An aqueous solution contains 0.10 M `H_2S` and 0.20 M HCl. If the equilibrium constants for the formation of HS from HS is `1.0 xx 10^(-7)` and that of `S^(2-)?` from `HS^(-)` ions is `1.2 xx 10^(-7)` then the concentration of `S^(2-)` ions in aqueous solution is

Options

A. `5 xx 10^(-19)`
B. `5 xx 10^(-8)`
C. `3 xx 10^(-20)`
D. `6 xx 10^(-21)`

Answer

To solve the problem, we need to analyze the dissociation of `H2S` in the presence of `HCl` and determine the concentration of `S^(2-)` ions in the solution.

### Step 1: Understand the dissociation of `H2S`
`H2S` can dissociate in two steps:
1. **First dissociation**:
\[ H_2S \rightleftharpoons H^+ + HS^- \]
The equilibrium constant for this reaction is denoted as \( K_{a1} \).

2. **Second dissociation**:
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]
The equilibrium constant for this reaction is denoted as \( K_{a2} \).

### Step 2: Write down the equilibrium constants
From the problem, we know:
- \( K_{a1} = 1.0 \times 10^{-7} \)
- \( K_{a2} = 1.2 \times 10^{-7} \)

### Step 3: Calculate the concentration of \( H^+ \) ions
Since we have `0.20 M` of `HCl`, which is a strong acid, it will completely dissociate:
\[ HCl \rightarrow H^+ + Cl^- \]
Thus, the concentration of \( H^+ \) from `HCl` is `0.20 M`.

### Step 4: Set up the equilibrium expressions
1. For the first dissociation of `H2S`:
\[ K_{a1} = \frac{[H^+][HS^-]}{[H2S]} \]
Substituting the known values:
\[ 1.0 \times 10^{-7} = \frac{(0.20)([HS^-])}{[H2S]} \]

2. For the second dissociation of `HS^-`:
\[ K_{a2} = \frac{[H^+][S^{2-}]}{[HS^-]} \]
Substituting the known values:
\[ 1.2 \times 10^{-7} = \frac{(0.20)([S^{2-}])}{[HS^-]} \]

### Step 5: Solve for the concentrations
From the first equilibrium expression, we can express `[HS^-]` in terms of `[H2S]`:
\[ [HS^-] = \frac{1.0 \times 10^{-7} \cdot [H2S]}{0.20} \]

Assuming that the initial concentration of `H2S` is `0.10 M`, we can substitute this value in:
\[ [HS^-] = \frac{1.0 \times 10^{-7} \cdot 0.10}{0.20} = 5.0 \times 10^{-8} M \]

Now, substituting `[HS^-]` into the second equilibrium expression:
\[ 1.2 \times 10^{-7} = \frac{(0.20)([S^{2-}])}{5.0 \times 10^{-8}} \]
Rearranging gives:
\[ [S^{2-}] = \frac{1.2 \times 10^{-7} \cdot 5.0 \times 10^{-8}}{0.20} \]
Calculating this gives:
\[ [S^{2-}] = 3.0 \times 10^{-15} M \]

### Final Answer
The concentration of \( S^{2-} \) ions in the aqueous solution is \( 3.0 \times 10^{-15} M \).

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