A sample of water from a river was analyzed for the presence of metal ions and the observations were recorded as given below
`{:("Reagent added","Observation"),("dil. HCl","No change"),("aq. " Na_(2)CO_(3),"White precipitate"),("Aq. " Na_(2)SO_(4),"No change"):}`
The water sample is likely to contain

Options

A. `Ba^(2+)`
B. `Cu^(2+)`
C. `Li^(+)`
D. `Mg^(2+)`

Answer

To solve the problem, we need to analyze the observations given for the reagents added to the water sample and determine which metal ion is likely present.

### Step-by-Step Solution:

1. **Analyze the first observation (dil. HCl)**:
- When dilute HCl is added, there is **no change** observed. This indicates that the metal ion present does not form a precipitate or a colored solution with HCl.
- Possible metal ions that could fit this observation include those that form soluble chlorides.

2. **Analyze the second observation (aq. Na2CO3)**:
- When aqueous sodium carbonate (Na2CO3) is added, a **white precipitate** is formed. This suggests that the metal ion reacts with carbonate ions to form an insoluble carbonate.
- Common metal ions that form white precipitates with carbonate include Ba²⁺, Mg²⁺, and Ca²⁺.

3. **Analyze the third observation (aq. Na2SO4)**:
- When aqueous sodium sulfate (Na2SO4) is added, there is **no change** observed. This indicates that the metal ion does not form a precipitate or colored solution with sulfate ions.
- Metal ions that typically do not form precipitates with sulfate include Mg²⁺ and Na⁺.

4. **Evaluate the possible metal ions**:
- **Ba²⁺**: Would react with HCl to form BaCl2, which is soluble, but would also form a white precipitate with Na2CO3. However, it would also react with Na2SO4 to form a precipitate (BaSO4), contradicting the observation.
- **Cu²⁺**: Would form a blue solution with Na2SO4, which contradicts the observation of no change.
- **Li⁺**: Would not produce a precipitate with Na2CO3 and would not fit the observations.
- **Mg²⁺**: Would remain soluble with HCl (no change), would form a white precipitate with Na2CO3 (MgCO3), and would remain soluble with Na2SO4 (MgSO4).

5. **Conclusion**:
- The only metal ion that fits all the observations is **Mg²⁺**. Therefore, the water sample is likely to contain **Mg²⁺**.

### Final Answer:
The water sample is likely to contain **Mg²⁺**.

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