Problem 108 - Chemistry The Science in Context

Question

Will lead(II) chloride precipitate if \(185 \mathrm{mL}\) of \(0.025 M\) sodium chloride is added to \(235 \mathrm{mL}\) of \(0.165 M\) lead (II) perchlorate at \(25^{\circ} \mathrm{C} ?\)

Step-by-Step Solution

Answer

Answer: No, lead(II) chloride will not precipitate under these conditions, as the solubility product (Q) is less than the solubility product constant (Ksp) for PbCl2 at 25°C.

Step 1: Calculate initial moles of chloride ions and lead ions

For sodium chloride, moles = (Volume × Concentration) moles of chloride ions = (185 mL × 0.025 M) moles of chloride ions = 4.625 mmol For lead(II) perchlorate, moles = (Volume × Concentration) moles of lead ions = (235 mL × 0.165 M) moles of lead ions = 38.775 mmol

Step 2: Perform the stoichiometry

The reaction between lead(II) perchlorate and sodium chloride is given as follows: Pb(ClO4)2 + 2 NaCl -> PbCl2(s) + 2 NaClO4 However, since chloride ions are in excess as compared to lead ions, no stoichiometry is required.

Step 3: Calculate the final concentrations after mixing

Total volume after mixing = (185 mL + 235 mL) = 420 mL Final concentration of Pb2+ = moles of lead ions / total volume Final concentration of Pb2+ = 38.775 mmol / 420 mL = 0.092 M Final concentration of Cl- = moles of chloride ions / total volume Final concentration of Cl- = 4.625 mmol / 420 mL = 0.011 M

Step 4: Calculate the solubility product for the reaction

Solubility product Q is given by Q = [Pb2+][Cl-]^2 Q = (0.092 M)(0.011 M)^2 = 1.1 × 10^-5

Step 5: Compare the reaction product with the solubility product constant

The solubility product constant (Ksp) for lead(II) chloride (PbCl2) at 25°C is 1.7 × 10^(-5). Since Q < Ksp (1.1 × 10^(-5) < 1.7 × 10^(-5)), no precipitate will form after mixing the two solutions.

Key Concepts

Solubility Product Constant (Ksp)

The solubility product constant, often denoted as \(K_{sp}\), is a useful value in chemistry that tells us how much of a compound can dissolve in water. For slightly soluble ionic compounds, it helps predict whether a precipitate will form in a solution.
In our problem, we're looking at lead(II) chloride, \(\text{PbCl}_2\). The \(K_{sp}\) of lead(II) chloride at 25°C is given as \(1.7 \times 10^{-5}\). This means that if the product of the ionic concentrations in the solution is less than \(K_{sp}\), the compound remains dissolved. If the value is greater, the compound will precipitate.
We calculated the reaction product (Q), which is similar to \(K_{sp}\), using the concentrations of \([\text{Pb}^{2+}]\) and \([\text{Cl}^-]^2\). In our scenario, \(Q = 1.1 \times 10^{-5}\), which is less than \(1.7 \times 10^{-5}\). Thus, no precipitate forms.

Stoichiometry

Stoichiometry is the chemistry tool that helps us understand the quantitative relationships between reactants and products in a chemical reaction. It is based on the balanced chemical equation.
For the given problem, the reaction is: \(\text{Pb(ClO}_4)_2 + 2\ \text{NaCl} \rightarrow \text{PbCl}_2(s) + 2\ \text{NaClO}_4\).
Here, each mole of lead(II) perchlorate reacts with two moles of sodium chloride to form one mole of lead(II) chloride and two moles of sodium perchlorate. However, we find that there's an excess of chloride ions, making additional stoichiometric calculations unnecessary in our specific case.

Concentration Calculations

Calculating concentrations is key to solving many chemistry problems, especially when determining if a precipitate will form. Concentration is usually expressed in molarity, which is moles of solute per liter of solution (M).
In the exercise, we find the initial moles for both chloride ions and lead ions using the volume and concentration. After mixing 185 mL of sodium chloride and 235 mL of lead(II) perchlorate, the total solution volume is 420 mL.
To get the new concentrations after mixing, we divide the moles of ions by the total volume of the solution. The final concentration of \(\text{Pb}^{2+}\) is \(0.092 \text{ M}\), and for \(\text{Cl}^-\), it is \(0.011 \text{ M}\). These concentrations are then used to calculate the reaction product (Q), which helps determine if a precipitate will form.