Problem 68 - Chemistry and Chemical Reactivity

Question

You have \(95 \mathrm{mL}\) of a solution that has a lead(II) concentration of 0.0012 M. Will PbCl_ precipitate when 1.20 g of solid \(\mathrm{NaCl}\) is added?

Step-by-Step Solution

Answer

PbCl_2 precipitates because Q > K_sp.

Step 1: Calculate moles of NaCl

First, determine the number of moles of NaCl in 1.20 g. The molar mass of NaCl is approximately 58.44 g/mol. Using the formula: \( ext{moles} = \frac{ ext{mass}}{ ext{molar mass}} \), the moles of NaCl are calculated as follows: \[ ext{moles of NaCl} = \frac{1.20 ext{ g}}{58.44 ext{ g/mol}} \approx 0.02054 ext{ mol} \].

Step 2: Find concentration of Cl^- ions

Since NaCl dissociates completely in solution, each mole of NaCl provides one mole of Cl^- ions. The total volume of the solution after adding NaCl remains approximately 95 mL or 0.095 L. The concentration of \( ext{Cl}^- \) ions is: \[ [ ext{Cl}^-] = \frac{0.02054 ext{ mol}}{0.095 ext{ L}} \approx 0.2162 ext{ M} \].

Step 3: Find concentration of Pb^{2+} ions

We are given that the concentration of \( ext{Pb}^{2+} \) is 0.0012 M.

Step 4: Calculate Q (reaction quotient)

To determine if a precipitate forms, calculate the reaction quotient \( Q \) by using the formula \( Q = [ ext{Pb}^{2+}][ ext{Cl}^-]^2 \):\[ Q = (0.0012 ext{ M})(0.2162 ext{ M})^2 \approx 5.61 \times 10^{-5} \].

Step 5: Compare Q with K_sp

The solubility product constant \( K_{sp} \) for \( ext{PbCl}_2 \) is approximately \( 1.7 \times 10^{-5} \). Since \( Q = 5.61 \times 10^{-5} > K_{sp} = 1.7 \times 10^{-5} \), precipitation of \( ext{PbCl}_2 \) occurs.

Key Concepts

Reaction Quotient (Q)

Understanding the Reaction Quotient, often denoted as \( Q \), is crucial in determining whether a precipitate will form in a solution. The reaction quotient is a snapshot of the product concentrations at any point during a chemical reaction. When dealing with precipitation reactions, it helps us predict the direction the reaction will proceed.
To calculate \( Q \), we use the formula: \[ Q = [\text{Pb}^{2+}][\text{Cl}^-]^2 \] Using the concentrations of the ions present in the solution. For this specific problem, it's crucial to note that \( Q \) takes into account the concentrations at present, not at equilibrium as \( K_{sp} \) does.

• If \( Q < K_{sp} \), the solution can dissolve more ions, and no precipitate forms.
• If \( Q = K_{sp} \), the solution is at equilibrium, and the system is saturated with maximum dissolved ions without precipitate.
• If \( Q > K_{sp} \), the solution contains excess ions, leading to the formation of a precipitate, as it cannot hold more dissolved ions.

In this example, since \( Q = 5.61 \times 10^{-5} \) is greater than \( K_{sp} = 1.7 \times 10^{-5} \), it indicates the precipitate \( \text{PbCl}_2 \) forms.

Solubility Product Constant (Ksp)

The solubility product constant, \( K_{sp} \), is a critical factor in precipitation calculations. It represents the maximum product of the ion concentrations in a saturated solution at equilibrium. Each type of salt, like \( \text{PbCl}_2 \), has its unique \( K_{sp} \) value, denoting how much of it can dissolve in a solution before reaching a saturated state.
For our reaction \( \text{PbCl}_2 \longrightarrow \text{Pb}^{2+} + 2\text{Cl}^- \), the \( K_{sp} \) expression is: \[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \] The value \( K_{sp} = 1.7 \times 10^{-5} \) implies that in a saturated solution, the product of these ion concentrations cannot exceed this constant.
Why is \( K_{sp} \) important? When we compare \( K_{sp} \) to the calculated \( Q \):

• \( K_{sp} \) helps predict the formation of a precipitate in a solution.
• It provides a threshold value to determine the solubility of an ionic compound.
• By understanding \( K_{sp} \), we can control reactions in terms of solubility and precipitation effectively.

Remember, \( K_{sp} \) is temperature-dependent, and changes in temperature may alter its value.

Ionic Concentration Calculations

Calculating ionic concentrations is a foundational aspect of understanding precipitation reactions. It allows us to use these concentrations to determine \( Q \), crucial for deciding if precipitation happens.
Let's break it down:

• Finding Moles: First, convert the grams of a solid like NaCl added to the moles using its molar mass. For NaCl, this is about 58.44 g/mol.
• Determine Ion Concentrations: When NaCl dissolves, it dissociates into Cl^- ions. The formula \[ [\text{Cl}^-] = \frac{\text{moles of NaCl}}{\text{total volume in L}} \] determines the molarity of the chloride ions.
• Given Ions: In our example, we started with a known concentration of \( \text{Pb}^{2+} = 0.0012 \text{ M} \).

These calculations are critical since they give the concentration required in the \( Q \) formula.
Remember that changes in volume or the addition of other substances can impact these concentrations significantly. Keeping all factors in mind ensures accurate and meaningful results when predicting precipitation.