The solubility products of `Al(OH)_(3)` and `Zn(OH)_(2)` are `8.5xx10^(-23)` and `1.8xx10^(-14)` respectively. If `NH_(4)OH` is added to a solution containing `Al^(3+)` and `Zn^(2+)` ions, then substance precipitated first is:

Options

A. `Al(OH)_(3)`
B. `Zn(OH)_(2)`
C. Both (a) and (b)
D. None of these

Explanation

To determine which substance precipitates first when `NH4OH` is added to a solution containing `Al^(3+)` and `Zn^(2+)` ions, we need to compare the solubility products (Ksp) of `Al(OH)3` and `Zn(OH)2`. The substance with the lower solubility product will precipitate first.

### Step-by-Step Solution:

1. **Identify the Ksp values**:
- Ksp of `Al(OH)3` = `8.5 × 10^(-23)`
- Ksp of `Zn(OH)2` = `1.8 × 10^(-14)`

2. **Write the dissociation equations**:
- For `Al(OH)3`:
\[ Al^{3+} + 3OH^{-} \rightleftharpoons Al(OH)3 \quad (Ksp = [Al^{3+}][OH^{-}]^3) \]
- For `Zn(OH)2`:
\[ Zn^{2+} + 2OH^{-} \rightleftharpoons Zn(OH)2 \quad (Ksp = [Zn^{2+}][OH^{-}]^2) \]

3. **Calculate the solubility (S) for both hydroxides**:
- For `Al(OH)3`:
\[ Ksp = [Al^{3+}][OH^{-}]^3 = S \cdot (3S)^3 = S \cdot 27S^3 = 27S^4 \]
\[ 8.5 × 10^{-23} = 27S^4 \implies S^4 = \frac{8.5 × 10^{-23}}{27} \implies S = \left(\frac{8.5 × 10^{-23}}{27}\right)^{1/4} \]
\[ S \approx 1.332 × 10^{-6} \text{ M} \]

- For `Zn(OH)2`:
\[ Ksp = [Zn^{2+}][OH^{-}]^2 = S \cdot (2S)^2 = S \cdot 4S^2 = 4S^3 \]
\[ 1.8 × 10^{-14} = 4S^3 \implies S^3 = \frac{1.8 × 10^{-14}}{4} \implies S = \left(\frac{1.8 × 10^{-14}}{4}\right)^{1/3} \]
\[ S \approx 1.65 × 10^{-5} \text{ M} \]

4. **Compare the solubility values**:
- Solubility of `Al(OH)3` = `1.332 × 10^{-6} M`
- Solubility of `Zn(OH)2` = `1.65 × 10^{-5} M`

5. **Determine which precipitates first**:
- Since `Al(OH)3` has a lower solubility than `Zn(OH)2`, it will precipitate first when `NH4OH` is added to the solution.

### Conclusion:
The substance that precipitates first is **Al(OH)3** (aluminum hydroxide).