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The molar solubility of Al(OH)(3) in 0.2...

The molar solubility of `Al(OH)_(3)` in 0.2 M NaOH solution is `x xx 10^(-22)"mol/L"`. Given that, solubility product of `Al(OH)_(3)=2.4xx10^(-24)`. What is numerical value of x?

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To solve the problem, we need to find the molar solubility of \( \text{Al(OH)}_3 \) in a 0.2 M NaOH solution and express it in the form \( x \times 10^{-22} \, \text{mol/L} \). We are also given the solubility product \( K_{sp} \) of \( \text{Al(OH)}_3 \) as \( 2.4 \times 10^{-24} \). ### Step-by-step Solution: 1. **Write the dissociation equation for \( \text{Al(OH)}_3 \)**: \[ \text{Al(OH)}_3 (s) \rightleftharpoons \text{Al}^{3+} (aq) + 3 \text{OH}^- (aq) \] 2. **Define the solubility**: Let the molar solubility of \( \text{Al(OH)}_3 \) be \( S \). Therefore, at equilibrium: - The concentration of \( \text{Al}^{3+} \) will be \( S \). - The concentration of \( \text{OH}^- \) will be \( 3S \). 3. **Consider the presence of \( \text{OH}^- \) from NaOH**: Since we have a 0.2 M NaOH solution, the initial concentration of \( \text{OH}^- \) ions is 0.2 M. Therefore, the total concentration of \( \text{OH}^- \) ions in the solution will be: \[ [\text{OH}^-] = 3S + 0.2 \] 4. **Apply the solubility product expression**: The solubility product \( K_{sp} \) for \( \text{Al(OH)}_3 \) is given by: \[ K_{sp} = [\text{Al}^{3+}][\text{OH}^-]^3 \] Substituting the concentrations: \[ K_{sp} = S \cdot (3S + 0.2)^3 \] 5. **Substitute the given \( K_{sp} \)**: We know \( K_{sp} = 2.4 \times 10^{-24} \): \[ 2.4 \times 10^{-24} = S \cdot (3S + 0.2)^3 \] 6. **Approximate \( (3S + 0.2) \)**: Since \( S \) is expected to be very small compared to 0.2, we can approximate: \[ 3S + 0.2 \approx 0.2 \] Thus, we can rewrite the equation as: \[ 2.4 \times 10^{-24} = S \cdot (0.2)^3 \] 7. **Calculate \( (0.2)^3 \)**: \[ (0.2)^3 = 0.008 \] Therefore, the equation becomes: \[ 2.4 \times 10^{-24} = S \cdot 0.008 \] 8. **Solve for \( S \)**: \[ S = \frac{2.4 \times 10^{-24}}{0.008} = 3 \times 10^{-22} \, \text{mol/L} \] 9. **Identify the value of \( x \)**: From the expression \( S = x \times 10^{-22} \), we find that \( x = 3 \). ### Final Answer: The numerical value of \( x \) is \( 3 \).

To solve the problem, we need to find the molar solubility of \( \text{Al(OH)}_3 \) in a 0.2 M NaOH solution and express it in the form \( x \times 10^{-22} \, \text{mol/L} \). We are also given the solubility product \( K_{sp} \) of \( \text{Al(OH)}_3 \) as \( 2.4 \times 10^{-24} \). ### Step-by-step Solution: 1. **Write the dissociation equation for \( \text{Al(OH)}_3 \)**: \[ \text{Al(OH)}_3 (s) \rightleftharpoons \text{Al}^{3+} (aq) + 3 \text{OH}^- (aq) \] ...
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