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Chapter 21: Problem 92

A 25.0 -mL sample of 0.050\(M\) barium nitrate solution was mixed with 25.0 mL of 0.050\(M\) sodium sulfate solution labeled with radioactive sulfur-35. The activity of the initial sodium sulfate solution was \(1.22 \times 10^{6} \mathrm{Bq} / \mathrm{mL}\) . After the resultant precipitate was removed by filtration, the remaining filtrate was found to have an activity of 250 \(\mathrm{Bq} / \mathrm{mL}\) . (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.

Short Answer

Expert verified
The balanced chemical equation for the reaction between barium nitrate and sodium sulfate is: \[Ba^{2+} (aq) + SO_4^{2-} (aq) \rightarrow BaSO_4 (s)\]. The solubility product constant, \(K_{sp}\), for the BaSO\(_4\) precipitate under the conditions of the experiment is \(9.83 \times 10^{-6}\).

Step by step solution

01

Write the balanced chemical equation

The reaction between barium nitrate and sodium sulfate can be represented as follows: \[Ba(NO_3)_2 (aq) + Na_2SO_4 (aq) \rightarrow 2NaNO_3 (aq) + BaSO_4 (s)\] The balanced chemical equation for this reaction is: \[Ba^{2+} (aq) + SO_4^{2-} (aq) \rightarrow BaSO_4 (s)\]
02

Calculate moles of solute

Based on the initial volume and concentration of solutions, we can calculate the moles of barium nitrate and sodium sulfate. Moles of Ba(NO\(_3\))\(_2\) = volume × concentration Moles of Ba(NO\(_3\))\(_2\) = 25.0 mL × 0.050 M Moles of Ba(NO\(_3\))\(_2\) = 0.00125 mol Moles of Na\(_2\)SO\(_4\) = volume × concentration Moles of Na\(_2\)SO\(_4\) = 25.0 mL × 0.050 M Moles of Na\(_2\)SO\(_4\) = 0.00125 mol
03

Determine concentration of ions in the filtrate

After the reaction, the remaining filtrate only contains radioactive sulfate ions (labeled with sulfur-35). Activity in the filtrate = 250 Bq/mL Initial activity of Na\(_2\)SO\(_4\) = \(1.22 \times 10^{6}\) Bq/mL Sulfate ions in the filtrate = \(\frac{250 \text{Bq/mL}}{1.22 \times 10^{6} \text{Bq/mL}} \times 0.50 \,\mathrm{M}\) = \(2.05 \times 10^{-4}\) M.
04

Calculate concentration of barium ions in the precipitate

Since the balanced chemical equation has a 1:1 ratio of barium and sulfate ions, the concentration of barium ions in the precipitate can be determined by subtracting the concentration of sulfate ions in the filtrate from the initial barium nitrate solution. Concentration of Ba\(^{2+}\) in the precipitate = Initial Concentration - Concentration in the filtrate Concentration of Ba\(^{2+}\) in the precipitate = 0.050 M - \(2.05\times 10^{-4}\) M Concentration of Ba\(^{2+}\) in the precipitate = \(4.795\times 10^{-2}\) M
05

Calculate K\(_{sp}\) of the precipitate

Now that we have the concentration of both ions in the precipitate, we can calculate the solubility product constant K\(_{sp}\) for BaSO\(_4\) by using the expression K\(_{sp} = [Ba^{2+}] \times [SO_4^{2-}]\). K\(_{sp} = (4.795\times 10^{-2}\,\mathrm{M}) \times (2.05\times 10^{-4}\,\mathrm{M})\) K\(_{sp} = 9.83 \times 10^{-6}\) So, the solubility product constant K\(_{sp}\) for the BaSO\(_4\) precipitate under the conditions of the experiment is \(9.83 \times 10^{-6}\).

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