A solution contains ` 0.10 M H_2S and 0.3 HCl` Calculate the concentration of `S^(2-) and HS^(-)` ions in the solution . For `H_2S , K_(a_1) = 1 xx 10^(-7) and K_(a_2) = 1.3 xx 10^(-13)`

Explanation

To solve the problem of calculating the concentrations of \( S^{2-} \) and \( HS^{-} \) ions in a solution containing \( 0.10 \, M \, H_2S \) and \( 0.30 \, M \, HCl \), we will follow these steps:

### Step 1: Write the dissociation equations for \( H_2S \)

The dissociation of \( H_2S \) occurs in two steps:

1. \( H_2S \rightleftharpoons H^+ + HS^- \) (with \( K_{a1} = 1.0 \times 10^{-7} \))
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) (with \( K_{a2} = 1.3 \times 10^{-13} \))

### Step 2: Determine the concentration of \( H^+ \) ions

Since \( HCl \) is a strong acid, it completely dissociates in solution. Therefore, the concentration of \( H^+ \) ions from \( HCl \) is:

\[ [H^+] = [HCl] = 0.30 \, M \]

### Step 3: Calculate the equilibrium constant \( K \) for the overall dissociation of \( H_2S \)

The overall dissociation of \( H_2S \) can be represented as:

\[ H_2S \rightleftharpoons 2H^+ + S^{2-} \]

The equilibrium constant \( K \) for this reaction is the product of the two dissociation constants:

\[ K = K_{a1} \times K_{a2} = (1.0 \times 10^{-7}) \times (1.3 \times 10^{-13}) = 1.3 \times 10^{-20} \]

### Step 4: Set up the expression for the equilibrium constant \( K \)

Using the equilibrium concentrations, we can express \( K \) as:

\[ K = \frac{[H^+]^2 [S^{2-}]}{[H_2S]} \]

Substituting the known values into this equation:

\[ 1.3 \times 10^{-20} = \frac{(0.30)^2 [S^{2-}]}{[H_2S]} \]

### Step 5: Determine the concentration of \( H_2S \) at equilibrium

Initially, the concentration of \( H_2S \) is \( 0.10 \, M \). However, some of it will dissociate. Let \( x \) be the amount of \( H_2S \) that dissociates. At equilibrium, the concentration of \( H_2S \) will be:

\[ [H_2S] = 0.10 - x \]

### Step 6: Substitute into the equilibrium expression

Now we can substitute into the equilibrium expression:

\[ 1.3 \times 10^{-20} = \frac{(0.30)^2 [S^{2-}]}{0.10 - x} \]

### Step 7: Solve for \( S^{2-} \)

Assuming \( x \) is very small compared to \( 0.10 \), we can approximate \( [H_2S] \approx 0.10 \):

\[ 1.3 \times 10^{-20} = \frac{(0.30)^2 [S^{2-}]}{0.10} \]

\[ 1.3 \times 10^{-20} = \frac{0.09 [S^{2-}]}{0.10} \]

\[ [S^{2-}] = \frac{1.3 \times 10^{-20} \times 0.10}{0.09} = 1.44 \times 10^{-19} \, M \]

### Step 8: Calculate the concentration of \( HS^{-} \)

Now we can use \( K_{a1} \) to find the concentration of \( HS^{-} \):

\[ K_{a1} = \frac{[H^+][HS^-]}{[H_2S]} \]

Substituting the known values:

\[ 1.0 \times 10^{-7} = \frac{(0.30)[HS^-]}{0.10} \]

Solving for \( [HS^-] \):

\[ [HS^-] = \frac{1.0 \times 10^{-7} \times 0.10}{0.30} = 3.33 \times 10^{-8} \, M \]

### Final Answer

- Concentration of \( S^{2-} \): \( 1.44 \times 10^{-19} \, M \)
- Concentration of \( HS^{-} \): \( 3.33 \times 10^{-8} \, M \)

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