The `pH` of solution, containing `0.1N` HCl and `0.1N` `CH_(3)COOH(K_(a)=2xx10^(-5))` is

Options

A. 1
B. `0.7`
C. `2`
D. `1.3`

Explanation

To find the pH of a solution containing 0.1N HCl and 0.1N CH₃COOH (acetic acid), we can follow these steps:

### Step 1: Understand the Contribution of Each Acid
HCl is a strong acid and will completely dissociate in solution, while CH₃COOH is a weak acid and will only partially dissociate.

### Step 2: Calculate the Contribution of HCl
Since HCl is a strong acid, the concentration of hydrogen ions [H⁺] from HCl will be equal to its normality:
\[ [H^+]_{HCl} = 0.1 \, N \]

### Step 3: Calculate the Contribution of CH₃COOH
For acetic acid (CH₃COOH), we use the dissociation constant \( K_a \) to find the concentration of hydrogen ions it contributes. The dissociation of acetic acid can be represented as:
\[ CH_3COOH \rightleftharpoons H^+ + CH_3COO^- \]

Using the formula for weak acids:
\[ K_a = \frac{[H^+][CH_3COO^-]}{[CH_3COOH]} \]

Assuming \( x \) is the concentration of \( H^+ \) produced by the dissociation of acetic acid:
\[ K_a = \frac{x^2}{0.1 - x} \]

Given \( K_a = 2 \times 10^{-5} \):
\[ 2 \times 10^{-5} = \frac{x^2}{0.1 - x} \]

Since \( K_a \) is very small, we can approximate \( 0.1 - x \approx 0.1 \):
\[ 2 \times 10^{-5} = \frac{x^2}{0.1} \]
\[ x^2 = 2 \times 10^{-6} \]
\[ x = \sqrt{2 \times 10^{-6}} \]
\[ x \approx 1.41 \times 10^{-3} \]

### Step 4: Calculate Total [H⁺] Concentration
Now, we can find the total concentration of hydrogen ions:
\[ [H^+]_{total} = [H^+]_{HCl} + [H^+]_{CH_3COOH} \]
\[ [H^+]_{total} = 0.1 + 1.41 \times 10^{-3} \approx 0.1 \, N \]

### Step 5: Calculate pH
Now we can calculate the pH using the formula:
\[ pH = -\log[H^+] \]
\[ pH = -\log(0.1) = 1 \]

### Conclusion
Thus, the pH of the solution is approximately 1.