`K_(sp)` value of `Al(OH)_(3)` and `Zn(OH)_(2)` are `8.5 xx 10^(-23)` and `1.8 xx 10^(-14)` respectively. If `NH_(4)OH` is added in a solution of `Al^(3+)` and `Zn^(2+)`, which will precipitate earlier

Options

A. `Al(OH)_(3)`
B. `Zn(OH)_(2)`
C. Both together
D. None

Explanation

To determine which hydroxide precipitates first when ammonium hydroxide (NH₄OH) is added to a solution containing Al³⁺ and Zn²⁺ ions, we need to compare the solubility product constants (Ksp) of aluminum hydroxide (Al(OH)₃) and zinc hydroxide (Zn(OH)₂).

### Step-by-Step Solution:

1. **Write the Ksp expressions for both hydroxides:**
- For aluminum hydroxide (Al(OH)₃):
\[ K_{sp} = [Al^{3+}][OH^-]^3 \]
- For zinc hydroxide (Zn(OH)₂):
\[ K_{sp} = [Zn^{2+}][OH^-]^2 \]

2. **Substitute the given Ksp values:**
- \( K_{sp} \) of Al(OH)₃ = \( 8.5 \times 10^{-23} \)
- \( K_{sp} \) of Zn(OH)₂ = \( 1.8 \times 10^{-14} \)

3. **Calculate the concentration of OH⁻ required for precipitation:**
- For Al(OH)₃:
\[ [OH^-]^3 = \frac{K_{sp}}{[Al^{3+}]} \]
- For Zn(OH)₂:
\[ [OH^-]^2 = \frac{K_{sp}}{[Zn^{2+}]} \]

4. **Assume initial concentrations of Al³⁺ and Zn²⁺:**
- Let’s assume \( [Al^{3+}] = 1 \, M \) and \( [Zn^{2+}] = 1 \, M \) for simplicity in calculations.

5. **Calculate [OH⁻] for both precipitates:**
- For Al(OH)₃:
\[ [OH^-]^3 = \frac{8.5 \times 10^{-23}}{1} = 8.5 \times 10^{-23} \]
\[ [OH^-] = (8.5 \times 10^{-23})^{1/3} \approx 2.04 \times 10^{-8} \, M \]
- For Zn(OH)₂:
\[ [OH^-]^2 = \frac{1.8 \times 10^{-14}}{1} = 1.8 \times 10^{-14} \]
\[ [OH^-] = (1.8 \times 10^{-14})^{1/2} \approx 4.24 \times 10^{-7} \, M \]

6. **Compare the [OH⁻] concentrations:**
- Al(OH)₃ precipitates at \( [OH^-] \approx 2.04 \times 10^{-8} \, M \)
- Zn(OH)₂ precipitates at \( [OH^-] \approx 4.24 \times 10^{-7} \, M \)

7. **Conclusion:**
Since the concentration of OH⁻ required to precipitate Al(OH)₃ is much lower than that required for Zn(OH)₂, **Al(OH)₃ will precipitate first** when NH₄OH is added to the solution.