A solution which is `10^(-3)M` each in `Mn^(2+), Fe^(2+), Zn^(2+)`, and `Hg^(2+)` it treated with `10^(-16)M` sulphide ion. If the `K_(sp)` of `MnS, FeS, ZnS`and `HgS` are `10^(-15), 10^(-23), 10^(-20)`,and `10^(-54)`, respectively, which one will precipitate first?

Options

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

Explanation

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 \(10^{-16} M\) sulfide ions, we need to compare the solubility product constants (\(K_{sp}\)) of the respective sulfides.

### Step-by-Step Solution:

1. **Identify the ions and their respective \(K_{sp}\) values:**
- \(MnS\): \(K_{sp} = 10^{-15}\)
- \(FeS\): \(K_{sp} = 10^{-23}\)
- \(ZnS\): \(K_{sp} = 10^{-20}\)
- \(HgS\): \(K_{sp} = 10^{-54}\)

2. **Understand the relationship between \(K_{sp}\) and precipitation:**
- The \(K_{sp}\) value indicates the solubility of the compound in water. A lower \(K_{sp}\) value means the compound is less soluble and will precipitate more readily when the ions are present in solution.

3. **Calculate the ion product (\(Q\)) for each sulfide:**
- The ion product \(Q\) is calculated as follows:
\[ Q = [M^{2+}][S^{2-}] \]
- Given that \([M^{2+}] = 10^{-3} M\) and \([S^{2-}] = 10^{-16} M\), we can calculate \(Q\) for each sulfide:
- 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 \(K_{sp}\) for each sulfide:**
- For \(MnS\): \(Q = 10^{-19}\) vs \(K_{sp} = 10^{-15}\) (not precipitating)
- For \(FeS\): \(Q = 10^{-19}\) vs \(K_{sp} = 10^{-23}\) (precipitating)
- For \(ZnS\): \(Q = 10^{-19}\) vs \(K_{sp} = 10^{-20}\) (precipitating)
- For \(HgS\): \(Q = 10^{-19}\) vs \(K_{sp} = 10^{-54}\) (precipitating)

5. **Determine which will precipitate first:**
- The precipitation occurs when \(Q\) exceeds \(K_{sp}\).
- Among the sulfides, \(HgS\) has the lowest \(K_{sp}\) value, indicating it is the least soluble and will precipitate first.

### Conclusion:
The sulfide that will precipitate first is \(HgS\).