Question

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

Short answer

We are asked to determine if a precipitate of \(\mathrm{PbF}_{2}(s)\) will form when a \(\mathrm{Pb(NO_3)_2}\) solution and a \(\mathrm{NaF}\) solution are mixed. We calculated the reaction quotient (\(Q_{sp}\)) as \(6.25 \times 10^{-11}\), which is less than the given solubility product constant (\(K_{sp} = 4 \times 10^{-8}\)). Since \(Q_{sp} < K_{sp}\), precipitation will not occur. Therefore, \(\mathrm{PbF}_{2}(s)\) will not precipitate in this mixture.

Step-by-step solution

Determine the moles of each ion

To begin, we'll calculate the moles of \(\mathrm{Pb}^{2+}\) and \(\mathrm{F}^-\) ions present in the initial solutions. Moles can be calculated using the formula: \(\text{moles} = \text{concentration} \times \text{volume}\). For \(\mathrm{Pb}^{2+}\) ions: Moles of \(\mathrm{Pb}^{2+} = (1.0 \times 10^{-2} \mathrm{M})(100.0 \mathrm{mL}) = 1.0 \times 10^{-3} \mathrm{mol}\) For \(\mathrm{F}^-\) ions: Moles of \(\mathrm{F}^- = (1.0 \times 10^{-3} \mathrm{M})(100.0 \mathrm{mL}) = 1.0 \times 10^{-4} \mathrm{mol}\)

Calculate the new concentrations

Next, we will calculate the new concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{F}^-\) ions in the mixture after the solutions are combined. Since both solutions have the same volume (100 mL), the combined volume is 200 mL. For \(\mathrm{Pb}^{2+}\) ions: Concentration of \(\mathrm{Pb}^{2+} = \frac{\text{moles of } \mathrm{Pb}^{2+}}{\text{total volume}} = \frac{1.0 \times 10^{-3} \mathrm{mol}}{200.0 \mathrm{mL}} = 5.0 \times 10^{-3} \mathrm{M}\) For \(\mathrm{F}^-\) ions: Concentration of \(\mathrm{F}^- = \frac{\text{moles of } \mathrm{F}^-}{\text{total volume}} = \frac{1.0 \times 10^{-4} \mathrm{mol}}{200.0 \mathrm{mL}} = 5.0 \times 10^{-4} \mathrm{M}\)

Calculate the reaction quotient (\(Q_{sp}\))

The reaction quotient, \(Q_{sp}\), can be calculated by taking the product of the concentrations of the ions involved in the reaction raised to their stoichiometric coefficients. In this case, the reaction is given by: \(\mathrm{PbF}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(aq) + 2\mathrm{F}^-(aq)\) So, \(Q_{sp} = [\mathrm{Pb}^{2+}][\mathrm{F}^-]^2\) Plugging in the values we found in step 2, we get: \(Q_{sp} = (5.0 \times 10^{-3}\mathrm{M})(5.0 \times 10^{-4}\mathrm{M})^2 = 6.25 \times 10^{-11}\)

Compare \(Q_{sp}\) to \(K_{sp}\)

Finally, we can determine if \(\mathrm{PbF}_{2}(s)\) will precipitate by comparing the calculated \(Q_{sp}\) value to the given \(K_{sp}\) value. If \(Q_{sp} > K_{sp}\), precipitation will occur. If \(Q_{sp} < K_{sp}\), precipitation will not occur. Given, \(K_{sp} = 4 \times 10^{-8}\) Comparing, we find that \(Q_{sp} = 6.25 \times 10^{-11} < K_{sp} = 4 \times 10^{-8}\). Since \(Q_{sp} < K_{sp}\), precipitation will not occur. Therefore, \(\mathrm{PbF}_{2}(s)\) will not precipitate in this mixture.

Key concepts

Chemical Equilibrium

Chemical equilibrium is a crucial concept in chemistry, especially when dealing with reactions in solution. It refers to the state of a reaction in which the rate of the forward reaction equals the rate of the reverse reaction. At this point, the concentrations of reactants and products remain constant over time, despite the continuous movement of molecules. An example of equilibrium in a precipitation reaction is when a solid salt, such as lead(II) fluoride (\textbf{PbF}\(_2\)), is in a dynamic balance with its dissolved ions (\textbf{Pb}^{2+} and \textbf{F}^- ions).

Equilibrium does not imply that the reactants and products are in equal amounts; it simply indicates that their ratios do not change over time. This is often represented using an equilibrium constant (\(K_{eq}\)), but in the case of sparingly soluble salts, we instead use the solubility product constant (\(K_{sp}\)). Understanding the concept of equilibrium enables chemists to predict the behavior of a system—including whether a compound will remain dissolved or form a precipitate under particular conditions.

Ionic Product

The ionic product concept is vital when discussing solubility equilibria. It is calculated similar to the solubility product constant, but instead of using the equilibrium concentrations, the ionic product uses the actual concentrations of ions in a solution at any given moment. For instance, if we consider a solution of \textbf{Pb(NO}\(_3\))\(_2\) and \textbf{NaF}, to predict whether lead(II) fluoride (\textbf{PbF}\(_2\)) will precipitate, we calculate the ionic product (Q\(_{sp}\)) by multiplying the concentrations of the lead ions (\textbf{Pb}^{2+}) and fluoride ions (\textbf{F}^-) present in solution. This is essential because the ionic product allows us to compare with the solubility product constant (\(K_{sp}\)) to determine if a precipitate will form. If Q\(_{sp}\) exceeds \(K_{sp}\), the system is super-saturated and precipitation is likely. If Q\(_{sp}\) is less than \(K_{sp}\), the solution remains unsaturated, and no precipitate will form.

Understanding the ionic product helps students predict whether changes in concentration will lead to precipitation or dissolution, providing insight into how concentration affects solubility.

Precipitation Reactions

Precipitation reactions are key in the field of analytical chemistry and have practical applications in water treatment, mineral extraction, and other chemical industries. A precipitation reaction occurs when ions in solution combine to form an insoluble compound or a 'precipitate'. This process takes place when the ionic product of the concentrations of the reacting ions exceeds the solubility product constant (\(K_{sp}\)) for the compound.

In the context of the exercise, \textbf{Pb(NO}\(_3\))\(_2\) and \textbf{NaF} are mixed, and we are asked to predict whether lead(II) fluoride (\textbf{PbF}\(_2\)) will precipitate. The solubility rules help to predict the likelihood of this occurrence generally, but knowing the solubility product constant allows for a quantitative analysis. By calculating and comparing \(K_{sp}\) and the reaction quotient (Q\(_{sp}\)), students can accurately determine the outcome of the reaction. This example helps in understanding the interplay between solubility, ion concentrations, and the principle of equilibrium, which are fundamental in predicting the results of reactions in aqueous solutions.