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 `0.1 M NaOH`. By how many units does the `pH` change from the start to the stage when the `HCl` is almost completely neutralized and acidic acid remains unreacted? `K_(a)` for acetic acid `=2xx10^(-5)`.

Explanation

To solve the problem step by step, we will analyze the titration of a mixture of HCl (a strong acid) and acetic acid (a weak acid) with NaOH (a strong base).

### Step 1: Determine Initial Conditions
We have a mixture of HCl and acetic acid (CH₃COOH), both at a concentration of 0.1 M, and a total volume of 20 mL. Since HCl is a strong acid, it will completely dissociate in solution, while CH₃COOH will only partially dissociate.

- **Initial concentration of H⁺ from HCl**:
\[ [H^+] = 0.1 \, M \]
- **Initial pH**:
\[ \text{pH} = -\log[H^+] = -\log(0.1) = 1 \]

### Step 2: Titration with NaOH
We are titrating this solution with 0.1 M NaOH. The volume of NaOH added will be equal to that of HCl to neutralize it completely.

- **Volume of NaOH added**: 20 mL (to neutralize 20 mL of HCl)
- **Total volume after titration**:
\[ 20 \, \text{mL (HCl)} + 20 \, \text{mL (NaOH)} = 40 \, \text{mL} \]

### Step 3: Calculate Concentration of Acetic Acid After Titration
After neutralizing HCl, we will have only the acetic acid remaining in the solution.

- **Moles of acetic acid initially**:
\[ \text{Moles of CH₃COOH} = 0.1 \, M \times 0.020 \, L = 0.002 \, \text{moles} \]
- **Concentration of acetic acid after titration**:
\[ [CH₃COOH] = \frac{0.002 \, \text{moles}}{0.040 \, L} = 0.05 \, M \]

### Step 4: Calculate Hydrogen Ion Concentration from Acetic Acid
Using the acid dissociation constant \( K_a \) for acetic acid, we can find the concentration of H⁺ ions produced from acetic acid.

- **Given**:
\[ K_a = 2 \times 10^{-5} \]
- **Using the formula**:
\[ [H^+] = \sqrt{K_a \times [CH₃COOH]} = \sqrt{2 \times 10^{-5} \times 0.05} \]

Calculating:
\[ [H^+] = \sqrt{1 \times 10^{-6}} = 10^{-3} \, M \]

### Step 5: Calculate Final pH
Now we can calculate the pH of the solution after HCl has been neutralized.

- **Final pH**:
\[ \text{pH} = -\log[H^+] = -\log(10^{-3}) = 3 \]

### Step 6: Calculate Change in pH
Finally, we need to find the change in pH from the initial state to the point where HCl is almost completely neutralized.

- **Change in pH**:
\[ \Delta \text{pH} = \text{Final pH} - \text{Initial pH} = 3 - 1 = 2 \]

### Conclusion
The change in pH from the start to the stage when HCl is almost completely neutralized is **2 units**.