Sulphide ion in alkaline solution reacts with solid sulphur to form polysulphide ions having formulae `S_2^(2-),S_3^(2-),S_4^(2-)` and so on. The equilibrium constant for the formation of `S_3^(2-)` is 132 `(K_2)` , both from S and `S^(2-)` What is the equilibrium constant for the formation of `S_3^(2-)` from `S_2^(2-)` and S ?

Options

A. 11
B. 12
C. 132
D. None of these

Answer

To solve the problem, we need to determine the equilibrium constant for the formation of \( S_3^{2-} \) from \( S_2^{2-} \) and \( S \). We are given the equilibrium constant \( K_2 \) for the formation of \( S_3^{2-} \) from \( S \) and \( S^{2-} \), which is 132.

### Step-by-Step Solution:

1. **Write the given reactions and their equilibrium constants:**
- The reaction for the formation of \( S_3^{2-} \):
\[ S + S^{2-} \rightleftharpoons S_3^{2-} \quad (K_2 = 132) \]
- The reaction for the formation of \( S_2^{2-} \):
\[ S + S^{2-} \rightleftharpoons S_2^{2-} \quad (K_1 = 12) \]

2. **Reverse the first reaction to find the equilibrium constant for the formation of \( S_2^{2-} \):**
- When we reverse the reaction for \( S_2^{2-} \), we get:
\[ S_2^{2-} \rightleftharpoons S + S^{2-} \]
- The equilibrium constant for this reversed reaction, \( K_3 \), is the reciprocal of \( K_1 \):
\[ K_3 = \frac{1}{K_1} = \frac{1}{12} \]

3. **Combine the reactions to find the desired equilibrium constant:**
- We want to find the equilibrium constant for the reaction:
\[ S_2^{2-} + S \rightleftharpoons S_3^{2-} \]
- To obtain this reaction, we can add the reversed reaction for \( S_2^{2-} \) to the reaction for \( S_3^{2-} \):
\[ S + S^{2-} \rightleftharpoons S_3^{2-} \quad (K_2 = 132) \]
\[ S_2^{2-} \rightleftharpoons S + S^{2-} \quad (K_3 = \frac{1}{12}) \]
- When we add these two reactions, the \( S^{2-} \) and \( S \) cancel out:
\[ S_2^{2-} + S \rightleftharpoons S_3^{2-} \]

4. **Calculate the equilibrium constant for the combined reaction:**
- The equilibrium constant for the overall reaction is the product of the individual constants:
\[ K = K_2 \times K_3 = 132 \times \frac{1}{12} = 11 \]

### Conclusion:
The equilibrium constant for the formation of \( S_3^{2-} \) from \( S_2^{2-} \) and \( S \) is \( K = 11 \).