What will be the value of pH of 0.01 `"mol dm"^(-3)CH_(3)COOH(K_(1)=1.74xx10^(-5))` ?

Options

A. 3.4
B. 3.6
C. 3.9
D. `3.0`

Explanation

To find the pH of a 0.01 mol/dm³ solution of acetic acid (CH₃COOH), we need to follow these steps:

### Step 1: Write the dissociation equation
Acetic acid (CH₃COOH) is a weak acid that dissociates in water as follows:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]

### Step 2: Set up the initial concentrations
Let the initial concentration of acetic acid (C) be 0.01 mol/dm³. At the start (t = 0):
- \([ \text{CH}_3\text{COOH} ] = C = 0.01\)
- \([ \text{CH}_3\text{COO}^- ] = 0\)
- \([ \text{H}^+ ] = 0\)

### Step 3: Define the change in concentration at equilibrium
Let \( \alpha \) be the degree of dissociation. At equilibrium, the concentrations will be:
- \([ \text{CH}_3\text{COOH} ] = C - C\alpha = C(1 - \alpha)\)
- \([ \text{CH}_3\text{COO}^- ] = C\alpha\)
- \([ \text{H}^+ ] = C\alpha\)

### Step 4: Write the expression for the acid dissociation constant (Kₐ)
The expression for the acid dissociation constant \( K_a \) is given by:
\[ K_a = \frac{[\text{CH}_3\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{COOH}]} \]
Substituting the equilibrium concentrations:
\[ K_a = \frac{(C\alpha)(C\alpha)}{C(1 - \alpha)} \]

### Step 5: Simplify the expression
Assuming that \( \alpha \) is very small compared to 1 (which is typical for weak acids), we can approximate \( 1 - \alpha \approx 1 \):
\[ K_a \approx \frac{C\alpha^2}{C} = C\alpha^2 \]
Thus, we can rearrange this to find \( \alpha \):
\[ \alpha = \sqrt{\frac{K_a}{C}} \]

### Step 6: Substitute the values
Given:
- \( K_a = 1.74 \times 10^{-5} \)
- \( C = 0.01 \, \text{mol/dm}^3 \)

Substituting these values:
\[ \alpha = \sqrt{\frac{1.74 \times 10^{-5}}{0.01}} = \sqrt{1.74 \times 10^{-3}} \approx 0.0417 \]

### Step 7: Calculate the concentration of H⁺ ions
The concentration of H⁺ ions at equilibrium is:
\[ [\text{H}^+] = C\alpha = 0.01 \times 0.0417 \approx 4.17 \times 10^{-4} \, \text{mol/dm}^3 \]

### Step 8: Calculate the pH
The pH is calculated using the formula:
\[ \text{pH} = -\log_{10}[\text{H}^+] \]
Substituting the value of \([\text{H}^+]\):
\[ \text{pH} = -\log_{10}(4.17 \times 10^{-4}) \approx 3.38 \]

### Step 9: Round to the appropriate number of significant figures
Rounding this value gives us:
\[ \text{pH} \approx 3.4 \]

### Final Answer
The pH of a 0.01 mol/dm³ solution of acetic acid is approximately 3.4.

---