A yellow precipitate is obtained when:

Options

A. lead acetate solution is treated with `K_(2)CrO_(4)`
B. `Pb(NO_(3))_(2)` solutino is treated with `K_(2)CrO_(4)`
C. `AgNO_(3)` solution treated with KI
D. `H_(2)S` is passed thorugh a solution of `CdSO_(4)`

Answer

To determine when a yellow precipitate is obtained, we can analyze the reactions provided in the options. Here’s a step-by-step breakdown of each reaction:

### Step 1: Analyzing Reaction A
**Reaction:** Lead acetate solution is treated with potassium chromate (K2CrO4).

- **Chemical Equation:**
\[ \text{Pb(CH}_3\text{COO)}_2 + \text{K}_2\text{CrO}_4 \rightarrow \text{PbCrO}_4 \downarrow + 2 \text{KCH}_3\text{COO} \]
- **Observation:** Lead(II) chromate (PbCrO4) is formed as a yellow precipitate.

### Step 2: Analyzing Reaction B
**Reaction:** Lead(II) nitrate solution is treated with potassium chromate (K2CrO4).

- **Chemical Equation:**
\[ \text{Pb(NO}_3\text{)}_2 + \text{K}_2\text{CrO}_4 \rightarrow \text{PbCrO}_4 \downarrow + 2 \text{KNO}_3 \]
- **Observation:** Again, lead(II) chromate (PbCrO4) precipitates as a yellow solid.

### Step 3: Analyzing Reaction C
**Reaction:** Silver nitrate solution is treated with potassium iodide (KI).

- **Chemical Equation:**
\[ \text{AgNO}_3 + \text{KI} \rightarrow \text{AgI} \downarrow + \text{KNO}_3 \]
- **Observation:** Silver iodide (AgI) is formed, which is also a yellow precipitate.

### Step 4: Analyzing Reaction D
**Reaction:** Hydrogen sulfide (H2S) is passed through a solution of cadmium sulfate (CdSO4).

- **Chemical Equation:**
\[ \text{CdSO}_4 + \text{H}_2\text{S} \rightarrow \text{CdS} \downarrow + \text{H}_2\text{SO}_4 \]
- **Observation:** Cadmium sulfide (CdS) precipitates as a yellow solid.

### Conclusion
In all four reactions, a yellow precipitate is formed:
- **Reaction A:** Lead chromate (PbCrO4)
- **Reaction B:** Lead chromate (PbCrO4)
- **Reaction C:** Silver iodide (AgI)
- **Reaction D:** Cadmium sulfide (CdS)

Thus, the answer is that a yellow precipitate is obtained in all four reactions.