The first ionization constant of `H_(2)S` is `9.1xx10^(-8)`. Calculate the concentration of `HS^(Θ)` ion in its `0.1 M` solution. How will this concentration be affected if the solution is `0.1 M` in `HCl` also? If the second dissociation constant if `H_(2)S` is `1.2xx10^(-13)`, calculate the concentration of `S^(2-)` under both conditions.

Explanation

To solve the problem step by step, we need to calculate the concentration of the \( HS^- \) ion in a \( 0.1 \, M \) solution of \( H_2S \), and then determine how this concentration changes when \( HCl \) is added. Finally, we will calculate the concentration of \( S^{2-} \) under both conditions.

### Step 1: Calculate the concentration of \( HS^- \) in \( 0.1 \, M \) \( H_2S \)

1. **Write the dissociation equation:**
\[ H_2S \rightleftharpoons H^+ + HS^- \]

2. **Set up the expression for the first ionization constant \( K_a \):**
\[ K_a = \frac{[H^+][HS^-]}{[H_2S]} \]
Given \( K_a = 9.1 \times 10^{-8} \).

3. **Assume initial concentration of \( H_2S \) is \( 0.1 \, M \) and let \( x \) be the concentration of \( H^+ \) and \( HS^- \) formed:**
\[ [H_2S] = 0.1 - x \quad \text{(approximately } 0.1 \text{ since } x \text{ is small)} \]
\[ [H^+] = x \quad \text{and} \quad [HS^-] = x \]

4. **Substituting into the \( K_a \) expression:**
\[ 9.1 \times 10^{-8} = \frac{x^2}{0.1} \]

5. **Solve for \( x \):**
\[ x^2 = 9.1 \times 10^{-8} \times 0.1 = 9.1 \times 10^{-9} \]
\[ x = \sqrt{9.1 \times 10^{-9}} \approx 9.54 \times 10^{-5} \, M \]

### Step 2: Effect of adding \( 0.1 \, M \) \( HCl \)

1. **In the presence of \( HCl \), the concentration of \( H^+ \) is already \( 0.1 \, M \).**
\[ H_2S \rightleftharpoons H^+ + HS^- \]

2. **Now, the \( K_a \) expression becomes:**
\[ 9.1 \times 10^{-8} = \frac{[H^+][HS^-]}{[H_2S]} \]
Here, \( [H^+] = 0.1 \, M \) and \( [H_2S] \approx 0.1 \, M \).

3. **Substituting into the \( K_a \) expression:**
\[ 9.1 \times 10^{-8} = \frac{0.1 \cdot [HS^-]}{0.1} \]
\[ [HS^-] = 9.1 \times 10^{-8} \, M \]

### Step 3: Calculate the concentration of \( S^{2-} \)

1. **For the first case (without \( HCl \)):**
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]
Given \( K_a = 1.2 \times 10^{-13} \).

2. **Set up the expression for \( K_a \):**
\[ K_a = \frac{[H^+][S^{2-}]}{[HS^-]} \]
Substituting \( [H^+] = 9.54 \times 10^{-5} \, M \) and \( [HS^-] = 9.54 \times 10^{-5} \, M \):
\[ 1.2 \times 10^{-13} = \frac{(9.54 \times 10^{-5})[S^{2-}]}{9.54 \times 10^{-5}} \]
\[ [S^{2-}] = 1.2 \times 10^{-13} \, M \]

3. **For the second case (with \( HCl \)):**
\[ K_a = \frac{[H^+][S^{2-}]}{[HS^-]} \]
Here, \( [H^+] = 0.1 \, M \) and \( [HS^-] = 9.1 \times 10^{-8} \, M \):
\[ 1.2 \times 10^{-13} = \frac{(0.1)[S^{2-}]}{9.1 \times 10^{-8}} \]
\[ [S^{2-}] = \frac{1.2 \times 10^{-13} \times 9.1 \times 10^{-8}}{0.1} = 1.092 \times 10^{-19} \, M \]

### Summary of Results
- Concentration of \( HS^- \) in \( 0.1 \, M \) \( H_2S \): \( 9.54 \times 10^{-5} \, M \)
- Concentration of \( HS^- \) in \( 0.1 \, M \) \( HCl \): \( 9.1 \times 10^{-8} \, M \)
- Concentration of \( S^{2-} \) in \( 0.1 \, M \) \( H_2S \): \( 1.2 \times 10^{-13} \, M \)
- Concentration of \( S^{2-} \) in \( 0.1 \, M \) \( HCl \): \( 1.092 \times 10^{-19} \, M \)