`0.1 M H_(2)S` has `K_(1)=10^(-5) & K_(2)=1.5xx10^(-12)`. What will be the concentration of `S^(-2)` in the solution.

Options

A. `~~10^(-8)`
B. `~~10^(-9)`
C. `~~1.5xx10^(-12)`
D. `1.2xx10^(-13)`

Explanation

To find the concentration of \( S^{2-} \) ions in a 0.1 M solution of \( H_2S \) with given dissociation constants \( K_1 = 10^{-5} \) and \( K_2 = 1.5 \times 10^{-12} \), we can follow these steps:

### Step 1: Write the Dissociation Reactions
The dissociation of \( H_2S \) occurs in two steps:
1. \( H_2S \rightleftharpoons H^+ + HS^- \) (with \( K_1 \))
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) (with \( K_2 \))

### Step 2: Set Up the Equilibrium Expressions
For the first dissociation, the equilibrium expression is:
\[ K_1 = \frac{[H^+][HS^-]}{[H_2S]} \]
For the second dissociation, the equilibrium expression is:
\[ K_2 = \frac{[H^+][S^{2-}]}{[HS^-]} \]

### Step 3: Define Variables
Let:
- \( x \) be the concentration of \( H^+ \) and \( HS^- \) produced from the first dissociation.
- \( y \) be the concentration of \( S^{2-} \) produced from the second dissociation.

At equilibrium:
- The concentration of \( H_2S \) will be \( 0.1 - x \).
- The concentration of \( H^+ \) will be \( x + y \).
- The concentration of \( HS^- \) will be \( x - y \).

### Step 4: Substitute into the Equilibrium Expressions
For the first dissociation:
\[ K_1 = \frac{x \cdot x}{0.1 - x} \implies K_1 = \frac{x^2}{0.1 - x} \]

For the second dissociation:
\[ K_2 = \frac{(x + y) \cdot y}{x - y} \]

### Step 5: Simplify Assumptions
Assuming \( x \) is much larger than \( y \) (since \( K_2 \) is very small), we can simplify:
- \( K_2 \approx \frac{x \cdot y}{x} = y \)

### Step 6: Solve for \( y \)
From the value of \( K_2 \):
\[ y = K_2 = 1.5 \times 10^{-12} \]

### Conclusion
The concentration of \( S^{2-} \) ions in the solution is:
\[ [S^{2-}] = 1.5 \times 10^{-12} \, M \]