Calculate the concentrations of various species in a 0.1 M `H_2S` saturated solution. `K_(1) =1 xx 10^(-7) and K_(2)=1.3 xx 10^(-13)`.

Answer

To calculate the concentrations of various species in a 0.1 M \( H_2S \) saturated solution, we will use the dissociation constants \( K_1 \) and \( K_2 \) provided for the dissociation of hydrogen sulfide.

### Step 1: Write the dissociation equations
The dissociation of \( H_2S \) in water can be represented by the following equations:

1. \( H_2S \rightleftharpoons H^+ + HS^- \) with \( K_1 = 1 \times 10^{-7} \)
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) with \( K_2 = 1.3 \times 10^{-13} \)

### Step 2: Define the initial concentrations
Let the initial concentration of \( H_2S \) be \( C = 0.1 \, M \).

At equilibrium, let \( \alpha \) be the degree of dissociation of \( H_2S \). Therefore, the concentrations at equilibrium will be:

- \( [H_2S] = C(1 - \alpha) \)
- \( [H^+] = C\alpha \)
- \( [HS^-] = C\alpha \)
- \( [S^{2-}] = 0 \) (initially, since it comes from the second dissociation)

### Step 3: Apply the first dissociation constant \( K_1 \)
Using the expression for \( K_1 \):

\[ K_1 = \frac{[H^+][HS^-]}{[H_2S]} = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} = \frac{C\alpha^2}{1 - \alpha} \]

Substituting \( K_1 = 1 \times 10^{-7} \):

\[ 1 \times 10^{-7} = \frac{0.1 \alpha^2}{1 - \alpha} \]

### Step 4: Simplify the equation
Assuming \( \alpha \) is small (which is reasonable since \( K_1 \) is small), we can approximate \( 1 - \alpha \approx 1 \):

\[ 1 \times 10^{-7} \approx 0.1 \alpha^2 \]

### Step 5: Solve for \( \alpha \)
Rearranging gives:

\[ \alpha^2 = \frac{1 \times 10^{-7}}{0.1} = 1 \times 10^{-6} \]

Taking the square root:

\[ \alpha = 1 \times 10^{-3} \]

### Step 6: Calculate \( [H^+] \) and \( [HS^-] \)
Now we can find the concentrations of \( H^+ \) and \( HS^- \):

\[ [H^+] = C\alpha = 0.1 \times 1 \times 10^{-3} = 1 \times 10^{-4} \, M \]
\[ [HS^-] = C\alpha = 0.1 \times 1 \times 10^{-3} = 1 \times 10^{-4} \, M \]

### Step 7: Apply the second dissociation constant \( K_2 \)
Now we consider the second dissociation:

Using the expression for \( K_2 \):

\[ K_2 = \frac{[H^+][S^{2-}]}{[HS^-]} = \frac{(1 \times 10^{-4})(x)}{1 \times 10^{-4}} \]

Where \( x \) is the concentration of \( S^{2-} \).

Substituting \( K_2 = 1.3 \times 10^{-13} \):

\[ 1.3 \times 10^{-13} = x \]

### Step 8: Final concentrations
Thus, the concentrations of the various species in the solution are:

- \( [H_2S] \approx 0.1 \, M \) (since \( \alpha \) is very small)
- \( [H^+] = 1 \times 10^{-4} \, M \)
- \( [HS^-] = 1 \times 10^{-4} \, M \)
- \( [S^{2-}] = 1.3 \times 10^{-13} \, M \)

### Summary of Results
- \( [H_2S] \approx 0.1 \, M \)
- \( [H^+] = 1 \times 10^{-4} \, M \)
- \( [HS^-] = 1 \times 10^{-4} \, M \)
- \( [S^{2-}] = 1.3 \times 10^{-13} \, M \)