Assertion (A): A solution contains `0.1M` each of `pB^(2+), Zn^(2+),Ni^(2+)`, ions. If `H_(2)S` is passed into this solution at `25^(@)C`.
`Pb^(2+), Ni^(2+), Zn^(2+)` will get precpitated simultanously.
Reason (R): `Pb^(2+)` and `Zn^(2+)` will get precipitated if the solution contains `0.1M HCI`.
`[K_(1) H_(2)S = 10^(-7), K_(2)H_(2)S = 10^(-14), K_(sp) PbS =3xx 10^(-29) K_(sp) NiS = 3 xx 10^(-19). K_(sp) ZnS = 10^(-25)]`

Options

A. If both (A) and (R) are correc, and (R) is the correct explanation of (A).
B. If both (A) and (R) are correct but (R) is not the correct explanation of (A).
C. If (A) is correct, but (R) is incorrect.
D. If (A) is incorrect, but (R) is correct.

Explanation

To solve the problem, we will analyze the assertion and reason provided, and calculate the solubility product (Ksp) values to determine if the ions will precipitate when H2S is passed into the solution.

### Step 1: Understand the Assertion
The assertion states that a solution contains `0.1 M` each of `Pb^(2+)`, `Zn^(2+)`, and `Ni^(2+)` ions. When `H2S` is passed into this solution at `25°C`, all three ions will precipitate simultaneously.

### Step 2: Analyze the Reason
The reason states that `Pb^(2+)` and `Zn^(2+)` will precipitate if the solution contains `0.1 M HCl`. We need to check if this is true and how it relates to the assertion.

### Step 3: Calculate the Concentration of Sulfide Ion (S^(2-))
The ionization of `H2S` can be represented as:
\[ H2S \rightleftharpoons 2H^+ + S^{2-} \]

The equilibrium constant for this reaction can be calculated as:
\[ K = \frac{[H^+]^2 [S^{2-}]}{[H2S]} \]

Given:
- \( K_1 = 10^{-7} \) (for the first dissociation)
- \( K_2 = 10^{-14} \) (for the second dissociation)

The overall equilibrium constant \( K \) for the dissociation of `H2S` is:
\[ K = K_1 \times K_2 = 10^{-7} \times 10^{-14} = 10^{-21} \]

### Step 4: Calculate [S^(2-)] Concentration
Assuming the concentration of `H2S` is approximately equal to the concentration of `H^+`, we can set:
\[ [H^+] \approx 10^{-7} \, M \]

Substituting into the equilibrium expression:
\[ 10^{-21} = (10^{-7})^2 \times [S^{2-}] \]
\[ [S^{2-}] = \frac{10^{-21}}{10^{-14}} = 10^{-40} \, M \]

### Step 5: Calculate Qsp for Each Salt
Now we will calculate the reaction quotient \( Q_{sp} \) for each sulfide:

1. **For PbS**:
\[ Q_{sp} = [Pb^{2+}][S^{2-}] = (0.1)(10^{-40}) = 10^{-50} \]
- \( K_{sp} \) for PbS = \( 3 \times 10^{-29} \)
- Since \( Q_{sp} < K_{sp} \), no precipitation occurs.

2. **For ZnS**:
\[ Q_{sp} = [Zn^{2+}][S^{2-}] = (0.1)(10^{-40}) = 10^{-50} \]
- \( K_{sp} \) for ZnS = \( 10^{-25} \)
- Since \( Q_{sp} < K_{sp} \), no precipitation occurs.

3. **For NiS**:
\[ Q_{sp} = [Ni^{2+}][S^{2-}] = (0.1)(10^{-40}) = 10^{-50} \]
- \( K_{sp} \) for NiS = \( 3 \times 10^{-19} \)
- Since \( Q_{sp} < K_{sp} \), no precipitation occurs.

### Step 6: Conclusion
From the calculations, we see that none of the ions precipitate when `H2S` is passed into the solution. Therefore, the assertion is incorrect.

### Step 7: Evaluate the Reason
The reason states that `Pb^(2+)` and `Zn^(2+)` will precipitate if the solution contains `0.1 M HCl`. However, we found that under the conditions stated, they do not precipitate. Thus, the reason is also incorrect.

### Final Answer
- Assertion (A) is incorrect.
- Reason (R) is incorrect.
- The correct option is D: A is incorrect and R is correct.