Problem 92 - Chemistry: The Central Science in SI Units, Global Edition

Question

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.

Step-by-Step Solution

Answer

The balanced chemical equation for the reaction is: AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq). After the analysis, the solubility product constant (\(K_{sp}\)) of AgCl under the given experimental conditions is approximately \(8.08\times10^{-16}\).

Step 1: (a) Write a balanced chemical equation for the reaction

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)

Step 2: (b) Calculate the initial concentration of ions

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

Step 3: (b) Calculate the initial activity of Cl⁻ ions

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 = \(6.15\times10^{6}\;Bq\)

Step 4: (b) Calculate the final activity of Cl⁻ ions in the filtrate

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

Step 5: (b) Determine the amount of Cl⁻ ions in the precipitate

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.

Step 6: (b) Calculate the concentration of Cl⁻ ions in the filtrate

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⁺] ≈ \(2.84\times10^{-8}\;M\)

Step 7: (b) Calculate the solubility product constant (\(K_{sp}\))

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 \(8.08\times10^{-16}\).

Key Concepts

Chemical Reaction Equations

Chemical reaction equations are a way to represent what happens when chemicals react. They use symbols and formulas to show which substances go in and which come out of a chemical reaction. In our specific exercise, we are looking at a reaction between silver nitrate

• Represented as AgNO₃
• It reacts with sodium chloride (NaCl)
• The result is a new substance, silver chloride (AgCl), which appears as a solid precipitate, and sodium nitrate (NaNO₃) which remains in solution.

An important part of writing chemical reaction equations is balancing them. Balancing ensures that the same number of each type of atom appears on both sides of the equation. For our reaction, the equation is:
\[\text{AgNO}_3(aq) + \text{NaCl}(aq) \rightarrow \text{AgCl}(s) + \text{NaNO}_3(aq)\]This balanced equation tells us that one molecule of AgNO₃ reacts with one molecule of NaCl to produce one molecule each of AgCl and NaNO₃.

Concentration Calculations

In chemistry, concentration calculations help us determine how much of a substance is present in a given volume of solution. This is crucial for understanding the proportions in mixtures and reactions. During our experiment:

• The concentration of silver nitrate and sodium chloride was provided as 0.188 M (molar, which means moles per liter).
• The volumes used were 2.5 mL each, leading to a combined solution volume of 5 mL.
• We needed to find the initial concentrations of
  • Silver ions (Ag⁺)
  • Chloride ions (Cl⁻) in the mixed solution.

We calculate moles using formula:
moles = volume (L) × concentration (M)
For both silver and chloride ions:
moles = 0.0025 L × 0.188 M = 4.7 × 10⁻⁴ mol.
When combined in a 5 mL solution, this gives an initial concentration of:
0.094 M (moles per liter) for both Ag⁺ and Cl⁻ ions. This calculation ensures we know the starting amounts of reactants in any chemical reaction or solution under study.

Radioactive Labeling

Radioactive labeling is a technique used to study the movement and transformation of substances by substituting some atoms in the substance with radioactive isotopes. Here, sodium chloride was labeled with radioactive chlorine-36. Such labeling allows us to track where the labeled atoms go, thanks to their measurable radiation.
Before the reaction, the sodium chloride solution had an initial activity of 2.46 × 10⁶ Bq/mL.
Key points:

• The solution was 2.5 mL in volume, leading to a total initial activity of 6.15 × 10⁶ Bq.
• After the reaction and removing the precipitate, remaining filtrate activity was 175 Bq/mL over a 5 mL volume, summing up to 875 Bq total.
• The significant drop in activity, calculated as the difference of initial and final:
  (6.15 × 10⁶ - 875) Bq, indicated that nearly all chloride ions had been precipitated as AgCl.

Understanding radioactive labeling means understanding **how to track and measure a specific element** even after it has gone through chemical transformations or physical processes.