A \(2.5-\mathrm{mL}\) sample of \(0.188 \mathrm{M}\) silver nitrate solution was mixed with \(2.5 \mathrm{~mL}\) of \(0.188 \mathrm{M}\) sodium chloride solution labeled with radioactive chlorine-36. The activity of the initial sodium chloride solution was \(2.46 \times 10^{6} \mathrm{~Bq} / \mathrm{mL}\). After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 175 Bq/mL. (a) Write a balanced chemical equation for the reaction that occurred. (b) Calculate the \(K_{s p}\) for the precipitate under the conditions of the experiment.

When silver nitrate (AgNO₃) reacts with sodium chloride (NaCl), a white precipitate of silver chloride (AgCl) is formed along with sodium nitrate (NaNO₃) in the solution. The balanced chemical equation is given by: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)

Firstly, we need to calculate the initial concentration of both Ag⁺ and Cl⁻ ions in the mixed solution. The total volume of the solution is 5 mL, in which 2.5 mL is from each of the two solutions. The number of moles of Ag⁺ in the silver nitrate solution can be calculated as follows: moles of Ag⁺ = volume × concentration moles of Ag⁺ = 0.0025 L × 0.188 mol/L = \(4.7\times10^{-4}\;mol\) Since there are equal volumes of the two solutions, the number of moles of Cl⁻ in the sodium chloride solution is equal to that of Ag⁺: moles of Cl⁻ = \(4.7\times10^{-4}\;mol\) Now, we can find the initial concentrations of Ag⁺ and Cl⁻ in the mixed solution: initial concentration of Ag⁺ and Cl⁻ = Total moles / Total volume initial concentration of Ag⁺ and Cl⁻ = \(4.7\times10^{-4}\;mol\)/(0.005 L) = 0.094 M

The initial activity of the Cl⁻ ions in the sodium chloride solution is given by: activity of Cl⁻ = \(2.46\times10^{6}\;Bq/mL\) Since the sodium chloride solution is 2.5 mL, the total initial activity of Cl⁻ ions is: Total initial activity = activity per mL × volume Total initial activity = \(2.46\times10^{6}\;Bq/mL\) × 2.5 mL = \(6.15\times10^{6}\;Bq\)

In the filtrate, the activity of Cl⁻ ions is given as 175 Bq/mL. Considering that the total volume of the solution is now 5 mL, the final total Cl⁻ ion activity can be found as: Total final activity = activity per mL × volume Total final activity = 175 Bq/mL × 5 mL = 875 Bq

As the activity of Cl⁻ ions is proportional to their concentration, we can calculate the ratio of precipitated Cl⁻ ions as follows: precipitated Cl⁻ ratio = (Total initial activity - Total final activity) / Total initial activity precipitated Cl⁻ ratio = ( \(6.15\times10^{6}\;Bq\) - 875 Bq) / \(6.15\times10^{6}\;Bq\) ≈ 1 Since this ratio is close to 1, it means that almost all the Cl⁻ ions in the solution have precipitated as AgCl.

We now know that almost all the Ag⁺ ions and Cl⁻ ions have reacted to form precipitate, and the concentration of Ag⁺ and Cl⁻ ions in the filtrate is very low. The concentration of Cl⁻ in the filtrate is proportional to its activity: [Cl⁻] = Total final activity / (Total initial activity × total volume) [Cl⁻] = 875 Bq / (\(6.15\times10^{6}\;Bq\) × 5 mL) ≈ \(2.84\times10^{-8}\;M\) Since we can assume that the concentration of Cl⁻ ions is equal to the concentration of Ag⁺ ions, we can write: [Ag⁺] ≈ \(2.84\times10^{-8}\;M\)

The solubility product constant can be calculated using the concentrations of Ag⁺ and Cl⁻ ions in the filtrate: \(K_{sp}\) = [Ag⁺][Cl⁻] \(K_{sp}\) ≈ \(2.84\times10^{-8}\;M\)^2 = \(8.08\times10^{-16}\) Thus, the solubility product constant of AgCl under the given experimental conditions is approximately \(8.08\times10^{-16}\).