An aqueous solution contains 0.10 `MH_2 S and 0.20 ` M HCl if the equilibrium constants for the formation of `HS^(-) " from " H_2S " is " 1.0 xx 10^(-7) and " that of " S^(2-) " from " HS^(-) ` ions is `1.2 xx 10 ^(-13)` then the concentration of `S^(2-)` ions in aqueous solution is
` 5 xx 10 ^(-8)`
` 3 xx 10 ^(-20)`
` 6 xx 10 ^(-21)`
` 5 xx 10 ^(-19)`

Options

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

Answer

To solve the problem, we need to determine the concentration of \( S^{2-} \) ions in an aqueous solution containing \( 0.10 \, M \, H_2S \) and \( 0.20 \, M \, HCl \). We are given the equilibrium constants for the dissociation reactions of \( H_2S \) and \( HS^- \).

### Step 1: Write the dissociation reactions and their equilibrium constants

1. The first dissociation of \( H_2S \):
\[ H_2S \rightleftharpoons H^+ + HS^- \]
The equilibrium constant \( K_1 \) for this reaction is given by:
\[ K_1 = \frac{[H^+][HS^-]}{[H_2S]} = 1.0 \times 10^{-7} \]

2. The second dissociation of \( HS^- \):
\[ HS^- \rightleftharpoons H^+ + S^{2-} \]
The equilibrium constant \( K_2 \) for this reaction is given by:
\[ K_2 = \frac{[H^+][S^{2-}]}{[HS^-]} = 1.2 \times 10^{-13} \]

### Step 2: Calculate the overall equilibrium constant

The overall reaction from \( H_2S \) to \( S^{2-} \) can be represented as:
\[ H_2S \rightleftharpoons 2H^+ + S^{2-} \]
The equilibrium constant \( K \) for this overall reaction can be expressed as:
\[ K = K_1 \times K_2 \]

Substituting the values of \( K_1 \) and \( K_2 \):
\[ K = (1.0 \times 10^{-7}) \times (1.2 \times 10^{-13}) = 1.2 \times 10^{-20} \]

### Step 3: Set up the equilibrium expression

At equilibrium, we can express the concentration of \( S^{2-} \) ions in terms of the concentrations of \( H^+ \) and \( H_2S \):
\[ K = \frac{[H^+]^2[S^{2-}]}{[H_2S]} \]

### Step 4: Substitute the known concentrations

From the problem, we know:
- The concentration of \( H_2S \) is \( 0.10 \, M \).
- The concentration of \( HCl \) is \( 0.20 \, M \), which contributes to the \( H^+ \) concentration.

Thus, the total concentration of \( H^+ \) ions is:
\[ [H^+] = 0.20 \, M \]

Now substituting the known values into the equilibrium expression:
\[ 1.2 \times 10^{-20} = \frac{(0.20)^2[S^{2-}]}{0.10} \]

### Step 5: Solve for \( [S^{2-}] \)

Rearranging the equation to solve for \( [S^{2-}] \):
\[ [S^{2-}] = \frac{1.2 \times 10^{-20} \times 0.10}{(0.20)^2} \]

Calculating the right-hand side:
\[ [S^{2-}] = \frac{1.2 \times 10^{-20} \times 0.10}{0.04} = \frac{1.2 \times 10^{-21}}{0.04} = 3.0 \times 10^{-20} \]

### Conclusion

The concentration of \( S^{2-} \) ions in the aqueous solution is:
\[ [S^{2-}] = 3.0 \times 10^{-20} \, M \]

### Final Answer
The correct option is \( 3 \times 10^{-20} \).

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