A certain mixture of HCl and `CH_3-COOH` is `0.1 M` in each of the acids.20 ml of this solution is titrated against 0.1M NaOH.By how many units does the pH changes form the start to the stage when the HCl is almost completely neutralised ? `K_a` for acetic acid=`1.8xx10^(-5)`.

Options

A. 2.03
B. 0.775
C. 2.325
D. 3.175

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the initial pH of the solution
The mixture contains 0.1 M HCl (a strong acid) and 0.1 M acetic acid (a weak acid). Since HCl is a strong acid, it completely dissociates in solution, contributing to the initial concentration of H⁺ ions.

- **Initial concentration of H⁺ from HCl**: 0.1 M
- **Initial concentration of H⁺ from acetic acid**: Negligible (since it is a weak acid)

Thus, the initial concentration of H⁺ ions is:
\[ [H^+] = 0.1 \, \text{M} \]

Using the formula for pH:
\[ \text{pH} = -\log[H^+] \]
\[ \text{pH} = -\log(0.1) = 1 \]

### Step 2: Calculate the total volume after titration
When 20 mL of the acid mixture is titrated with 20 mL of 0.1 M NaOH, the total volume becomes:
\[ \text{Total Volume} = 20 \, \text{mL (acid)} + 20 \, \text{mL (NaOH)} = 40 \, \text{mL} \]

### Step 3: Determine the concentration of acetic acid after mixing
Initially, we have 20 mL of 0.1 M acetic acid. After mixing, the concentration of acetic acid in the total volume of 40 mL can be calculated using the dilution formula:
\[ M_1V_1 = M_2V_2 \]
Where:
- \( M_1 = 0.1 \, \text{M} \) (initial concentration)
- \( V_1 = 20 \, \text{mL} \) (initial volume)
- \( V_2 = 40 \, \text{mL} \) (final volume)

Calculating \( M_2 \):
\[ M_2 = \frac{M_1V_1}{V_2} = \frac{0.1 \times 20}{40} = 0.05 \, \text{M} \]

### Step 4: Calculate the concentration of H⁺ ions from acetic acid
Using the dissociation constant \( K_a \) for acetic acid:
\[ K_a = 1.8 \times 10^{-5} \]
The concentration of H⁺ ions from acetic acid can be calculated using the formula:
\[ [H^+] = \sqrt{K_a \times [CH_3COOH]} \]
Substituting the values:
\[ [H^+] = \sqrt{1.8 \times 10^{-5} \times 0.05} \]
\[ [H^+] = \sqrt{9.0 \times 10^{-7}} \]
\[ [H^+] \approx 9.48 \times 10^{-4} \, \text{M} \]

### Step 5: Calculate the pH after neutralization of HCl
Now, we can calculate the pH after HCl is almost completely neutralized:
\[ \text{pH} = -\log(9.48 \times 10^{-4}) \]
Calculating this gives:
\[ \text{pH} \approx 3.03 \]

### Step 6: Calculate the change in pH
The change in pH from the initial pH to the pH after neutralization is:
\[ \Delta \text{pH} = \text{pH}_{\text{final}} - \text{pH}_{\text{initial}} \]
\[ \Delta \text{pH} = 3.03 - 1 = 2.03 \]

### Final Answer
The change in pH from the start to the stage when HCl is almost completely neutralized is **2.03**.

---