Question

A solution contains \(2.0 \times 10^{-4} M \mathrm{Ag}^{+}(a q)\) a n d \(1.5 \times 10^{-3} M \mathrm{Pb}^{2+}(a q)\). I f N a I i s a d d e d , w i l l A g I (\(K_{s p}=8.3 \times 10^{-17}\)) or \(\mathrm{PbI}_{2}\left(K_{s p}=7.9 \times 10^{-9}\right)\) precipitate first? Specify the concentration of \(\mathrm{I}^{-}(a q)\) needed to begin precipitation.

Short answer

The concentration of \(I^{-}(aq)\) needed to initiate precipitation is \(4.15 × 10^{-13} M\). The first precipitate formed is AgI as it has a lower threshold concentration of I⁻ ions to initiate precipitation.

Step-by-step solution

Write the balanced equation and Ksp expression for each salt

The balanced equations for the dissolution of AgI and PbI₂ are: \[AgI \longleftrightarrow Ag^+(aq) + I^-(aq)\] and \[PbI_2 \longleftrightarrow Pb^{2+}(aq) + 2I^-(aq)\] Now let's write the Ksp expressions for each salt. For AgI, the Ksp expression is \(K_{sp} = [Ag^+][I^-]\) For PbI₂, the Ksp expression is \(K_{sp} = [Pb^{2+}][I^-]^2\)

Calculate the concentration of I⁻(aq) for each salt at the start of precipitation

We need to find the concentration of I⁻(aq) that satisfies the Ksp expressions using the initial concentrations of the cations and their respective Ksp values. For AgI, we have \(8.3 × 10^{-17} = (2.0 × 10^{-4})([I^-]_{AgI})\) Now solve for \([I^-]_{AgI}\) : \([I^-]_{AgI} = \frac{8.3 × 10^{-17}}{2.0 × 10^{-4}} = 4.15 × 10^{-13} M\) For PbI₂, we have \(7.9 × 10^{-9} = (1.5 × 10^{-3})([I^-]_{PbI_2})^2\) Now solve for \([I^-]_{PbI_2}\) : \([I^-]_{PbI_2} = \sqrt{\frac{7.9 × 10^{-9}}{1.5 × 10^{-3}}} = 7.27 × 10^{-4} M\)

Compare the concentration of I⁻(aq) required for precipitation

Comparing the calculated I⁻(aq) concentrations: \([I^-]_{AgI} = 4.15 × 10^{-13} M\) \([I^-]_{PbI_2} = 7.27 × 10^{-4} M\) As \([I^-]_{AgI}\) is smaller than \([I^-]_{PbI_2}\), AgI precipitates first.

Specify the concentration of I⁻(aq) needed to begin precipitation

So, the concentration of I⁻(aq) needed to initiate precipitation is \(4.15 \times 10^{-13}M\). The first precipitate formed is AgI as it has a lower threshold concentration of I⁻ ions to initiate precipitation.

Key concepts

Chemical Equilibrium

In chemistry, chemical equilibrium refers to a state where the rate of the forward reaction equals the rate of the reverse reaction. This means that the concentrations of reactants and products remain constant over time. Equilibrium does not mean that the reactants and products are present in equal amounts, but rather that their ratios do not change with time.

When a salt dissolves in water, it dissociates into its constituent ions. For example, the dissolution of silver iodide (AgI) can be represented by the following equilibrium expression:\[\[\begin{align*}AgI_{(s)} &\longleftrightarrow Ag^+_{(aq)} + I^-_{(aq)}\frac{}{}\end{align*}\]\]The symbol '\longleftrightarrow' indicates that the dissolution reaction can proceed in both forward and reverse directions and will reach an equilibrium state. The solubility product constant (Ksp) is a special type of equilibrium constant that applies to these dissolution reactions. It represents the maximum product of the ionic concentrations ([Ag^+] and [I^-] in this case) when the solution is saturated and at equilibrium.

Precipitation Reactions

A precipitation reaction occurs when two soluble salts react in solution to form an insoluble salt, or precipitate. The formation of a precipitate can happen when the product of the concentrations of the ions exceeds the solubility product constant (Ksp). At this point, the solution becomes supersaturated, and the excess ions combine to form a solid that settles out of the solution.

For example, in our exercises, we discussed the potential precipitation of AgI and PbI2 when iodide (I^-) is added to a solution. Since AgI has a much smaller Ksp value than PbI2, it will precipitate at a much lower concentration of I^- ions. This difference in Ksp values reflects the differing solubilities of the two compounds; lower Ksp values indicate less soluble substances, which will precipitate more readily.

Ionic Product

The ionic product of a salt is the product of the molar concentrations of its ions, each raised to the power of its coefficient in the balanced chemical equation. It is similar to the solubility product constant (Ksp), but while Ksp is a fixed value characteristic of a substance at a given temperature, the ionic product can change with varying concentrations of ions in the solution. If the ionic product becomes greater than the Ksp, a precipitate will form because the solution can no longer hold the excess of ions in solution.

In the context of our exercise, the ionic product for AgI is simply the product of [Ag^+] and [I^-]. For PbI2, the ionic product would be the concentration of [Pb^{2+}] times the square of the concentration of [I^-], since PbI2 produces two moles of iodide for every mole of lead (II) ion. Thus by comparing the ionic product to Ksp, we can predict whether a precipitate will form and which precipitate will form first upon the addition of I- ions to the mixture of Ag+ and Pb2+.