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

Explanation

To find the concentration of \( S^{2-} \) ions in the given solution containing \( 0.10 \, M \, H_2S \) and \( 0.20 \, M \, HCl \), we will follow these steps:

### Step 1: Write the Equilibrium Reactions
The dissociation of \( H_2S \) can be represented in two steps:
1. \( H_2S \rightleftharpoons H^+ + HS^- \) with equilibrium constant \( K_1 = 1.0 \times 10^{-7} \)
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) with equilibrium constant \( K_2 = 1.2 \times 10^{-13} \)

### Step 2: Determine 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:
- \( [H^+] = 0.20 \, M \)

### Step 3: Apply the First Equilibrium
For the first equilibrium reaction:
\[ K_1 = \frac{[H^+][HS^-]}{[H_2S]} \]
Substituting the known values:
\[ 1.0 \times 10^{-7} = \frac{(0.20)([HS^-])}{(0.10)} \]

### Step 4: Solve for \( [HS^-] \)
Rearranging the equation gives:
\[ [HS^-] = \frac{1.0 \times 10^{-7} \times 0.10}{0.20} \]
\[ [HS^-] = \frac{1.0 \times 10^{-8}}{0.20} = 5.0 \times 10^{-8} \, M \]

### Step 5: Apply the Second Equilibrium
For the second equilibrium reaction:
\[ K_2 = \frac{[H^+][S^{2-}]}{[HS^-]} \]
Substituting the known values:
\[ 1.2 \times 10^{-13} = \frac{(0.20)([S^{2-}])}{(5.0 \times 10^{-8})} \]

### Step 6: Solve for \( [S^{2-}] \)
Rearranging the equation gives:
\[ [S^{2-}] = \frac{1.2 \times 10^{-13} \times (5.0 \times 10^{-8})}{0.20} \]
\[ [S^{2-}] = \frac{6.0 \times 10^{-21}}{0.20} = 3.0 \times 10^{-20} \, M \]

### Final Answer
Thus, the concentration of \( S^{2-} \) ions in the aqueous solution is approximately \( 3.0 \times 10^{-20} \, M \).

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