Problem 59 - Chemistry and Chemical Reactivity

Question

You have a solution that has a lead(II) concentration of \(0.0012 \mathrm{M}\) \begin{equation}\mathrm{PbCl}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Pb}^{2+}(\mathrm{aq})+2 \mathrm{Cl}^{-}(\mathrm{aq})\end{equation} 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

\( \mathrm{PbCl}_2 \) will not precipitate.

Step 1: Write the solubility product expression

For the given equilibrium, the solubility product, \( K_{sp} \), is expressed as: \[ K_{sp} = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2 \] This expression relates the concentration of lead ions and chloride ions when \( \mathrm{PbCl}_2 \) is at equilibrium with its ions in a saturated solution.

Step 2: Identify known values

Given: Concentration of \( \mathrm{Pb}^{2+} \) is \( 0.0012 \mathrm{M} \) and concentration of \( \mathrm{Cl}^{-} \) is \( 0.010 \mathrm{M} \).

Step 3: Calculate the reaction quotient \( Q \)

The reaction quotient \( Q \) is used to predict the direction of the reaction. Calculate \( Q \) with the formula: \( Q = [\mathrm{Pb}^{2+}][\mathrm{Cl}^{-}]^2 \). Substitute the given values into the equation: \[ Q = (0.0012)(0.010)^2 \] \[ Q = 1.2 \times 10^{-7} \]

Step 4: Compare \( Q \) with \( K_{sp} \) of \( \mathrm{PbCl}_2 \)

The \( K_{sp} \) for \( \mathrm{PbCl}_2 \) is known or can be looked up as \( 1.6 \times 10^{-5} \). Compare \( Q \) and \( K_{sp} \): If \( Q < K_{sp} \), no precipitation occurs. If \( Q > K_{sp} \), precipitation will occur.

Step 5: Determine if precipitation occurs

Since \( Q = 1.2 \times 10^{-7} \) and \( K_{sp} = 1.6 \times 10^{-5} \), and \( Q < K_{sp} \), the solution is unsaturated and \( \mathrm{PbCl}_2 \) will not precipitate.

Key Concepts

Equilibrium

In a chemical reaction, equilibrium refers to the state where the rate of the forward reaction equals the rate of the backward reaction. At this point, the concentrations of the reactants and products remain constant over time, assuming the system remains undisturbed. In the context of solubility, equilibrium is crucial because it defines the concentrations of ions in a saturated solution. For example, with \( \mathrm{PbCl}_2(s) \rightleftarrows \mathrm{Pb}^{2+} (aq) + 2\mathrm{Cl}^- (aq) \), equilibrium indicates a balance between the solid \(\mathrm{PbCl}_2\) and its dissolved ions. This balance is temporary and can be affected by changing conditions, such as concentration or temperature.
Throughout this process, equilibrium is guided by the solubility product constant (\(K_{sp}\)), which is specific to each compound. The \(K_{sp}\) value helps predict whether a solid will precipitate or remain dissolved under given conditions. It's an essential concept in understanding how saturation and unsaturation of solutions work.

Precipitation Reaction

A precipitation reaction occurs when two solutions containing soluble salts combine to form an insoluble solid, known as a precipitate. This happens when the product of the concentrations of the ions in solution exceeds the solubility product (\(K_{sp}\)) of the compound they can form.
Consider \(\mathrm{PbCl}_2\) as an example. When solutions of lead(II) ions and chloride ions mix, a precipitate of \(\mathrm{PbCl}_2\) will form if the combined ion product exceeds \(K_{sp}\). If \(Q\), the reaction quotient calculated, is smaller than \(K_{sp}\), no precipitation occurs since the solution remains unsaturated.

• If \(Q > K_{sp}\), precipitation will occur, as the system surpasses saturation point, forming solid \(\mathrm{PbCl}_2\).
• If \(Q < K_{sp}\), like in our case, the solution stays unsaturated and \(\mathrm{PbCl}_2\) will not precipitate.

This process is essential in various applications, such as water treatment and analytical chemistry, where controlling the formation or dissolution of precipitates is fundamental.

Chemical Concentrations

Chemical concentrations play a pivotal role in reactions, determining how chemicals interact in a solution. Concentration is often expressed in molarity, which is the number of moles of solute per liter of solution (M). In equilibrium and precipitation reactions, concentration dictates whether or not a reaction will occur.
For our reaction with \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^-\) ions, knowing their concentrations allows us to predict the system's behavior. With \(C(\mathrm{Pb}^{2+}) = 0.0012 \mathrm{M}\) and \(C(\mathrm{Cl}^-) = 0.010 \mathrm{M}\), these values are inserted into the equation for \(Q\).
This calculation assesses whether the ion product exceeds \(K_{sp}\), directing whether \(\mathrm{PbCl}_2\) will precipitate. Hence, precise control and understanding of chemical concentrations enable chemists to predict and manipulate the outcome of reactions.