Question

Show that in qualitative cation analysis group \(1,\) if you obtain \(1.00 \mathrm{mL}\) of saturated \(\mathrm{PbCl}_{2}(\mathrm{aq})\) at \(25^{\circ} \mathrm{C}\), sufficient \(\mathrm{Pb}^{2+}\) should be present to produce a precipitate of \(\mathrm{PbCrO}_{4}(\mathrm{s}) .\) Assume that you use \(1 \mathrm{drop}\) \((0.05 \mathrm{mL})\) of \(1.0 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}\) for the test.

Short answer

Based on the calculated values for the concentrations of the ions involved and the calculated ion product (Q), if \(Q > Ksp\) for \(PbCrO_{4}\), then the precipitate of \(PbCrO_{4}(s)\) will indeed form.

Step-by-step solution

Write the Balanced Equation

Write down the balanced equation for the precipitation reaction. The formula would be \(Pb^{2+}(aq) + CrO_{4}^{2−}(aq) → PbCrO_{4}(s)\)

Calculate Initial Concentrations

Calculate the molar concentration of each ion. It can be derived using M1V1 = M2V2 where M1 is the initial molarity of K2CrO4, V1 is the initial volume of K2CrO4, M2 is the final molarity of Pb2+ and CrO4 2−, and V2 is the final volume of the mixture. From the given volume and molarity for the ions, calculate the initial concentration of \(Pb^{2+}\) and \(CrO_{4}^{2−}\). Hence, initial concentration of \(Pb^{2+}\) = (1 ml / 1.05 ml) * solubility of \(PbCl_{2}\) = \(x M\) and initial concentration of \(CrO_{4}^{2−}\) = (0.05 ml/ 1.05 ml) *1.0 M \(= y M\)

Calculate Ion Product and Compare with Ksp

Calculate the Ion product Q by multiplying the concentrations of the ions as per the stoichiometry of the balanced equation from step 1. Using the formula for the ion product, \(Q = [Pb^{2+}] * [CrO_{4}^{2−}]\), calculate Q. Compare Ion product (Q) with solubility product constant (Ksp) of \(PbCrO_{4}\). If \(Q > Ksp\), it indicates that the solution is supersaturated with ions and a precipitate will form. To prove the original exercise, Q should come out as greater than the Ksp.

Key concepts

Precipitation Reaction

When discussing chemical processes, **precipitation reactions** are important to understand. A precipitation reaction occurs when two soluble salts react in aqueous solution to form one or more insoluble products, called precipitates. This process happens because the product, often a solid, does not dissolve in water at room temperature.
In the given example, a precipitation reaction occurs between lead ions ( Pb^{2+}) and chromate ions ( CrO_4^{2-}) to form lead chromate ( PbCrO_4), which is a yellow solid. Precipitation reactions are crucial in identifying and separating ions in qualitative cation analysis, especially in group 1, where you identify simple cations.

• Balanced Reaction: For the reaction, the balanced chemical equation is: Pb^{2+}(aq) + CrO_4^{2−}(aq) → PbCrO_4(s).
• Visual Confirmation: The formation of a solid precipitate provides a clear visual indication that the reaction has occurred.

Solubility Product Constant (Ksp)

The **solubility product constant**, abbreviated as **Ksp**, is a crucial concept in predicting the extent to which a compound will dissolve and the potential for precipitation. Ksp is an equilibrium constant that measures the solubility of a solid in a solvent, typically water. It tells us the level at which ions in solution can exist without forming a precipitate.
Specifically, for a partially soluble compound like lead chromate ( PbCrO_4), the Ksp is defined for the equilibrium: PbCrO_4(s) ⇌ Pb^{2+}(aq) + CrO_4^{2−}(aq).

• Significance: A small Ksp value indicates low solubility, meaning the ions readily form a solid precipitate rather than remaining in solution.
• Ksp Value Usage: Using the Ksp, we can calculate to what extent precipitation occurs when a solution is mixed, as in the exercise.

Ion Product (Q)

**Ion product**, represented by **Q**, is a measure of the concentrations of ions in a solution before equilibrium is reached. While the Ksp relates to the maximum product of the ion concentrations at equilibrium, Q is used to predict whether further precipitation will occur.
The ion product for a reaction is calculated in the same way as Ksp, but it's used to understand the solution's condition relative to solubility equilibrium. In the problem, the ion product for lead chromate formation is expressed as: Q = [Pb^{2+}] * [CrO_4^{2−}]. After calculating Q, compare it with the Ksp to predict precipitate formation.

• If Q < Ksp, the solution is unsaturated, and no precipitate will form.
• If Q = Ksp, the solution is at equilibrium, and it is saturated.
• If Q > Ksp, the solution is supersaturated, and a precipitate will form.

Supersaturation

**Supersaturation** is an intriguing condition of a solution where it contains more solute than it can theoretically hold at a given temperature and pressure. This occurs when Q exceeds the Ksp, indicating a build-up of ion concentration that exceeds what can dissolve.
In practical terms, if you continue adding solute to a saturated solution without some physical change (like cooling, crystallization), the solution might become supersaturated. In the problem described in the exercise, the calculation of Q leads to a value greater than Ksp for PbCrO_4, showing that the solution is supersaturated and thus will form a precipitate.

• Formation of Precipitate: The excess ions in a supersaturated solution will try to come out of the solution by forming solid precipitates.
• Visual Proof: The formation of a solid is clear evidence that the solution has moved past saturation and into supersaturation.