(5) Silver phosphate, Ag,PO4, is an ionic compound with a K, of 2.6 ×x 10". Calculate the solubility of this compound in (a) pure water. (b) 0.20 mol L² Na;PO, solution.
(5) Silver phosphate, Ag,PO4, is an ionic compound with a K, of 2.6 ×x 10". Calculate the solubility of this compound in (a) pure water. (b) 0.20 mol L² Na;PO, solution.
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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![**Silver Phosphate Solubility Calculation**
Silver phosphate, \( \text{Ag}_3\text{PO}_4 \), is an ionic compound with a solubility product constant (\( K_{sp} \)) of \( 2.6 \times 10^{-18} \). This exercise requires calculating the solubility of silver phosphate in two different scenarios:
(a) **In Pure Water:**
- Determine the solubility of Ag\(_3\)PO\(_4\) when dissolved in pure water.
(b) **In a 0.20 mol L\(^{-1}\) Na\(_3\)PO\(_4\) Solution:**
- Calculate the solubility of Ag\(_3\)PO\(_4\) in a solution already containing 0.20 mol L\(^{-1}\) of Na\(_3\)PO\(_4\).
**Guidance for Calculation:**
1. **Understand the Dissociation:**
- Silver phosphate dissociates as follows:
\[
\text{Ag}_3\text{PO}_4 (s) \rightarrow 3\text{Ag}^+ (aq) + \text{PO}_4^{3-} (aq)
\]
2. **Expression for \( K_{sp} \):**
- The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ag}^+]^3 [\text{PO}_4^{3-}]
\]
3. **Set Up the Equilibria:**
- Use stoichiometry to express ion concentrations in terms of solubility (s) and substitute into the \( K_{sp} \) expression.
- Solve for solubility (s) in both pure water and the Na\(_3\)PO\(_4\) solution, considering the common ion effect.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F01e5e66e-0e79-4c86-85a9-0b0202abd934%2F241a910d-8cb2-43e2-91c7-66311897e39f%2Fm1wdwi9_processed.png&w=3840&q=75)
Transcribed Image Text:**Silver Phosphate Solubility Calculation**
Silver phosphate, \( \text{Ag}_3\text{PO}_4 \), is an ionic compound with a solubility product constant (\( K_{sp} \)) of \( 2.6 \times 10^{-18} \). This exercise requires calculating the solubility of silver phosphate in two different scenarios:
(a) **In Pure Water:**
- Determine the solubility of Ag\(_3\)PO\(_4\) when dissolved in pure water.
(b) **In a 0.20 mol L\(^{-1}\) Na\(_3\)PO\(_4\) Solution:**
- Calculate the solubility of Ag\(_3\)PO\(_4\) in a solution already containing 0.20 mol L\(^{-1}\) of Na\(_3\)PO\(_4\).
**Guidance for Calculation:**
1. **Understand the Dissociation:**
- Silver phosphate dissociates as follows:
\[
\text{Ag}_3\text{PO}_4 (s) \rightarrow 3\text{Ag}^+ (aq) + \text{PO}_4^{3-} (aq)
\]
2. **Expression for \( K_{sp} \):**
- The \( K_{sp} \) expression is:
\[
K_{sp} = [\text{Ag}^+]^3 [\text{PO}_4^{3-}]
\]
3. **Set Up the Equilibria:**
- Use stoichiometry to express ion concentrations in terms of solubility (s) and substitute into the \( K_{sp} \) expression.
- Solve for solubility (s) in both pure water and the Na\(_3\)PO\(_4\) solution, considering the common ion effect.
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