The pH of a mixture of 0.01 M HCI and 0.1 M `CH_3COOH` is approximately

Options

A. 1
B. 2
C. 4
D. 7

Answer

To find the pH of a mixture of 0.01 M HCl and 0.1 M CH₃COOH, we can follow these steps:

### Step 1: Identify the Contributions of H⁺ Ions
- HCl is a strong acid and completely dissociates in solution. Therefore, the concentration of H⁺ ions from HCl is:
\[ [H^+]_{HCl} = 0.01 \, \text{M} \]

- CH₃COOH (acetic acid) is a weak acid and does not completely dissociate. We need to calculate the contribution of H⁺ ions from CH₃COOH using its dissociation constant (Kₐ).

### Step 2: Calculate the Contribution of H⁺ Ions from CH₃COOH
- The dissociation of acetic acid can be represented as:
\[ CH₃COOH \rightleftharpoons CH₃COO^- + H^+ \]
- The dissociation constant (Kₐ) for acetic acid is given as \(1.8 \times 10^{-5}\).

- Let \(x\) be the concentration of H⁺ ions produced from the dissociation of acetic acid. The equilibrium concentrations will be:
- \([CH₃COOH] = 0.1 - x\)
- \([CH₃COO^-] = x\)
- \([H^+] = x\)

- The expression for Kₐ is:
\[ K_a = \frac{[CH₃COO^-][H^+]}{[CH₃COOH]} = \frac{x \cdot x}{0.1 - x} \approx \frac{x^2}{0.1} \quad (\text{assuming } x \text{ is small}) \]

- Setting this equal to Kₐ:
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1} \]
\[ x^2 = 1.8 \times 10^{-6} \]
\[ x = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \, \text{M} \]

### Step 3: Calculate Total H⁺ Concentration
- The total concentration of H⁺ ions in the mixture is the sum of the contributions from HCl and CH₃COOH:
\[ [H^+]_{total} = [H^+]_{HCl} + [H^+]_{CH₃COOH} = 0.01 + 1.34 \times 10^{-3} \]
\[ = 0.01 + 0.00134 = 0.01134 \, \text{M} \]

### Step 4: Calculate pH
- The pH is calculated using the formula:
\[ pH = -\log[H^+] \]
\[ pH = -\log(0.01134) \approx 2.24 \]

### Step 5: Approximate pH
- Since we are asked for an approximate value, we can round this to:
\[ \text{pH} \approx 2 \]

### Final Answer
The pH of the mixture of 0.01 M HCl and 0.1 M CH₃COOH is approximately **2**.

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