A solution which is `10^(-3)` M each in `Mn^(2+),Fe^(2+),Zn^(2+) and Hg^(2+)` is treated with `10^(-16)` M sulphide ion. If `K_(sp) "of " MnS , FeS, ZnS and HgS " are " 10^(-13),10^(-18),10^(-24) and 10^(-53)` respectively. Which one will precipitate first ?

Options

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

Answer

To determine which sulfide will precipitate first from a solution containing \(10^{-3} \, M\) of \(Mn^{2+}\), \(Fe^{2+}\), \(Zn^{2+}\), and \(Hg^{2+}\) when treated with \(10^{-16} \, M\) of sulfide ions, we will follow these steps:

### Step 1: Calculate the Ionic Product (IP)
The ionic product for each metal sulfide can be calculated using the formula:

\[ IP = [M^{2+}][S^{2-}] \]

Given that the concentration of each metal ion \([M^{2+}]\) is \(10^{-3} \, M\) and the concentration of sulfide ion \([S^{2-}]\) is \(10^{-16} \, M\), we can substitute these values into the equation:

\[ IP = (10^{-3})(10^{-16}) = 10^{-19} \]

### Step 2: Compare the Ionic Product with the Solubility Product (Ksp)
Next, we will compare the calculated ionic product with the solubility product (\(K_{sp}\)) values of the sulfides:

- \(K_{sp} \, of \, MnS = 10^{-13}\)
- \(K_{sp} \, of \, FeS = 10^{-18}\)
- \(K_{sp} \, of \, ZnS = 10^{-24}\)
- \(K_{sp} \, of \, HgS = 10^{-53}\)

### Step 3: Determine Precipitation Conditions
A precipitate will form when the ionic product exceeds the solubility product (\(IP > K_{sp}\)). We will check this condition for each sulfide:

1. **MnS**:
- \(IP = 10^{-19}\)
- \(K_{sp} = 10^{-13}\)
- \(10^{-19} < 10^{-13}\) (No precipitation)

2. **FeS**:
- \(IP = 10^{-19}\)
- \(K_{sp} = 10^{-18}\)
- \(10^{-19} < 10^{-18}\) (No precipitation)

3. **ZnS**:
- \(IP = 10^{-19}\)
- \(K_{sp} = 10^{-24}\)
- \(10^{-19} > 10^{-24}\) (Precipitation occurs)

4. **HgS**:
- \(IP = 10^{-19}\)
- \(K_{sp} = 10^{-53}\)
- \(10^{-19} > 10^{-53}\) (Precipitation occurs)

### Step 4: Determine Which Precipitates First
Among the sulfides that precipitate (ZnS and HgS), we need to determine which one will precipitate first. The one with the lower \(K_{sp}\) value will precipitate first:

- \(K_{sp} \, of \, ZnS = 10^{-24}\)
- \(K_{sp} \, of \, HgS = 10^{-53}\)

Since \(10^{-53} < 10^{-24}\), \(HgS\) will precipitate first.

### Conclusion
The sulfide that will precipitate first is **HgS**.