Calculate the `pH` of a solution of given mixture.
`(2g CH_(3)COOH +3g CH_(3) COONa)` in `100mL` of mixture.

Explanation

To calculate the pH of the given mixture of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa), we can use the Henderson-Hasselbalch equation:

\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

### Step-by-Step Solution:

**Step 1: Calculate the number of moles of CH₃COOH (acetic acid).**

- Given mass of acetic acid (CH₃COOH) = 2 g
- Molar mass of acetic acid = 60 g/mol

\[ \text{Moles of CH₃COOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \, \text{g}}{60 \, \text{g/mol}} = 0.0333 \, \text{mol} \]

**Step 2: Calculate the concentration of CH₃COOH in the solution.**

- Volume of the solution = 100 mL = 0.1 L

\[ [\text{CH₃COOH}] = \frac{\text{moles}}{\text{volume}} = \frac{0.0333 \, \text{mol}}{0.1 \, \text{L}} = 0.333 \, \text{M} \]

**Step 3: Calculate the number of moles of CH₃COONa (sodium acetate).**

- Given mass of sodium acetate (CH₃COONa) = 3 g
- Molar mass of sodium acetate = 82 g/mol

\[ \text{Moles of CH₃COONa} = \frac{3 \, \text{g}}{82 \, \text{g/mol}} = 0.0366 \, \text{mol} \]

**Step 4: Calculate the concentration of CH₃COONa in the solution.**

\[ [\text{CH₃COONa}] = \frac{\text{moles}}{\text{volume}} = \frac{0.0366 \, \text{mol}}{0.1 \, \text{L}} = 0.366 \, \text{M} \]

**Step 5: Use the Henderson-Hasselbalch equation to calculate pH.**

- Given pKa of acetic acid = 4.74

\[ \text{pH} = 4.74 + \log \left( \frac{[\text{CH₃COONa}]}{[\text{CH₃COOH}]} \right) \]

Substituting the concentrations:

\[ \text{pH} = 4.74 + \log \left( \frac{0.366}{0.333} \right) \]

**Step 6: Calculate the logarithmic term.**

\[ \log \left( \frac{0.366}{0.333} \right) = \log(1.099) \approx 0.041 \]

**Step 7: Final calculation of pH.**

\[ \text{pH} = 4.74 + 0.041 = 4.781 \]

### Final Answer:
The pH of the solution is approximately **4.78**.

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