A 1 litre solution containing `NH_(4)Cl` and `NH_(4)OH` has hydroxide ion ion concentration of `10^-6)` mol//litre. Which of the following hydroxides could be precipitated when the solution is added to 1 litre solution of 0.1 M metal ions?
(I) `Ba(OH)_(2)(K_(sp)=5xx10^(-3))` , (II) `Ni(OH)_(2)(K_(sp)=1.6xx10^(-16))`
(III) `Mn(OH)_(2) (K_(sp)=2xx10^(-13))` , (IV) `Fe(OH)_(2) (K_(sp)=8xx10^(-16))`

Options

A. I,II,IV
B. IV
C. II and IV
D. II,III,IV

Answer

To solve the problem, we need to determine which of the given metal hydroxides can precipitate when a solution containing NH₄Cl and NH₄OH is mixed with a 0.1 M solution of metal ions, given that the hydroxide ion concentration is \(10^{-6}\) mol/L.

### Step-by-Step Solution:

1. **Understanding the Solution**:
The solution contains NH₄Cl and NH₄OH, which together create a buffer solution. The concentration of hydroxide ions \([OH^-]\) is given as \(10^{-6}\) mol/L.

2. **Mixing Solutions**:
When we mix 1 L of the buffer solution with 1 L of a 0.1 M metal ion solution, the concentration of metal ions will be halved because the total volume becomes 2 L. Therefore, the concentration of metal ions \([M^{2+}]\) after mixing will be:
\[ [M^{2+}] = \frac{0.1 \, \text{mol/L}}{2} = 0.05 \, \text{mol/L} \]

3. **Precipitation Condition**:
For a metal hydroxide \(M(OH)_2\) to precipitate, the reaction quotient \(Q\) must be greater than the solubility product constant \(K_{sp}\). The expression for \(Q\) is:
\[ Q = [M^{2+}][OH^-]^2 \]

4. **Calculating \(Q\)**:
Substituting the values we have:
\[ Q = (0.05) \times (10^{-6})^2 = 0.05 \times 10^{-12} = 5 \times 10^{-14} \]

5. **Comparing \(Q\) with \(K_{sp}\)**:
We need to compare \(Q\) with the \(K_{sp}\) values for the given metal hydroxides:
- (I) \(Ba(OH)_2\) \(K_{sp} = 5 \times 10^{-3}\)
- (II) \(Ni(OH)_2\) \(K_{sp} = 1.6 \times 10^{-16}\)
- (III) \(Mn(OH)_2\) \(K_{sp} = 2 \times 10^{-13}\)
- (IV) \(Fe(OH)_2\) \(K_{sp} = 8 \times 10^{-16}\)

6. **Determining Precipitation**:
- For \(Ba(OH)_2\): \(Q = 5 \times 10^{-14} < K_{sp} = 5 \times 10^{-3}\) (No precipitation)
- For \(Ni(OH)_2\): \(Q = 5 \times 10^{-14} > K_{sp} = 1.6 \times 10^{-16}\) (Precipitation occurs)
- For \(Mn(OH)_2\): \(Q = 5 \times 10^{-14} < K_{sp} = 2 \times 10^{-13}\) (No precipitation)
- For \(Fe(OH)_2\): \(Q = 5 \times 10^{-14} > K_{sp} = 8 \times 10^{-16}\) (Precipitation occurs)

7. **Final Answer**:
The hydroxides that could be precipitated are \(Ni(OH)_2\) and \(Fe(OH)_2\). Therefore, the correct options are (II) and (IV).