI: \(Q_{AgI} = (2.0 \times 10^{-4})([I^{-}]) \geq 8.3 \times 10^{-17}\), \([I^{-}] \geq \frac{8.3 \times 10^{-17}}{2.0 \times 10^{-4}} = 4.15 \times 10^{-13}\) For PbI₂: \(Q_{PbI_2} = (1.5 \times 10^{-3})([I^{-}])^2 \geq 7.9 \times 10^{-9}\), \([I^{-}] \geq \sqrt{\frac{7.9 \times 10^{-9}}{1.5 \times 10^{-3}}} = 7.3 \times 10^{-4}\)

Determine which precipitate forms first and the iodide concentration required

Based on the values calculated in Step 3, we can now determine which precipitate forms first and the minimum iodide concentration required to begin precipitation. Since AgI has a smaller iodide concentration required for precipitation, AgI will precipitate first. Minimum concentration of \(I^-\) required to begin precipitation: \([I^{-}] = 4.15 \times 10^{-13} M\)

Key concepts

AgI Precipitation

AgI, or silver iodide, precipitation involves the formation of a solid compound from ions in a solution. This occurs when the product of the concentrations of the ions exceeds the solubility product constant, known as the solubility product, or \(K_{sp}\). For AgI, the dissociation is described by:

• \(AgI_{(s)} \rightleftharpoons Ag_{(aq)}^{+} + I_{(aq)}^{-}\)

The solubility product expression is \(K_{sp, AgI} = [Ag^{+}][I^{-}]\).When sodium iodide (NaI) is added to a solution containing silver ions \((Ag^+)\), silver iodide will begin to precipitate when \([Ag^+][I^-] \geq K_{sp}\). In this scenario, with \(K_{sp} = 8.3 \times 10^{-17}\), precipitation of AgI starts at a very low iodide concentration, meaning it generally occurs first before other possible precipitates like lead(II) iodide (PbI₂), given the same iodide increase.

Reaction Quotient (Q)

The reaction quotient, \(Q\), is similar to the equilibrium constant, \(K\), but it applies to non-equilibrium conditions. It helps us understand whether a reaction will proceed forward or backward to reach equilibrium. For a solution, \(Q\) predicts if precipitation will occur:

• \(Q < K_{sp}\): The solution is unsaturated, and no precipitation occurs.
• \(Q = K_{sp}\): The solution is saturated; it is at the point of precipitation equilibrium.
• \(Q > K_{sp}\): The solution is supersaturated, and precipitation occurs to reduce ion concentration.

In the exercise, the \(Q\) values for AgI and PbI₂ are calculated using initial ion concentrations, indicating which compound will precipitate first as iodide ions increase. AgI precipitates before PbI₂ because its required \(Q\), based on \(K_{sp}\), is exceeded at a lower \([I^-]\). This shows that the order of \(Q\) meeting or exceeding \(K_{sp}\) dictates the sequence of precipitation.

Iodide Concentration

The iodide concentration \([I^-]\) plays a crucial role in determining when a precipitate will form in a solution, specifically for the salts AgI and PbI₂. The goal is to find the concentration threshold at which the product of the ion concentrations meets or exceeds the solubility product constant, \(K_{sp}\). For silver iodide (AgI), the minimum concentration of \([I^-]\) required to begin precipitation is calculated as:

• \([I^-] \geq \frac{K_{sp}}{[Ag^+]} = \frac{8.3 \times 10^{-17}}{2.0 \times 10^{-4}} = 4.15 \times 10^{-13} \text{ M}\)

This very low concentration of iodide means AgI will precipitate first before PbI₂ in a scenario where iodide is gradually added. For PbI₂, a higher iodide concentration of \(7.3 \times 10^{-4} \text{ M}\) is necessary for precipitation. Thus, knowledge of \([I^-]\) enables careful control of the precipitation process in solutions containing multiple ion species.