Problem 65 - Chemistry and Chemical Reactivity

Question

You have a solution that has a lead(II) ion concentration of 0.0012 M. If enough soluble chloride-containing salt is added so that the \(\mathrm{Cl}^{-}\) concentration is 0.010 \(\mathrm{M},\) will \(\mathrm{PbCl}_{2}\) precipitate?

Step-by-Step Solution

Answer

No, \(\text{PbCl}_2\) will not precipitate because \(Q < K_{sp}\).

Step 1: Understand the problem

We have a solution containing lead(II) ions with a concentration of 0.0012 M and chloride ions with a concentration of 0.010 M. We need to determine if \(\text{PbCl}_2\) will precipitate by comparing the ionic product to the solubility product constant \(K_{sp}\) for \(\text{PbCl}_2\).

Step 2: Write the solubility equilibrium expression

The equilibrium reaction for lead(II) chloride dissolving in water is:\[\text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq)\]The solubility product expression \(K_{sp}\) is:\[K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2\]

Step 3: Calculate the ionic product

To determine whether precipitation occurs, calculate the ionic product (Q):\[Q = [\text{Pb}^{2+}][\text{Cl}^-]^2\]Substitute the given concentrations into the expression:\[Q = (0.0012)(0.010)^2\]\[Q = 0.0012 \times 0.0001 = 0.00000012\, \text{or}\, 1.2 \times 10^{-7}\]

Step 4: Compare with the solubility product constant

The solubility product constant \(K_{sp}\) for \(\text{PbCl}_2\) is approximately \(1.7 \times 10^{-5}\). Compare \(Q\) to \(K_{sp}\):- If \(Q > K_{sp}\), precipitation occurs.- If \(Q < K_{sp}\), no precipitation occurs.- If \(Q = K_{sp}\), the system is at equilibrium.In this case, \(Q (1.2 \times 10^{-7}) < K_{sp}(1.7 \times 10^{-5})\), so no precipitation occurs.

Key Concepts

Solubility Product Constant

The solubility product constant, often represented as \( K_{sp} \), is a key concept in understanding the solubility equilibrium of sparingly soluble salts. It is a type of equilibrium constant that applies to the dissolution of such salts into their constituent ions in a saturated solution.
For lead(II) chloride (\( \text{PbCl}_2 \)), the dissolving process in water yields lead ions \( \text{Pb}^{2+} \) and chloride ions \( \text{Cl}^{-} \). The solubility product expression is given by:

• \(K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2\)

This expression allows us to predict whether a compound will dissolve in a solution or if it will precipitate.
When the ionic product of the ions exceeds the \( K_{sp} \), precipitation occurs.

Ionic Product

The ionic product \( Q \) is similar to the \( K_{sp} \) but represents a dynamic picture of the ion concentrations in a specific solution at any given moment. It is calculated in the same way as the \( K_{sp} \):

• \(Q = [\text{Pb}^{2+}][\text{Cl}^-]^2\)

In our example, we substitute the given concentrations: \(0.0012 \, \text{M} \) for \([\text{Pb}^{2+}]\) and \(0.010 \, \text{M} \) for \([\text{Cl}^-]\) to compute \( Q \).
Doing the math:

• \(Q = 0.0012 \times (0.010)^2 = 1.2 \times 10^{-7}\)

Understanding \( Q \) allows us to quickly determine if the solution is supersaturated (\( Q > K_{sp} \)), saturated (\( Q = K_{sp} \)), or unsaturated (\( Q < K_{sp} \)).

Precipitation

Precipitation is the process wherein ions in solution join together to form a solid, called a precipitate. This happens when the product of the concentrations of the ions in solution exceeds the \( K_{sp} \) of the compound.
In our scenario, we compare \( Q \), the ionic product, with \( K_{sp} \) for lead(II) chloride. The rule is quite simple:

• If \( Q > K_{sp} \), precipitation will occur because the solution is supersaturated.
• If \( Q < K_{sp} \), no precipitation happens, as the solution is still unsaturated.
• If \( Q = K_{sp} \), the solution is at equilibrium, and no net change occurs.

For this specific example, since \( Q (1.2 \times 10^{-7}) < K_{sp} (1.7 \times 10^{-5}) \), no precipitation occurs, indicating that lead(II) chloride remains dissolved.

Lead(II) Chloride

Lead(II) chloride, represented by \( \text{PbCl}_2 \), is an ionic compound consisting of lead ions (\( \text{Pb}^{2+} \)) and chloride ions (\( \text{Cl}^- \)). It’s known for having poor solubility in water.
The compound dissolves according to the equation:

• \( \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq) \)

In solutions, the small amount that dissolves follows the solubility product principle guided by its \( K_{sp} \). Understanding its behavior in solutions is crucial in predicting if it will precipitate under varying conditions.
Importantly, by manipulating concentrations of ions in the solution, one can control the balance between dissolution and precipitation, which finds applications in various fields such as chemical synthesis and environmental science.