Problem 51 - Chemistry An Atoms First Approach

Question

\(100.0 \mathrm{mL}\) of \(1.0 \times 10^{-2}\) \(M\) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{mL}\) of \(1.0 \times 10^{-3} \mathrm{M}\) NaF. Will \(\mathrm{PbF}_{2}(s)\) \(\left(K_{\mathrm{sp}}=4 \times 10^{-8}\right)\) precipitate?

Step-by-Step Solution

Answer

The final concentrations of Pb2+ and F- ions after mixing the solutions are \(5.0 \times 10^{-3} \text{M}\) and \(5.0 \times 10^{-4} \text{M}\), respectively. The reaction quotient (Q) is calculated as \(1.25 \times 10^{-9}\). Since Q < Ksp (\(4 \times 10^{-8}\)), no precipitation of PbF2 will occur upon mixing the given solutions.

Step 1: Calculate Initial Concentrations

First, let's calculate the initial concentrations of Pb2+ and F- ions in the final mixed solution. To do this, we will use the following formula: \(C_{f} = \frac{C_i V_i}{V_f}\) Where \(C_f\) is the final concentration, \(C_i\) is the initial concentration, \(V_i\) is the initial volume, and \(V_f\) is the final volume. For Pb2+ ions: \(C_{f(Pb^{2+})} = \frac{(1.0 \times 10^{-2} \text{M})(100.0 \text{mL})}{(100.0 \text{mL} + 100.0 \text{mL})}\) For F- ions: \(C_{f(F^{-})} = \frac{(1.0 \times 10^{-3} \text{M})(100.0 \text{mL})}{(100.0 \text{mL} + 100.0 \text{mL})}\)

Step 2: Calculate Final Concentrations

Now, let's calculate the final concentrations of Pb2+ and F- ions: \(C_{f(Pb^{2+})} = \frac{(1.0 \times 10^{-2} \text{M})(100.0 \text{mL})}{200.0 \text{mL}} = 5.0 \times 10^{-3} \text{M}\) \(C_{f(F^{-})} = \frac{(1.0 \times 10^{-3} \text{M})(100.0 \text{mL})}{200.0 \text{mL}} = 5.0 \times 10^{-4} \text{M}\)

Step 3: Calculate the Reaction Quotient (Q)

Now, we will calculate the reaction quotient (Q) using the formula: \(Q = [Pb^{2+}][F^-]^2\) Where [Pb2+] is the concentration of Pb2+ ions and [F-] is the concentration of F- ions. \(Q = (5.0 \times 10^{-3} \text{M})(5.0 \times 10^{-4} \text{M})^2\)

Step 4: Determine if Precipitation Occurs

Now let's determine if precipitation occurs by comparing the reaction quotient (Q) to the Ksp of PbF2: If Q > Ksp, precipitation occurs. If Q < Ksp, no precipitation occurs. \(Q = (5.0 \times 10^{-3} \text{M})(5.0 \times 10^{-4} \text{M})^2 = 1.25 \times 10^{-9}\) Since \(Q = 1.25 \times 10^{-9}\) and \(K_{\text{sp}} = 4 \times 10^{-8}\), we have: \(Q < K_{\text{sp}}\) Since the reaction quotient (Q) is less than the solubility product constant (Ksp) of PbF2, no precipitation will occur upon mixing the given solutions.

Key Concepts

Precipitation Reaction

Understanding precipitation reactions is foundational to grasping many chemical processes that involve the formation of a solid from a solution. A precipitation reaction occurs when two soluble salts are mixed in a solution and result in the formation of an insoluble salt, which we refer to as the precipitate. This occurs when the product of the concentration of the ions exceeds the solubility product constant, denoted by Ksp.

For example, mixing solutions containing lead(II) nitrate and sodium fluoride could potentially lead to the formation of lead(II) fluoride (PbF2), an insoluble compound. Whether PbF2 precipitates depends on the reaction quotient, which compares the concentrations of the ions in the solution to the Ksp. If this quotient is higher than the Ksp value, the ions will exceed their solubility limit and precipitate from the solution.

Reaction Quotient

The reaction quotient (Q) helps predict whether a reaction will proceed forward or reverse to reach equilibrium. For a precipitation reaction, Q is the product of the concentrations of the ionic species involved in the reaction raised to the power of their stoichiometric coefficients in the balanced equation.

In the context of precipitation, if the value of Q is greater than Ksp (the solubility product constant), it indicates that the system has too many ions in solution and that a precipitate will form to reduce those concentrations, pushing the reaction towards equilibrium. Conversely, when Q is less than Ksp, the system can still dissolve more ions, and precipitation will not occur.

Initial and Final Concentration Calculations

The calculation of initial and final concentrations in a reaction mixture is crucial for predicting whether a precipitate will form. The initial concentrations are determined based on the molarities of the reactants before mixing. These concentrations change once the solutions are combined, leading to the final concentrations used to calculate the reaction quotient, Q.

To find these final concentrations, you can use the dilution formula:
\(C_{f} = \frac{C_i V_i}{V_f}\)

Here, \(C_f\) is the final concentration, \(C_i\) is the initial concentration, \(V_i\) is the initial volume, and \(V_f\) is the total final volume of the mixed solutions. These calculations are fundamental for assessing the solubility and predicting the formation of a precipitate in chemical reactions.