A chemist opened a cupboard to find four bottles containing water solutions, each of which has lost its label. Bottles 1,2,3 contained colourless solutions, Whilst bottle 4 contained a blue solution. The labels from the bottles were lying scattered ont he floor of the cupboard.
They were
copper (II) sulphate
sodium carbonate
lead nitrate
hydrochloric acid
By mixing sammples of teh contents of the bottles, in pairs, the chemist made the following observations:
(i) Bottle 1 + Bottle 2 white precipitate
(ii) Bottle 1 + bottle 3 white precipitate
(iii) Bottle 1 + Bottle 4 white precipitate
(iv) Bottle 2 +Bottle 3 colourless gas evolved
(v) Bottle 2+ Bottle 4 no visible reaction
(vi) bottle 3 +bottle 4 blue precipitate
Q. Chemical formula of white ppt. formed in observation (iii) is:

Options

A. `PbCl_(2)`
B. `PbCO_(3)`
C. `CuCO_(3)`
D. `PbSO_(4)`

Explanation

To solve the problem, we need to analyze the observations made by the chemist when mixing the solutions from the bottles. Let's go through the steps systematically:

### Step 1: Identify the contents of each bottle based on the observations.

1. **Observation (i)**: Bottle 1 + Bottle 2 → White precipitate
2. **Observation (ii)**: Bottle 1 + Bottle 3 → White precipitate
3. **Observation (iii)**: Bottle 1 + Bottle 4 → White precipitate
4. **Observation (iv)**: Bottle 2 + Bottle 3 → Colourless gas evolved
5. **Observation (v)**: Bottle 2 + Bottle 4 → No visible reaction
6. **Observation (vi)**: Bottle 3 + Bottle 4 → Blue precipitate

### Step 2: Analyze the observations.

- **Observation (iv)** indicates that when Bottle 2 (likely hydrochloric acid) is mixed with Bottle 3 (likely sodium carbonate), a colorless gas (carbon dioxide) is evolved. This suggests that Bottle 2 is hydrochloric acid (HCl) and Bottle 3 is sodium carbonate (Na2CO3).

- **Observation (v)** shows that when Bottle 2 (HCl) is mixed with Bottle 4 (which must be copper(II) sulfate, CuSO4), there is no visible reaction. This suggests that Bottle 4 is not a strong base or does not react with HCl.

- **Observation (vi)** indicates that when Bottle 3 (sodium carbonate) is mixed with Bottle 4 (copper(II) sulfate), a blue precipitate forms. This suggests that Bottle 4 is indeed copper(II) sulfate (CuSO4), as it forms a blue precipitate with sodium carbonate.

### Step 3: Determine the contents of Bottle 1.

- Since Bottle 1 reacts with both Bottle 2 and Bottle 4 to form white precipitates, and considering that Bottle 2 is HCl and Bottle 4 is CuSO4, we can deduce that Bottle 1 must be lead nitrate (Pb(NO3)2).

### Step 4: Identify the white precipitate formed in Observation (iii).

- In Observation (iii), when Bottle 1 (lead nitrate) is mixed with Bottle 4 (copper(II) sulfate), the reaction can be written as:

\[ \text{Pb(NO}_3\text{)}_2 + \text{CuSO}_4 \rightarrow \text{PbSO}_4 \text{ (white precipitate)} + \text{Cu(NO}_3\text{)}_2 \]

- The white precipitate formed in this reaction is lead sulfate (PbSO4).

### Conclusion

The chemical formula of the white precipitate formed in Observation (iii) is **PbSO4** (lead sulfate).

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