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 will follow these steps:

### Step 1: Understand the dissociation of acids
- HCl is a strong acid and dissociates completely in water:
\[ \text{HCl} \rightarrow \text{H}^+ + \text{Cl}^- \]
Therefore, the concentration of \(\text{H}^+\) from HCl is 0.01 M.

- CH₃COOH (acetic acid) is a weak acid and partially dissociates:
\[ \text{CH}_3\text{COOH} \rightleftharpoons \text{CH}_3\text{COO}^- + \text{H}^+ \]
The dissociation constant \(K_a\) for acetic acid is approximately \(1.8 \times 10^{-5}\).

### Step 2: Calculate the contribution of \(\text{H}^+\) from CH₃COOH
- The concentration of CH₃COOH is 0.1 M. We can use the formula for weak acid dissociation:
\[ K_a = \frac{[\text{H}^+][\text{CH}_3\text{COO}^-]}{[\text{CH}_3\text{COOH}]} \]
Assuming \(x\) is the concentration of \(\text{H}^+\) produced from CH₃COOH, we can write:
\[ K_a = \frac{x^2}{0.1 - x} \approx \frac{x^2}{0.1} \quad (\text{since } x \text{ is small}) \]
\[ 1.8 \times 10^{-5} = \frac{x^2}{0.1} \]
\[ x^2 = 1.8 \times 10^{-6} \]
\[ x = \sqrt{1.8 \times 10^{-6}} \approx 0.00134 \text{ M} \]

### Step 3: Calculate total \(\text{H}^+\) concentration
- The total concentration of \(\text{H}^+\) in the mixture is the sum of contributions from both acids:
\[ [\text{H}^+]_{\text{total}} = [\text{H}^+]_{\text{HCl}} + [\text{H}^+]_{\text{CH}_3\text{COOH}} = 0.01 + 0.00134 \approx 0.01134 \text{ M} \]

### Step 4: Calculate the pH
- The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
\[ \text{pH} = -\log(0.01134) \approx 1.87 \]

### Step 5: Approximate the pH
- Since the question asks for an approximate value, we can round this to:
\[ \text{pH} \approx 1.87 \quad \text{(which is approximately 2)} \]

### Final Answer
The pH of the mixture of 0.01 M HCl and 0.1 M CH₃COOH is approximately **1.87**.

---