An aqueous solution contains `0.10 M H_(2)S` and 0.20 M HCl. If the equilibrium constants for the formation of `HS^(–)` from `H_(2)S` is `1.0xx10^(–7)` and that of `S^(2-)` from `HS^(–)` ions is `1.2xx10^(–13)` then the concentration of `S^(2-)` ions in aqueous solution is

Options

A. `5xx10^(-19)`
B. `5xx10^(-8)`
C. `3xx10^(-20)`
D. `6xx10^(-21)`

Answer

To solve the problem, we need to find the concentration of \( S^{2-} \) ions in the given solution containing \( H_2S \) and \( HCl \). Let's break down the steps:

### Step 1: Write the dissociation reactions
1. The first dissociation of \( H_2S \):
\[ H_2S \rightleftharpoons H^+ + HS^- \]
The equilibrium constant for this reaction is given as \( K_a1 = 1.0 \times 10^{-7} \).

2. The second dissociation of \( HS^- \):
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]
The equilibrium constant for this reaction is given as \( K_a2 = 1.2 \times 10^{-13} \).

### Step 2: Set up the equilibrium expressions
For the first dissociation:
\[ K_a1 = \frac{[H^+][HS^-]}{[H_2S]} \]
For the second dissociation:
\[ K_a2 = \frac{[H^+][S^{2-}]}{[HS^-]} \]

### Step 3: Initial concentrations
Given:
- Initial concentration of \( H_2S = 0.10 \, M \)
- Initial concentration of \( HCl = 0.20 \, M \)

Since \( HCl \) is a strong acid, it will completely dissociate, contributing \( 0.20 \, M \) of \( H^+ \) ions.

### Step 4: Calculate the concentration of \( H^+ \)
The total concentration of \( H^+ \) ions will be:
\[ [H^+] = 0.20 \, M + \text{(contribution from } H_2S) \]
Assuming \( H_2S \) contributes a negligible amount due to its weak dissociation, we can approximate:
\[ [H^+] \approx 0.20 \, M \]

### Step 5: Substitute into the first equilibrium expression
Let \( x \) be the concentration of \( HS^- \) produced from the dissociation of \( H_2S \):
\[ K_a1 = \frac{(0.20)(x)}{(0.10 - x)} \approx \frac{(0.20)(x)}{0.10} \]
Setting this equal to \( 1.0 \times 10^{-7} \):
\[ 1.0 \times 10^{-7} = \frac{0.20x}{0.10} \]
\[ 1.0 \times 10^{-7} = 2x \]
\[ x = \frac{1.0 \times 10^{-7}}{2} = 5.0 \times 10^{-8} \, M \]
Thus, \( [HS^-] \approx 5.0 \times 10^{-8} \, M \).

### Step 6: Substitute into the second equilibrium expression
Now, we can find \( [S^{2-}] \):
Let \( y \) be the concentration of \( S^{2-} \) produced from the dissociation of \( HS^- \):
\[ K_a2 = \frac{(0.20)(y)}{(5.0 \times 10^{-8})} \]
Setting this equal to \( 1.2 \times 10^{-13} \):
\[ 1.2 \times 10^{-13} = \frac{(0.20)(y)}{5.0 \times 10^{-8}} \]
\[ y = \frac{1.2 \times 10^{-13} \times 5.0 \times 10^{-8}}{0.20} \]
\[ y = \frac{6.0 \times 10^{-21}}{0.20} = 3.0 \times 10^{-20} \, M \]

### Final Answer
The concentration of \( S^{2-} \) ions in the aqueous solution is:
\[ [S^{2-}] = 3.0 \times 10^{-20} \, M \]