What is the molar solubility of `Al(OH)_(3)` in 0.2 M NaOH solution? Given that, solubility product of `Al(OH)_(3) = 2.4 xx 10^(-2)`

Explanation

To find the molar solubility of \( Al(OH)_3 \) in a 0.2 M NaOH solution, we can follow these steps:

### Step 1: Write the dissociation equation for \( Al(OH)_3 \)
The dissociation of aluminum hydroxide in water can be represented as:
\[ Al(OH)_3 (s) \rightleftharpoons Al^{3+} (aq) + 3 OH^{-} (aq) \]

### Step 2: Define the solubility product constant (\( K_{sp} \))
The solubility product constant for \( Al(OH)_3 \) is given by:
\[ K_{sp} = [Al^{3+}][OH^{-}]^3 \]
Given that \( K_{sp} = 2.4 \times 10^{-2} \).

### Step 3: Set up the expression for solubility in the presence of NaOH
Let \( S \) be the molar solubility of \( Al(OH)_3 \) in the 0.2 M NaOH solution. In this case, the concentration of \( OH^{-} \) ions from NaOH is 0.2 M. The dissociation of \( Al(OH)_3 \) will contribute additional \( OH^{-} \) ions:
\[ [OH^{-}] = 0.2 + 3S \]
However, since \( S \) is expected to be very small compared to 0.2, we can approximate:
\[ [OH^{-}] \approx 0.2 \]

### Step 4: Substitute into the \( K_{sp} \) expression
Now we can substitute the concentrations into the \( K_{sp} \) expression:
\[ K_{sp} = [Al^{3+}][OH^{-}]^3 = S(0.2)^3 \]
This simplifies to:
\[ K_{sp} = S \cdot 0.008 \]
Where \( (0.2)^3 = 0.008 \).

### Step 5: Solve for \( S \)
Now we can set this equal to the \( K_{sp} \):
\[ 2.4 \times 10^{-2} = S \cdot 0.008 \]
To find \( S \), we rearrange the equation:
\[ S = \frac{2.4 \times 10^{-2}}{0.008} \]
Calculating this gives:
\[ S = 3.0 \, M \]

### Final Answer
The molar solubility of \( Al(OH)_3 \) in a 0.2 M NaOH solution is \( 3.0 \, M \).

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