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What is the molar solubility of silver chloride (Ksp = 1.8 x 10 10) in a 0.103 M sodium chloride solution?

What is the molar solubility of silver chloride (Ksp = 1.8 x 10 10) in a 0.103 M sodium chloride solution?

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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**Question: Calculating Molar Solubility** What is the molar solubility of silver chloride (\(K_{sp} = 1.8 \times 10^{-10}\)) in a 0.103 M sodium chloride solution? **Explanation:** This question requires understanding the concept of molar solubility and the common ion effect in solubility equilibria. The solubility product constant (\(K_{sp}\)) of silver chloride provides a measure of its solubility in pure water, which can be altered in the presence of a common ion (in this case, chloride from sodium chloride). The presence of 0.103 M sodium chloride reduces the solubility of silver chloride due to the Le Chatelier's principle, affecting the equilibrium of dissociation. By applying the common ion effect and using the \(K_{sp}\) value, one can calculate the new solubility of silver chloride in this solution.
Transcribed Image Text:**Question: Calculating Molar Solubility** What is the molar solubility of silver chloride (\(K_{sp} = 1.8 \times 10^{-10}\)) in a 0.103 M sodium chloride solution? **Explanation:** This question requires understanding the concept of molar solubility and the common ion effect in solubility equilibria. The solubility product constant (\(K_{sp}\)) of silver chloride provides a measure of its solubility in pure water, which can be altered in the presence of a common ion (in this case, chloride from sodium chloride). The presence of 0.103 M sodium chloride reduces the solubility of silver chloride due to the Le Chatelier's principle, affecting the equilibrium of dissociation. By applying the common ion effect and using the \(K_{sp}\) value, one can calculate the new solubility of silver chloride in this solution.
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