A saturated solution of `H_2S` in water has concetration of approximately 0.10 M. What is the pH of this soluton and equilibrium concentrations of `H_2S,HS^(-)` and `S^(2-)` ? Hydrogen sulphide is a diprotic acid and its dissociation constants are `K_(a_1)=9.1xx10^(-8) ,K_(a_2)=1.3xx10^(-3) "mol L"^(-1)` respectively.

Explanation

To solve the problem, we need to find the pH of a saturated solution of hydrogen sulfide (H₂S) and the equilibrium concentrations of H₂S, HS⁻, and S²⁻. The dissociation constants given are Kₐ₁ = 9.1 × 10⁻⁸ and Kₐ₂ = 1.3 × 10⁻³ mol L⁻¹.

### Step 1: Write the dissociation reactions
Hydrogen sulfide (H₂S) is a diprotic acid and dissociates in two steps:
1. H₂S ⇌ H⁺ + HS⁻ (Kₐ₁)
2. HS⁻ ⇌ H⁺ + S²⁻ (Kₐ₂)

### Step 2: Set up the equilibrium expression for the first dissociation
For the first dissociation, we can write the equilibrium expression:
\[ K_{a_1} = \frac{[H^+][HS^-]}{[H_2S]} \]
Let x be the concentration of H⁺ and HS⁻ produced from the first dissociation. The initial concentration of H₂S is 0.10 M.

At equilibrium:
- [H₂S] = 0.10 - x
- [H⁺] = x
- [HS⁻] = x

Substituting into the equilibrium expression gives:
\[ 9.1 \times 10^{-8} = \frac{x \cdot x}{0.10 - x} \]
\[ 9.1 \times 10^{-8} = \frac{x^2}{0.10 - x} \]

### Step 3: Solve for x
Assuming x is small compared to 0.10 M, we can approximate:
\[ 9.1 \times 10^{-8} = \frac{x^2}{0.10} \]
\[ x^2 = 9.1 \times 10^{-8} \times 0.10 \]
\[ x^2 = 9.1 \times 10^{-9} \]
\[ x = \sqrt{9.1 \times 10^{-9}} \]
\[ x \approx 9.5 \times 10^{-5} \, \text{M} \]

### Step 4: Calculate the pH
The concentration of H⁺ ions is approximately:
\[ [H^+] = x = 9.5 \times 10^{-5} \]
Now, we can calculate the pH:
\[ \text{pH} = -\log[H^+] = -\log(9.5 \times 10^{-5}) \]
\[ \text{pH} \approx 4.02 \]

### Step 5: Find equilibrium concentrations of H₂S and HS⁻
- [H₂S] at equilibrium:
\[ [H_2S] = 0.10 - x = 0.10 - 9.5 \times 10^{-5} \approx 0.10 \, \text{M} \]
- [HS⁻] at equilibrium:
\[ [HS^-] = x = 9.5 \times 10^{-5} \, \text{M} \]

### Step 6: Set up the equilibrium expression for the second dissociation
For the second dissociation:
\[ K_{a_2} = \frac{[H^+][S^{2-}]}{[HS^-]} \]
Let y be the concentration of S²⁻ produced from the second dissociation. At equilibrium:
- [HS⁻] = 9.5 × 10⁻⁵ - y
- [H⁺] = 9.5 × 10⁻⁵ + y
- [S²⁻] = y

Substituting into the equilibrium expression gives:
\[ 1.3 \times 10^{-3} = \frac{(9.5 \times 10^{-5} + y) \cdot y}{9.5 \times 10^{-5} - y} \]

Assuming y is small compared to 9.5 × 10⁻⁵, we can approximate:
\[ 1.3 \times 10^{-3} \approx \frac{(9.5 \times 10^{-5}) \cdot y}{9.5 \times 10^{-5}} \]
\[ y \approx 1.3 \times 10^{-3} \]

### Step 7: Calculate the equilibrium concentrations
- [S²⁻] = y = 1.3 × 10⁻³ M
- [HS⁻] = 9.5 × 10⁻⁵ - y ≈ 9.5 × 10⁻⁵ (since y is much larger)
- [H₂S] remains approximately 0.10 M.

### Final Results
- pH ≈ 4.02
- [H₂S] ≈ 0.10 M
- [HS⁻] ≈ 9.5 × 10⁻⁵ M
- [S²⁻] ≈ 1.3 × 10⁻³ M