Approximate p.H,0.1M aqueous `H_(2)S` solution when `K_(1)` and `K_(2)` for `H_(2)S` at `25^(@)C` are `1xx10^(-7)` and `1.3xx10^(-13)` respectively:-

Options

A. 4
B. 5
C. 6
D. 8

Answer

To approximate the pH of a 0.1 M aqueous solution of \( H_2S \), we will follow these steps:

### Step 1: Understand the dissociation of \( H_2S \)
The dissociation of \( H_2S \) in water can be represented as:
\[ H_2S \rightleftharpoons H^+ + HS^- \]
This is the first dissociation, and it has a dissociation constant \( K_1 \).

### Step 2: Write the expression for \( K_1 \)
The expression for the first dissociation constant \( K_1 \) is given by:
\[ K_1 = \frac{[H^+][HS^-]}{[H_2S]} \]
Where:
- \([H^+]\) is the concentration of hydrogen ions,
- \([HS^-]\) is the concentration of hydrogen sulfide ions,
- \([H_2S]\) is the concentration of undissociated hydrogen sulfide.

### Step 3: Set up the equilibrium concentrations
Assuming that \( \alpha \) is the degree of dissociation of \( H_2S \), we can express the concentrations at equilibrium:
- Initial concentration of \( H_2S \) = 0.1 M
- At equilibrium:
- \([H_2S] = 0.1 - \alpha\)
- \([H^+] = \alpha\)
- \([HS^-] = \alpha\)

### Step 4: Substitute into the \( K_1 \) expression
Substituting the equilibrium concentrations into the \( K_1 \) expression gives:
\[ K_1 = \frac{\alpha \cdot \alpha}{0.1 - \alpha} \]
Since \( K_1 \) is much larger than \( K_2 \), we can assume that \( \alpha \) is small compared to 0.1, allowing us to simplify:
\[ K_1 \approx \frac{\alpha^2}{0.1} \]

### Step 5: Solve for \( \alpha \)
Given \( K_1 = 1 \times 10^{-7} \):
\[ 1 \times 10^{-7} = \frac{\alpha^2}{0.1} \]
\[ \alpha^2 = 1 \times 10^{-7} \times 0.1 \]
\[ \alpha^2 = 1 \times 10^{-8} \]
\[ \alpha = \sqrt{1 \times 10^{-8}} = 1 \times 10^{-4} \]

### Step 6: Calculate the concentration of \( H^+ \)
The concentration of \( H^+ \) ions is:
\[ [H^+] = 0.1 \times \alpha = 0.1 \times 1 \times 10^{-4} = 1 \times 10^{-5} \, \text{M} \]

### Step 7: Calculate the pH
Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
Substituting the value of \([H^+]\):
\[ \text{pH} = -\log(1 \times 10^{-5}) = 5 \]

### Final Answer
The approximate pH of the 0.1 M aqueous \( H_2S \) solution is **5**.

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