The Al `(OH)_(3)` is involved in the following two equilibria,
`Al(OH)_(3)(s)hArrAl^(3+)(aq)+3OH^(-)(aq),K_(sp)`
`Al(OH)_(3)(s)+OH^(-)(aq.)hArrAl(OH)_(4)^(-)(aq),K_(c)`
Which of the following relationship is correct at which solubility is minimum?

Options

A. `[OH^(-)]=((K_(sp))/K_(c))^(1//3)`
B. `[OH^(-)]=((K_(c))/(K_(sp)))^(1//4)`
C. `[OH^(-)]=sqrt(((K_(sp))/(K_(c)))^(1//4))`
D. None of these

Answer

To solve the problem involving the equilibria of Aluminum hydroxide \((Al(OH)_3)\), we need to analyze the two given reactions and derive the relationship that indicates the condition for minimum solubility.

### Step-by-Step Solution:

1. **Understanding the Equilibria**:
- The first equilibrium reaction is:
\[ Al(OH)_3(s) \rightleftharpoons Al^{3+}(aq) + 3OH^{-}(aq) \quad (K_{sp}) \]
- The second equilibrium reaction is:
\[ Al(OH)_3(s) + OH^{-}(aq) \rightleftharpoons Al(OH)_4^{-}(aq) \quad (K_c) \]

2. **Writing the Expressions for \(K_{sp}\) and \(K_c\)**:
- For the first equilibrium, the solubility product constant \(K_{sp}\) can be expressed as:
\[ K_{sp} = [Al^{3+}][OH^{-}]^3 \]
- For the second equilibrium, the equilibrium constant \(K_c\) can be expressed as:
\[ K_c = \frac{[Al(OH)_4^{-}]}{[OH^{-}]} \]

3. **Combining the Reactions**:
- If we add the two reactions, we can derive a new relationship:
\[ 2Al(OH)_3(s) \rightleftharpoons Al^{3+}(aq) + Al(OH)_4^{-}(aq) \]
- This indicates that the dissolved Aluminum hydroxide is present as both \(Al^{3+}\) and \(Al(OH)_4^{-}\).

4. **Expressing Solubility**:
- The total solubility of \(Al(OH)_3\) can be expressed in terms of the concentrations of \(Al^{3+}\) and \(Al(OH)_4^{-}\):
\[ S = [Al^{3+}] + [Al(OH)_4^{-}] \]

5. **Finding the Condition for Minimum Solubility**:
- To find the condition for minimum solubility, we can differentiate the solubility with respect to the hydroxide ion concentration \([OH^{-}]\) and set the derivative to zero:
\[ \frac{dS}{d[OH^{-}]} = 0 \]
- This leads to the equation:
\[ -\frac{3K_{sp}}{[OH^{-}]^4} + K_c = 0 \]
- Rearranging gives:
\[ [OH^{-}]^4 = \frac{3K_{sp}}{K_c} \]
- Taking the fourth root:
\[ [OH^{-}] = \left(\frac{3K_{sp}}{K_c}\right)^{1/4} \]

6. **Conclusion**:
- The relationship for the hydroxide ion concentration at which the solubility of \(Al(OH)_3\) is minimum is:
\[ [OH^{-}] = \left(\frac{3K_{sp}}{K_c}\right)^{1/4} \]
- Since none of the provided options match this derived relationship, the correct answer is "none of these".