Approximate pH of `0.01M` 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

Explanation

To approximate the pH of a 0.01 M aqueous solution of \( H_2S \), we will follow these steps:

### Step 1: Write the dissociation equations of \( H_2S \)
The dissociation of \( H_2S \) in water can be represented in two steps:
1. \( H_2S \rightleftharpoons H^+ + HS^- \) (First dissociation)
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) (Second dissociation)

### Step 2: Identify the equilibrium constants
The equilibrium constants for the dissociations are given as:
- \( K_1 = 1 \times 10^{-7} \) (for the first dissociation)
- \( K_2 = 1.3 \times 10^{-13} \) (for the second dissociation)

### Step 3: Set up the expression for the first dissociation
For the first dissociation, we can set up the equilibrium expression:
\[ K_1 = \frac{[H^+][HS^-]}{[H_2S]} \]
Assuming \( x \) is the concentration of \( H^+ \) produced from the dissociation, we can express the concentrations at equilibrium:
- \([H^+] = x\)
- \([HS^-] = x\)
- \([H_2S] = 0.01 - x \approx 0.01\) (since \( x \) will be very small compared to 0.01)

Substituting these into the equilibrium expression gives:
\[ K_1 = \frac{x^2}{0.01} \]

### Step 4: Solve for \( x \)
Substituting \( K_1 \) into the equation:
\[ 1 \times 10^{-7} = \frac{x^2}{0.01} \]
Rearranging gives:
\[ x^2 = 1 \times 10^{-7} \times 0.01 = 1 \times 10^{-9} \]
Taking the square root:
\[ x = \sqrt{1 \times 10^{-9}} = 1 \times 10^{-4} \]

### Step 5: Calculate the pH
The concentration of \( H^+ \) ions is \( 1 \times 10^{-4} \) M. The pH is calculated using the formula:
\[ \text{pH} = -\log[H^+] \]
Substituting the value:
\[ \text{pH} = -\log(1 \times 10^{-4}) = 4 \]

### Final Answer
The approximate pH of the 0.01 M aqueous \( H_2S \) solution is **4**.

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