In equalitative analysis, cations of graph `II` as well as group `IV` both are precipitated in the form of sulphides. Due to low value of `K_(sp)` of group `II` sulphides, group reagent is `H_(2)S` in the presence of dil. `HC1`, and due to high value of `K_(sp)` of group `IV` sulphides, group reagent is `H_(2)S` in the presence of `NH_(4)OH` and `NH_(4)C1`. In a solution containing `0.1M` each of `Sn^(2+), Cd^(2+)`, and `Ni^(2+)` ions, `H_(2)S`gas is passed.
`K_(sp) of SnS = 8 xx 10^(-29), K_(sp) of CdS = 10^(-28), K_(sp) of NiS - 3 xx 10^(-21)`
`K_(1) of H_(2)S = 1 xx 10^(-7), K_(2) of H_(2)S = 1 xx 10^(-14)`
If `0.1M HC1` is mixed in the solution containing only `0.1 M Cd^(2+)` ions and saturated with `H_(2)S`, then `[Cd^(2+)]` remaining in the solution after `CdS` stopes to precipitate is:

Options

A. `10^(-8)`
B. `8.2 xx 10^(-9)`
C. `5.6 xx 10^(-6)`
D. `5.6 xx 10^(-10)`

Explanation

To solve the problem step by step, we will follow the principles of solubility product (Ksp) and the equilibrium of the sulfide ions in the presence of H2S.

### Step 1: Write the Ksp expression for CdS
The solubility product (Ksp) for cadmium sulfide (CdS) can be expressed as:

\[ K_{sp} = [Cd^{2+}][S^{2-}] \]

Given that \( K_{sp} \) for CdS is \( 1 \times 10^{-28} \).

### Step 2: Determine the concentration of sulfide ions \([S^{2-}]\)
To find the concentration of sulfide ions, we need to consider the dissociation of H2S in water:

\[ H_2S \rightleftharpoons H^+ + HS^- \]
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]

The equilibrium constants for these reactions are given as:
- \( K_1 = [H^+][HS^-]/[H_2S] = 1 \times 10^{-7} \)
- \( K_2 = [H^+][S^{2-}]/[HS^-] = 1 \times 10^{-14} \)

### Step 3: Set up the equilibrium expressions
Assuming that the concentration of H2S is large enough that it does not change significantly, we can express the concentration of sulfide ions \([S^{2-}]\) in terms of the concentration of \(H^+\):

From the second dissociation of H2S, we can write:

\[ K_2 = \frac{[H^+][S^{2-}]}{[HS^-]} \]

We can also express \([HS^-]\) in terms of \([H^+]\) using \(K_1\):

\[ [HS^-] = \frac{K_1[H_2S]}{[H^+]} \]

Substituting this into the expression for \(K_2\):

\[ K_2 = \frac{[H^+][S^{2-}]}{\frac{K_1[H_2S]}{[H^+]}} \]

This simplifies to:

\[ K_2 = \frac{[H^+]^2[S^{2-}]}{K_1[H_2S]} \]

### Step 4: Calculate \([S^{2-}]\) in the presence of HCl
When \(0.1M\) HCl is added, it will provide a significant concentration of \(H^+\) ions, which will shift the equilibrium to the left, reducing the concentration of sulfide ions.

Assuming \( [H^+] = 0.1M \):

Substituting into the equation for \(K_2\):

\[ 1 \times 10^{-14} = \frac{(0.1)^2[S^{2-}]}{1 \times 10^{-7}[H_2S]} \]

Solving for \([S^{2-}]\):

\[ [S^{2-}] = \frac{K_2 \cdot K_1 \cdot [H_2S]}{(0.1)^2} \]

### Step 5: Calculate \([Cd^{2+}]\) remaining after precipitation
Now, substituting \([S^{2-}]\) back into the Ksp expression for CdS:

\[ K_{sp} = [Cd^{2+}][S^{2-}] \]

Rearranging gives:

\[ [Cd^{2+}] = \frac{K_{sp}}{[S^{2-}]} \]

Substituting the values:

\[ [Cd^{2+}] = \frac{1 \times 10^{-28}}{[S^{2-}]} \]

### Final Calculation
After calculating \([S^{2-}]\) from the previous steps, we can find the remaining concentration of \([Cd^{2+}]\).

### Conclusion
The concentration of \(Cd^{2+}\) remaining in the solution after \(CdS\) stops precipitating can be calculated using the derived equations.