Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![### Solubility Product Constant Calculation for Calcium Hydroxide
#### Problem Statement:
A saturated solution of \( \text{Ca(OH)}_2 \) at 20°C has a concentration of 0.0233 M.
**Question:** What is the \( K_{sp} \) for calcium hydroxide solution at 20°C?
#### Explanation:
To calculate the solubility product constant (\( K_{sp} \)) for calcium hydroxide:
1. **Dissociation of \( \text{Ca(OH)}_2 \):**
\[
\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^-
\]
2. **Solubility and Ion Concentration:**
Let the solubility of \( \text{Ca(OH)}_2 \) be S M. Therefore,
\[
[\text{Ca}^{2+}] = S = 0.0233 \, \text{M}
\]
and
\[
[\text{OH}^-] = 2S = 2 \times 0.0233 = 0.0466 \, \text{M}
\]
3. **Expression for \( K_{sp} \):**
\[
K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2
\]
Substituting the values,
\[
K_{sp} = (0.0233) \times (0.0466)^2
\]
4. **Calculation:**
Calculate the value:
\[
K_{sp} = 5.07 \times 10^{-5}
\]
5. **Representation:**
\[
K_{sp (\text{Ca(OH)}_2)} = [ \text{5.07} ] \times 10^{-5}
\]
**Interactive Component:**
- **Coefficient (green):** Enter `5.07`
- **Exponent (yellow):** Enter `-5`
- **Enter Button:** Click to verify your answer.
This exercise helps in understanding the concept of solubility product constants, the dissociation of compounds in solutions, and applying stoichiometric conversions to find the necessary ion concentrations](https://content.bartleby.com/qna-images/question/81a43bbd-f504-4d06-af9d-cc35582a0d5e/cb6e9e71-22b7-4825-8499-0d6ded42045a/vv5jakf_processed.jpeg)
Transcribed Image Text:### Solubility Product Constant Calculation for Calcium Hydroxide
#### Problem Statement:
A saturated solution of \( \text{Ca(OH)}_2 \) at 20°C has a concentration of 0.0233 M.
**Question:** What is the \( K_{sp} \) for calcium hydroxide solution at 20°C?
#### Explanation:
To calculate the solubility product constant (\( K_{sp} \)) for calcium hydroxide:
1. **Dissociation of \( \text{Ca(OH)}_2 \):**
\[
\text{Ca(OH)}_2 \rightleftharpoons \text{Ca}^{2+} + 2\text{OH}^-
\]
2. **Solubility and Ion Concentration:**
Let the solubility of \( \text{Ca(OH)}_2 \) be S M. Therefore,
\[
[\text{Ca}^{2+}] = S = 0.0233 \, \text{M}
\]
and
\[
[\text{OH}^-] = 2S = 2 \times 0.0233 = 0.0466 \, \text{M}
\]
3. **Expression for \( K_{sp} \):**
\[
K_{sp} = [\text{Ca}^{2+}][\text{OH}^-]^2
\]
Substituting the values,
\[
K_{sp} = (0.0233) \times (0.0466)^2
\]
4. **Calculation:**
Calculate the value:
\[
K_{sp} = 5.07 \times 10^{-5}
\]
5. **Representation:**
\[
K_{sp (\text{Ca(OH)}_2)} = [ \text{5.07} ] \times 10^{-5}
\]
**Interactive Component:**
- **Coefficient (green):** Enter `5.07`
- **Exponent (yellow):** Enter `-5`
- **Enter Button:** Click to verify your answer.
This exercise helps in understanding the concept of solubility product constants, the dissociation of compounds in solutions, and applying stoichiometric conversions to find the necessary ion concentrations
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