Question

A \(25-\mathrm{mL}\) sample of \(0.050 \mathrm{M} \mathrm{AgNO}_{3}\) is mixed with \(25 \mathrm{~mL}\) of \(0.050 \mathrm{M}\) NaI labeled with I-131. The NaI solution has an activity of \(1.25 \times 10^{10}\) counts/min\cdotmL. The following reaction takes place: $$ \mathrm{Ag}^{+}(a q)+\mathrm{I}^{-}(a q) \longrightarrow \mathrm{AgI}(s) $$ After the precipitate is removed by filtration, what is the activity of \(\mathrm{I}^{-}\) in the filtrate? The \(K_{\mathrm{sp}}\) for \(\mathrm{AgI}\) is \(1 \times 10^{-16} .\)

Short answer

Answer: The activity of I⁻ ions in the filtrate is approximately \(1.25 × 10^{10}\) counts/min·mL.

Step-by-step solution

Determine initial concentrations of Ag⁺ and I⁻ ions

We first need to find the initial concentrations of Ag⁺ and I⁻ ions in the mixed solution. To do this, we can use the formula: Concentration = (Volume₁ × Concentration₁ + Volume₂ × Concentration₂) / Total Volume For Ag⁺ ions, the initial concentration is of AgNO₃ solution, Concentration of Ag⁺ = (25 mL × 0.050 M) / (25 mL + 25 mL) = (25×0.050)/(50) = 0.025 M For I⁻ ions, the initial concentration is of NaI solution, Concentration of I⁻ = (25 mL × 0.050 M) / (25 mL + 25 mL) = (25×0.050)/(50) = 0.025 M

Calculate remaining concentration of I⁻ using Ksp

Now that we have initial concentrations of Ag⁺ and I⁻ ions, we can use the Ksp expression for AgI to calculate the remaining concentration of I⁻ ions. Ksp = [Ag⁺][I⁻] Substitute the given Ksp value and concentrations into the equation: \(1 × 10^{-16} = [0.025 - x][0.025 - x]\) Solve for x, which represents the change in concentration of Ag⁺ and I⁻ as AgI precipitates: \(x^2 - 0.05x + 1 × 10^{-16} = 0\) We notice that \(10^{-16}\) is a very small value, so we can assume that x will be also very small and the change in concentrations of Ag⁺ and I⁻ ions will be negligible. Hence, we can simplify the equation to: \(x \approx 0.05x\) Therefore, x ≈ \(2 × 10^{-16}\) M

Calculate activity of I⁻ in the filtrate

Now we need to find the activity of I⁻ in the filtrate. The activity is given per mL of NaI solution, and we need to calculate the activity for the remaining I⁻ ions in the filtrate. We are given the initial activity of I⁻ and the initial concentration of I⁻ ions, so using ratios, we can calculate the final activity: Initial Activity of I⁻ / Initial Concentration of I⁻ = Final Activity of I⁻ / (Initial Concentration of I⁻ - x) Plug in the values: \((1.25 × 10^{10} \text{counts/min·mL}) / 0.025 = \text{Final Activity of I⁻} / (0.025 - 2 × 10^{-16})\) Now, calculate the final activity: Final Activity of I⁻ = \((1.25 × 10^{10} \text{counts/min·mL} × (0.025 - 2 × 10^{-16})) / 0.025\) Final Activity of I⁻ ≈ \(1.25 × 10^{10}\) counts/min·mL So, the activity of I⁻ ions in the filtrate is approximately \(1.25 × 10^{10}\) counts/min·mL.