Question

A solution is \(0.010 \mathrm{M}\) in both \(\mathrm{CrO}_{4}^{2-}\) and \(\mathrm{SO}_{4}^{2-}\). To this solution, \(0.50 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(\text { aq })\) is slowly added. (a) Which anion will precipitate first from solution? (b) What is \(\left[\mathrm{Pb}^{2+}\right]\) at the point at which the second anion begins to precipitate? (c) Are the two anions effectively separated by this fractional precipitation?

Short answer

a) The anion, \(\mathrm{CrO}_{4}^{2-}\) will precipitate first. b) The concentration of \(\left[\mathrm{Pb}^{2+}\right]\) is \(2.6 \times 10^{-6}\) M when the second anion begins to precipitate. c) Yes, the two anions are effectively separated by this fractional precipitation.

Step-by-step solution

determine the anion that precipitates first

To do this, use the solubility product constants of \(PbCrO_4\) and \(PbSO_4\), which are \(1.8 \times 10^{-14}\) and \(2.6 \times 10^{-8}\) respectively. The smaller solubility product constant belongs to \(PbCrO_4\), meaning it will precipitate first. This is because a smaller solubility product constant indicates a lower solubility in solution, and thus the compound will precipitate out of solution faster.

Find the concentration [Pb2+] at the start of the second anion's precipitation

Here, the solubility product constant of the second precipitate is used. Given that \(PbSO_4\) is the second to precipitate and its solubility product constant is \(K_{sp2} = 2.6 \times 10^{-8} = [Pb^{2+}][SO_4^{2-}]\), and given that the original concentration of \(SO_4^{2-}\) is 0.01M, the equation simplifies to \(2.6 \times 10^{-8} = [Pb^{2+}]\times (0.01)\). Solving for \([Pb^{2+}]\) gives \([Pb^{2+}] = 2.6 \times 10^{-6}\).

Determine if the two anions are effectively separated

If the concentration of \([Pb^{2+}]\) at the start of \(SO_4^{2-}\)'s precipitation is less than the initial \(CrO_4^{2-}\) concentration (that is \(2.6 \times 10^{-6} < 0.01\)), then complete precipitation of \(\mathrm{CrO}_{4}^{2-}\) occurs before \(\mathrm{SO}_{4}^{2-}\) begins to precipitate. This indicates that the two anions are effectively separated by fractional precipitation, which is indeed the case in this scenario.

Key concepts

Solubility Product Constant

The solubility product constant, often represented as \(K_{sp}\), is a measure of how readily a compound will dissolve in water. For any sparingly soluble ionic compound, it is calculated by multiplying the molar concentrations of the ions in solution, each raised to the power of its coefficient in the balanced equation. Here’s a simple breakdown:

• For the compound \(PbCrO_4\), which dissociates into \(Pb^{2+}\) and \(CrO_4^{2-}\), its solubility product constant is described by the equation: \(K_{sp1} = [Pb^{2+}][CrO_4^{2-}]\).
• Similarly, for \(PbSO_4\), which dissociates into \(Pb^{2+}\) and \(SO_4^{2-}\), its equation is \(K_{sp2} = [Pb^{2+}][SO_4^{2-}]\).

In the context of precipitation reactions, the smaller the \(K_{sp}\), the less soluble the compound is, and therefore it will precipitate sooner as you add a common ion such as \(Pb^{2+}\). This knowledge is key in understanding fractional precipitation and the sequence in which reactions will occur.

Anion Separation

Anion separation by fractional precipitation depends on the differences in solubility of the various ions in a mixture. By carefully adding a precipitating agent, one anion can be removed from the solution while leaving others behind. This is predicated on the differences in the solubility product constants.

• Anions with lower \(K_{sp}\) values will precipitate first; in our case, \(CrO_4^{2-}\) precipitates ahead of \(SO_4^{2-}\) because \(PbCrO_4\) has a lower \(K_{sp}\) value than \(PbSO_4\).
• Once \(CrO_4^{2-}\) precipitates, it can be effectively separated from \(SO_4^{2-}\) present in solution.

This method allows for the selective removal of specific ions from a mixture, making it an essential concept in analytical chemistry especially when dealing with complex mixtures.

Precipitation Order

The order of precipitation is crucial in chemical analysis and separation techniques. It is determined largely by the solubility product constants of the respective compounds.
In our exercise, the precipitation order is deduced by comparing the \(K_{sp}\) values of the compounds involved:

• \(PbCrO_4\) with a \(K_{sp}\) of \(1.8 \times 10^{-14}\) precipitates first because it is less soluble than \(PbSO_4\), which has a \(K_{sp}\) of \(2.6 \times 10^{-8}\).
• By observing the solubility thresholds, the precise amount of \(Pb^{2+}\) added helps control which anion precipitates when, thereby allowing for effective separation of \(CrO_4^{2-}\) from \(SO_4^{2-}\).

This clear order helps in designing procedures to effectively separate distinct ions in a solution, leveraging the predictable outcomes based on solubility differences.