An aqueous solution contains `Ni^(2+), Co^(2+), Pb^(2+)` ions at equal concentrations. The solubility product of NiS, PbS and CoS in water at `25^@C` are respectively given below. Indicate which of these ions will be precipitated first and last when sulphide concentration is progressively increased from zero ?`[K_(sp) of NiS=3xx10^(-19), K_(sp) of CoS=4xx10^(-21), K_(sp) of PbS=3xx10^(-28)`

Options

A. NiS and PbS
B. NiS and CoS
C. CoS and NiS
D. PbS and NiS

Explanation

To determine which of the ions \( Ni^{2+}, Co^{2+}, Pb^{2+} \) will precipitate first and last when the concentration of sulfide ions (\( S^{2-} \)) is progressively increased from zero, we can follow these steps:

### Step 1: Write the solubility product expressions
The solubility product (\( K_{sp} \)) for each sulfide can be expressed as follows:

- For \( NiS \):
\[ K_{sp} = [Ni^{2+}][S^{2-}] \]

- For \( CoS \):
\[ K_{sp} = [Co^{2+}][S^{2-}] \]

- For \( PbS \):
\[ K_{sp} = [Pb^{2+}][S^{2-}] \]

### Step 2: Rearrange to find the sulfide ion concentration for precipitation
To find the concentration of sulfide ions required for precipitation, we can rearrange the \( K_{sp} \) expressions:

- For \( NiS \):
\[ [S^{2-}] = \frac{K_{sp}}{[Ni^{2+}]} \]

- For \( CoS \):
\[ [S^{2-}] = \frac{K_{sp}}{[Co^{2+}]} \]

- For \( PbS \):
\[ [S^{2-}] = \frac{K_{sp}}{[Pb^{2+}]} \]

### Step 3: Substitute the given \( K_{sp} \) values
Given:
- \( K_{sp} \) of \( NiS = 3 \times 10^{-19} \)
- \( K_{sp} \) of \( CoS = 4 \times 10^{-21} \)
- \( K_{sp} \) of \( PbS = 3 \times 10^{-28} \)

Assuming equal concentrations for \( Ni^{2+}, Co^{2+}, Pb^{2+} \) (let's denote this concentration as \( C \)), we can calculate:

- For \( NiS \):
\[ [S^{2-}]_{Ni} = \frac{3 \times 10^{-19}}{C} \]

- For \( CoS \):
\[ [S^{2-}]_{Co} = \frac{4 \times 10^{-21}}{C} \]

- For \( PbS \):
\[ [S^{2-}]_{Pb} = \frac{3 \times 10^{-28}}{C} \]

### Step 4: Compare the sulfide concentrations
Since \( C \) is the same for all three ions, we can directly compare the \( K_{sp} \) values to determine which sulfide will precipitate first:

1. \( PbS \) has the lowest \( K_{sp} \) ( \( 3 \times 10^{-28} \) ), so it will precipitate first.
2. \( CoS \) has a higher \( K_{sp} \) ( \( 4 \times 10^{-21} \) ), so it will precipitate second.
3. \( NiS \) has the highest \( K_{sp} \) ( \( 3 \times 10^{-19} \) ), so it will precipitate last.

### Conclusion
- **First to precipitate:** \( PbS \)
- **Last to precipitate:** \( NiS \)