A solution which is `10^(-3)` M each in `Mn^(2+),Fe^(2+),Zn^(2+)andHg^(2+)` is treated with `10^(-16)M` sulphide ion. If `K_(sp)` od MnS, ZnS and HgS are `10^(-15),10^(-25),10^(-20)and10^(-54)` respectively, which one will precipitate first ?

Options

A. FeS
B. MnS
C. HgS
D. ZnS

Explanation

To solve the problem, we need to determine which sulfide will precipitate first when a solution containing \(10^{-3} M\) of \(Mn^{2+}, Fe^{2+}, Zn^{2+},\) and \(Hg^{2+}\) ions is treated with \(10^{-16} M\) of sulfide ions (\(S^{2-}\)). We will use the solubility product constant (\(K_{sp}\)) values provided for each sulfide to find out which will precipitate first.

### Step-by-Step Solution:

1. **Identify the relevant ions and their \(K_{sp}\) values**:
- \(K_{sp}\) of \(MnS = 10^{-15}\)
- \(K_{sp}\) of \(FeS = 10^{-23}\)
- \(K_{sp}\) of \(ZnS = 10^{-25}\)
- \(K_{sp}\) of \(HgS = 10^{-54}\)

2. **Determine the concentration of sulfide ions**:
- The concentration of sulfide ions is given as \(10^{-16} M\).

3. **Calculate the ion product (\(Q\)) for each sulfide**:
- The ion product \(Q\) for a precipitate can be calculated using the formula:
\[ Q = [M^{2+}][S^{2-}] \]
- For each metal ion, we will substitute \( [M^{2+}] = 10^{-3} M \) and \( [S^{2-}] = 10^{-16} M \).

- For \(MnS\):
\[ Q_{MnS} = (10^{-3})(10^{-16}) = 10^{-19} \]

- For \(FeS\):
\[ Q_{FeS} = (10^{-3})(10^{-16}) = 10^{-19} \]

- For \(ZnS\):
\[ Q_{ZnS} = (10^{-3})(10^{-16}) = 10^{-19} \]

- For \(HgS\):
\[ Q_{HgS} = (10^{-3})(10^{-16}) = 10^{-19} \]

4. **Compare \(Q\) with \(K_{sp}\)**:
- A precipitate will form when \(Q\) exceeds \(K_{sp}\).
- For \(MnS\), \(Q = 10^{-19}\) and \(K_{sp} = 10^{-15}\) (no precipitate).
- For \(FeS\), \(Q = 10^{-19}\) and \(K_{sp} = 10^{-23}\) (no precipitate).
- For \(ZnS\), \(Q = 10^{-19}\) and \(K_{sp} = 10^{-25}\) (no precipitate).
- For \(HgS\), \(Q = 10^{-19}\) and \(K_{sp} = 10^{-54}\) (precipitate will form).

5. **Conclusion**:
- Since \(HgS\) has the lowest \(K_{sp}\), it will precipitate first when the sulfide ions are added to the solution.

### Final Answer:
**The sulfide that will precipitate first is \(HgS\).**