`K_(1) and K_(2)\ of H_(2)`S are `10^(-3) and 10^(-12)` respectively what is `[S^(-2)]` in 0.1 M `H_(2) ` S buffered at `p(H)` =2

Options

A. `10^(-10)` M
B. `10^(-12)` M
C. `10^(-11)` M
D. `10^(-8)` M

Answer

To solve the problem, we need to determine the concentration of \( [S^{2-}] \) in a 0.1 M \( H_2S \) solution buffered at \( pH = 2 \). We will use the dissociation constants \( K_1 \) and \( K_2 \) of \( H_2S \), which are given as \( 10^{-3} \) and \( 10^{-12} \) respectively.

### Step-by-Step Solution:

1. **Understanding the Dissociation of \( H_2S \)**:
The dissociation of \( H_2S \) can be represented as:
\[ H_2S \rightleftharpoons H^+ + HS^- \]
The first dissociation constant \( K_1 \) is given by:
\[ K_1 = \frac{[H^+][HS^-]}{[H_2S]} \]

2. **Setting up the Equilibrium Expression**:
Let the initial concentration of \( H_2S \) be \( C = 0.1 \, M \). At equilibrium, if \( x \) is the amount that dissociates, we have:
\[ [H_2S] = C - x = 0.1 - x \]
\[ [H^+] = x \]
\[ [HS^-] = x \]
Thus, the expression for \( K_1 \) becomes:
\[ K_1 = \frac{x^2}{0.1 - x} \]

3. **Using the Given \( pH \)**:
Since the solution is buffered at \( pH = 2 \):
\[ [H^+] = 10^{-2} \, M \]
Therefore, we can set \( x = 10^{-2} \).

4. **Substituting \( x \) into the \( K_1 \) Expression**:
Substituting \( x = 10^{-2} \) into the \( K_1 \) expression:
\[ K_1 = \frac{(10^{-2})^2}{0.1 - 10^{-2}} = \frac{10^{-4}}{0.1 - 0.01} = \frac{10^{-4}}{0.09} \]
\[ K_1 \approx 1.11 \times 10^{-3} \]
This is consistent with the given \( K_1 \) value.

5. **Dissociation of \( HS^- \)**:
The second dissociation of \( HS^- \) is:
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]
The expression for \( K_2 \) is:
\[ K_2 = \frac{[H^+][S^{2-}]}{[HS^-]} \]
At equilibrium, if \( y \) is the amount that dissociates:
\[ [HS^-] = x - y = 10^{-2} - y \]
\[ [S^{2-}] = y \]
Thus, we have:
\[ K_2 = \frac{(10^{-2})(y)}{(10^{-2} - y)} \]

6. **Assuming \( y \) is Small**:
Given that \( K_2 = 10^{-12} \), we can assume \( y \) is small compared to \( 10^{-2} \):
\[ K_2 \approx \frac{(10^{-2})(y)}{10^{-2}} = y \]
Therefore:
\[ y = K_2 = 10^{-12} \]

7. **Final Concentration of \( S^{2-} \)**:
Thus, the concentration of \( [S^{2-}] \) in the solution is:
\[ [S^{2-}] = 10^{-12} \, M \]

### Conclusion:
The concentration of \( [S^{2-}] \) in the 0.1 M \( H_2S \) solution buffered at \( pH = 2 \) is \( 10^{-12} \, M \).