A solution containing `NH_(4)CI` and `NH_(4)OH` has `[overset(Theta)OH] = 10^(-6) molL^(-1)`, which of the following hydroxides would be precipitated when this solution in added in equal volume to a solution containing `0.1M` of metal ions?

Options

A. `Mg(OH)_(2),(K_(sp)=3xx10^(-11))`
B. `Fe(OH)_(2)(K_(sp)=8xx10^(-16))`
C. `Cd(OH)_(2)(K_(sp)=8xx10^(-6))`
D. `AgOH(K_(sp)=5xx10^(-3))`

Explanation

To solve the problem, we need to determine which hydroxides will precipitate when a solution containing \( NH_4Cl \) and \( NH_4OH \) is mixed with a solution of metal ions at a concentration of \( 0.1 \, M \). The concentration of hydroxide ions \( [OH^-] \) in the initial solution is given as \( 10^{-6} \, mol \, L^{-1} \).

### Step-by-Step Solution:

1. **Identify Initial Concentrations**:
- The concentration of hydroxide ions \( [OH^-] = 10^{-6} \, mol \, L^{-1} \).
- The concentration of metal ions \( [M^{n+}] = 0.1 \, mol \, L^{-1} \).

2. **Calculate New Concentrations After Mixing**:
- When the two solutions are mixed in equal volumes, the concentrations will be halved.
- New concentration of \( [OH^-] = \frac{10^{-6}}{2} = 5 \times 10^{-7} \, mol \, L^{-1} \).
- New concentration of \( [M^{n+}] = \frac{0.1}{2} = 0.05 \, mol \, L^{-1} \).

3. **Determine Ionic Product (IP) for Each Hydroxide**:
- The ionic product for a metal hydroxide \( M(OH)_2 \) is given by:
\[ IP = [M^{n+}][OH^-]^2 \]
- For \( M(OH)_2 \) (e.g., \( Mg(OH)_2 \)):
\[ IP_{Mg(OH)_2} = [0.05][(5 \times 10^{-7})^2] = 0.05 \times 25 \times 10^{-14} = 1.25 \times 10^{-15} \]

4. **Compare with Ksp Values**:
- Given \( Ksp \) for \( Mg(OH)_2 = 3 \times 10^{-11} \).
- Since \( IP < Ksp \), no precipitation occurs for \( Mg(OH)_2 \).

5. **Repeat for Other Hydroxides**:
- For \( Fe(OH)_2 \):
\[ IP_{Fe(OH)_2} = [0.05][(5 \times 10^{-7})^2] = 1.25 \times 10^{-15} \]
- Given \( Ksp \) for \( Fe(OH)_2 = 8 \times 10^{-16} \).
- Since \( IP > Ksp \), precipitation occurs for \( Fe(OH)_2 \).

- For \( Cd(OH)_2 \):
\[ IP_{Cd(OH)_2} = [0.05][(5 \times 10^{-7})^2] = 1.25 \times 10^{-15} \]
- Given \( Ksp \) for \( Cd(OH)_2 = 8 \times 10^{-6} \).
- Since \( IP < Ksp \), no precipitation occurs for \( Cd(OH)_2 \).

- For \( AgOH \):
\[ IP_{AgOH} = [0.05][(5 \times 10^{-7})] = 0.05 \times 5 \times 10^{-7} = 2.5 \times 10^{-8} \]
- Given \( Ksp \) for \( AgOH = 5 \times 10^{-3} \).
- Since \( IP < Ksp \), no precipitation occurs for \( AgOH \).

6. **Conclusion**:
- The only hydroxide that will precipitate is \( Fe(OH)_2 \).

### Final Answer:
The hydroxide that will precipitate when the solution is mixed is \( Fe(OH)_2 \).