At 22 °C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.20. What is the Ksp of the salt at 22 °C? Ksp = 2.3 x10-12 Incorrect
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At 22 ∘C, an excess amount of a generic metal hydroxide, M(OH)2, is mixed with pure water. The resulting equilibrium solution has a pH of 10.20. What is the ?sp of the salt at 22 C?
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### Solubility Product Constant Calculation for M(OH)₂
**Question**: At 22°C, an excess amount of a generic metal hydroxide, M(OH)₂, is mixed with pure water. The resulting equilibrium solution has a pH of 10.20. What is the \( K_{sp} \) of the salt at 22°C?
**Given Answer**:
\[ K_{sp} = 2.3 \times 10^{-12} \]
**Status**: Incorrect
#### Explanation:
When determining the solubility product constant ( \( K_{sp} \) ) for the metal hydroxide \( M(OH)_2 \), follow these steps:
1. **Determine the concentration of hydroxide ions (\( OH^{-} \))**:
- pH = 10.20
- \( pOH = 14 - pH = 14 - 10.20 = 3.80 \)
- \( [OH^{-}] = 10^{-pOH} = 10^{-3.80} \)
2. **Express the dissociation in water and calculate ion concentrations**:
- The generic metal hydroxide \( M(OH)_2 \) dissociates in water as:
\[ M(OH)_2 \rightarrow M^{2+} + 2OH^{-} \]
- If \( s \) is the solubility of \( M(OH)_2 \), then:
\[ [OH^{-}] = 2s \]
3. **Calculate the solubility \( s \)**:
- Set \( 2s \) equal to the hydroxide ion concentration:
\[ 2s = 10^{-3.80} \]
- Solve for \( s \).
4. **Determine the \( K_{sp} \)**:
- \( K_{sp} = [M^{2+}][OH^{-}]^2 \)
- Substitute the values found for \( [M^{2+}] \) and \( [OH^{-}] \):
\[ K_{sp} = s \times (2s)^2 = 4s^3 \]
#### Important Points:
- The calculation has to be exact and precise.
- Ensure that all steps and mathematical conversions are properly followed.
This method should accurately determine the \( K_{sp} \) of the generic metal](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85650e2f-17ad-4aa4-9d56-acd675fbc675%2F768314df-2352-4759-a47a-65a8bc7573ba%2F3idtu6t_processed.png&w=3840&q=75)
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