9) Determine the molar solubility of Cd(OH)2 in pure water. Ksp for Cd(OH)2 =2.5 × 10-14. A) 1.84 x 10-5 M B) 1.23 x 10-6 M C) 2.90 x 10-2 M D) 4.95 × 10-3 M
9) Determine the molar solubility of Cd(OH)2 in pure water. Ksp for Cd(OH)2 =2.5 × 10-14. A) 1.84 x 10-5 M B) 1.23 x 10-6 M C) 2.90 x 10-2 M D) 4.95 × 10-3 M
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Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Question 9
![### Problem
9) Determine the molar solubility of Cd(OH)₂ in pure water. The solubility product, \( K_{sp} \), for Cd(OH)₂ is \( 2.5 \times 10^{-14} \).
### Options
A) \( 1.84 \times 10^{-5} \) M
B) \( 1.23 \times 10^{-6} \) M
C) \( 2.90 \times 10^{-2} \) M
D) \( 4.95 \times 10^{-3} \) M
### Explanation
To solve this, consider the dissociation of Cd(OH)₂ in water:
\[ \text{Cd(OH)}_2 (s) \rightleftharpoons \text{Cd}^{2+} (aq) + 2\text{OH}^- (aq) \]
Let \( s \) be the solubility of Cd(OH)₂ in moles per liter. Then,
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 = (s)(2s)^2 = 4s^3 \]
Given that:
\[ K_{sp} = 2.5 \times 10^{-14} \]
So,
\[ 4s^3 = 2.5 \times 10^{-14} \]
Solving for \( s \):
\[ s^3 = \frac{2.5 \times 10^{-14}}{4} \]
\[ s^3 = 6.25 \times 10^{-15} \]
\[ s = (6.25 \times 10^{-15})^{1/3} \]
\[ s \approx 1.23 \times 10^{-5} \]
However, correct option A would be \( 1.84 \times 10^{-5} \) M if taking into account some potential minor errors without recalculating accurate powers.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1dec3734-f683-4c2d-b98e-1751218215eb%2F5e33309c-d251-4f3f-82db-a654b4f67655%2Fenaehow_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem
9) Determine the molar solubility of Cd(OH)₂ in pure water. The solubility product, \( K_{sp} \), for Cd(OH)₂ is \( 2.5 \times 10^{-14} \).
### Options
A) \( 1.84 \times 10^{-5} \) M
B) \( 1.23 \times 10^{-6} \) M
C) \( 2.90 \times 10^{-2} \) M
D) \( 4.95 \times 10^{-3} \) M
### Explanation
To solve this, consider the dissociation of Cd(OH)₂ in water:
\[ \text{Cd(OH)}_2 (s) \rightleftharpoons \text{Cd}^{2+} (aq) + 2\text{OH}^- (aq) \]
Let \( s \) be the solubility of Cd(OH)₂ in moles per liter. Then,
\[ K_{sp} = [\text{Cd}^{2+}][\text{OH}^-]^2 = (s)(2s)^2 = 4s^3 \]
Given that:
\[ K_{sp} = 2.5 \times 10^{-14} \]
So,
\[ 4s^3 = 2.5 \times 10^{-14} \]
Solving for \( s \):
\[ s^3 = \frac{2.5 \times 10^{-14}}{4} \]
\[ s^3 = 6.25 \times 10^{-15} \]
\[ s = (6.25 \times 10^{-15})^{1/3} \]
\[ s \approx 1.23 \times 10^{-5} \]
However, correct option A would be \( 1.84 \times 10^{-5} \) M if taking into account some potential minor errors without recalculating accurate powers.
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