What is the `pH` of the solution when `0.20 mol` of `HCI` is added to `1L` of a solution containing
a. `1M` each of acetic acid and acetate ion.
b. `0.1M`each of aceta acid and acetate ion.
Assume the total volume is `1L. K_(a)` for acetic acid is `1.8 xx 10^(-5)`.

Explanation

To solve the problem, we will calculate the pH of the solution after adding 0.20 mol of HCl to two different buffer solutions containing acetic acid and acetate ion.

### Part a: 1M Acetic Acid and 1M Acetate Ion

1. **Initial Concentrations**:
- Acetic Acid (CH₃COOH) = 1 M
- Acetate Ion (CH₃COO⁻) = 1 M

2. **Adding HCl**:
- When 0.20 mol of HCl is added, it dissociates completely into H⁺ and Cl⁻ ions.
- Thus, [H⁺] = 0.20 M.

3. **Reaction**:
- H⁺ from HCl will react with acetate ion (CH₃COO⁻) to form acetic acid (CH₃COOH).
- The reaction can be represented as:
\[ \text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH} \]

4. **Changes in Concentration**:
- Acetic Acid will increase by 0.20 M.
- Acetate Ion will decrease by 0.20 M.
- New concentrations:
- Acetic Acid = 1 M + 0.20 M = 1.20 M
- Acetate Ion = 1 M - 0.20 M = 0.80 M

5. **Calculating pKa**:
- Given \( K_a \) for acetic acid = \( 1.8 \times 10^{-5} \)
- Calculate \( pK_a \):
\[ pK_a = -\log(1.8 \times 10^{-5}) \approx 4.744 \]

6. **Using the Henderson-Hasselbalch Equation**:
- The pH is calculated using the formula:
\[ pH = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
- Substitute the values:
\[ pH = 4.744 + \log\left(\frac{0.80}{1.20}\right) \]
- Calculate the logarithm:
\[ \log\left(\frac{0.80}{1.20}\right) = \log(0.6667) \approx -0.176 \]
- Finally, calculate pH:
\[ pH = 4.744 - 0.176 \approx 4.568 \]

### Part b: 0.1M Acetic Acid and 0.1M Acetate Ion

1. **Initial Concentrations**:
- Acetic Acid (CH₃COOH) = 0.1 M
- Acetate Ion (CH₃COO⁻) = 0.1 M

2. **Adding HCl**:
- Again, [H⁺] = 0.20 M from HCl.

3. **Reaction**:
- H⁺ will react with acetate ion (CH₃COO⁻):
\[ \text{CH}_3\text{COO}^- + \text{H}^+ \rightarrow \text{CH}_3\text{COOH} \]

4. **Changes in Concentration**:
- Acetic Acid will increase by 0.1 M (the amount of acetate ion available).
- Acetate Ion will decrease by 0.1 M.
- New concentrations:
- Acetic Acid = 0.1 M + 0.1 M = 0.2 M
- Acetate Ion = 0.1 M - 0.1 M = 0 M

5. **Final Concentration of H⁺**:
- Since all acetate ions are consumed, the remaining H⁺ concentration is:
\[ [H^+] = 0.20 M - 0.10 M = 0.10 M \]

6. **Calculating pH**:
- pH is calculated as:
\[ pH = -\log(0.10) = 1 \]

### Final Answers:
- **Part a**: pH ≈ 4.568
- **Part b**: pH = 1