Question

You have 95 mL of a solution that has a lead(II) concentration of \(0.0012 \mathrm{M}\). Will \(\mathrm{PbCl}_{2}\) precipitate when \(1.20 \mathrm{g}\) of solid NaCl is added?

Short answer

Yes, \(\text{PbCl}_2\) will precipitate since the ion product exceeds the solubility product constant.

Step-by-step solution

Calculate Moles of NaCl Added

First, calculate the number of moles of NaCl. The molar mass of NaCl is approximately \(58.44 \text{ g/mol}\). Therefore, the moles of NaCl added is given by the formula: \[moles = \frac{mass}{molar \ mass} = \frac{1.20 \text{ g}}{58.44 \text{ g/mol}} \approx 0.0205 \text{ mol}\]

Calculate Molarity of Cl^- Ions After NaCl Addition

Determine the total volume of the solution after the addition of NaCl. Since volume change is negligible, it's still \(95 \text{ mL} = 0.095 \text{ L}\). The molarity of \(\text{Cl}^-\) in solution is:\[\text{Molarity of } \text{Cl}^- = \frac{0.0205 \text{ mol}}{0.095 \text{ L}} \approx 0.216 \text{ M} \]

Calculate Ion Product [IP] for PbCl2

The ion product \([\text{IP}]\) for \(\text{PbCl}_2\) is calculated using the initial concentration of \(\text{Pb}^{2+}\) ions \(0.0012 \text{ M}\) and the calculated molarity of \(\text{Cl}^-\):\[\text{IP} = [\text{Pb}^{2+}] [\text{Cl}^-]^2 = 0.0012 \times (0.216)^2 \approx 5.60 \times 10^{-5} \text{ M}^3\]

Compare Ion Product to Solubility Product Constant [Ksp]

The solubility product constant \(K_{\text{sp}}\) for \(\text{PbCl}_2\) is \(1.7 \times 10^{-5} \text{ M}^3\). Compare the ion product to \(K_{\text{sp}}\): \[\text{IP} = 5.60 \times 10^{-5} > K_{\text{sp}} = 1.7 \times 10^{-5} \]Since \(\text{IP}\) is greater than \(K_{\text{sp}}\), precipitation of \(\text{PbCl}_2\) will occur.

Key concepts

Lead(II) Chloride

Lead(II) chloride, or \(\text{PbCl}_2\), is an important compound often encountered in precipitation reactions. It is an ionic compound that consists of lead ions \(\text{Pb}^{2+}\) and chloride ions \(\text{Cl}^-\). Lead(II) chloride is known for its low solubility in water, which means it tends to form a solid precipitate when mixed in solutions that reach a certain concentration of ions. Understanding its behavior in solutions involves examining its dissolution in water, which can be represented by the equilibrium:\[\text{PbCl}_2 (s) \leftrightarrow \text{Pb}^{2+} (aq) + 2\text{Cl}^- (aq)\]This equation shows that for every one mole of \(\text{PbCl}_2\) dissolved, one mole of \(\text{Pb}^{2+}\) and two moles of \(\text{Cl}^-\) ions are produced in the solution. This stoichiometry is crucial when calculating whether a precipitation reaction will occur after adding more chloride ions to the solution.

Solubility Product Constant

The solubility product constant, denoted as \(K_{\text{sp}}\), is a key concept in understanding precipitation reactions. It is an equilibrium constant that reflects the level of saturation in a solution when a salt like \(\text{PbCl}_2\) is barely dissolved. Crucially, it depends on the particular ions involved in the dissolved compound. For \(\text{PbCl}_2\), the solubility product expression is defined as:\[K_{\text{sp}} = [\text{Pb}^{2+}][\text{Cl}^-]^2\]In this expression, \([\text{Pb}^{2+}]\) and \([\text{Cl}^-]^2\) represent the molar concentrations of these ions in a saturated solution. The value for \(\text{PbCl}_2\) is typically \(1.7 \times 10^{-5} \text{ M}^3\) at room temperature. When the ion product, \([\text{Pb}^{2+}][\text{Cl}^-]^2\), exceeds \(K_{\text{sp}}\), the solution is supersaturated with respect to the compound, and precipitation occurs as the system seeks equilibrium.

Molarity Calculations

Molarity is a measure of concentration that describes the number of moles of a solute per liter of solution, often denoted as \(\text{M}\). In the context of precipitation reactions, knowing the molarity of participating ions is essential to predicting if a precipitate will form.For example, the molarity of chloride ions \(\text{Cl}^-\) after adding a certain amount of \(\text{NaCl}\) to the solution is calculated by: \[\text{Molarity} = \frac{\text{moles of } \text{Cl}^-}{\text{total volume of solution (L)}}\]In the given problem, after adding \(1.20 \text{ g}\) of \(\text{NaCl}\) (which is about \(0.0205 \text{ moles}\)) to a \(95 \text{ mL}\) solution, the resulting \(\text{Cl}^-\) molarity is \(0.216 \text{ M}\). To achieve this, you calculate the moles of solute from its mass and molar mass, then use the solution’s volume to find the molarity. This step is vital to determining whether the added ions will cause precipitation.

Ion Product Comparison

The comparison of the ion product \([\text{IP}]\) to the known solubility product constant \([K_{\text{sp}}]\) is a crucial step in predicting precipitation. The ion product is calculated using the concentrations of the ions that make up the potentially precipitating compound. For \(\text{PbCl}_2\), the ion product is determined by:\[\text{IP} = [\text{Pb}^{2+}] [\text{Cl}^-]^2\]After adding \(\text{NaCl}\) to our solution, this substitution involves: - The initial \(\text{Pb}^{2+}\) from the solution, here \(0.0012 \text{ M}\).- The calculated molarity of \(\text{Cl}^-\) after addition, discovered as \(0.216 \text{ M}\).By multiplying as per the formula, if your \(\text{IP} = 5.60 \times 10^{-5} \text{ M}^3\) is greater than \(K_{\text{sp}} = 1.7 \times 10^{-5} \text{ M}^3\), it indicates that the ionic concentrations surpass the solubility threshold, guiding the precipitation of \(\text{PbCl}_2\). This process showcases how equilibrium principles apply to real-world chemical scenarios.