`18mL` of mixture of `CH_(3)COOH` and `CH_(3)COONa` required `6mL` of `0.1M NaOH` for neutralisation of the acid `12mL` of `0.1M HCl` for reaction with salt, separately. If `pK_(a)` of the acid is `4.75`, what is the `pH` of the mixture

Options

A. `4.5`
B. `4.6`
C. `4.75`
D. `5.05`

Explanation

To solve the problem, we need to determine the pH of a mixture of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa) after neutralization with NaOH and HCl. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the millimoles of NaOH used for neutralizing the acid
- Volume of NaOH = 6 mL
- Molarity of NaOH = 0.1 M

Using the formula:
\[ \text{Millimoles of NaOH} = \text{Volume (mL)} \times \text{Molarity (M)} \]
\[ \text{Millimoles of NaOH} = 6 \, \text{mL} \times 0.1 \, \text{M} = 0.6 \, \text{mmol} \]

### Step 2: Calculate the millimoles of HCl used for neutralizing the salt
- Volume of HCl = 12 mL
- Molarity of HCl = 0.1 M

Using the same formula:
\[ \text{Millimoles of HCl} = 12 \, \text{mL} \times 0.1 \, \text{M} = 1.2 \, \text{mmol} \]

### Step 3: Identify the millimoles of acetic acid and sodium acetate
From the neutralization reactions:
- Millimoles of CH₃COOH = Millimoles of NaOH = 0.6 mmol
- Millimoles of CH₃COONa = Millimoles of HCl = 1.2 mmol

### Step 4: Calculate the concentrations of CH₃COOH and CH₃COONa in the mixture
The total volume of the mixture is 18 mL.

- Concentration of CH₃COOH:
\[ \text{Concentration of CH₃COOH} = \frac{0.6 \, \text{mmol}}{18 \, \text{mL}} = \frac{0.6}{18} \, \text{M} = 0.0333 \, \text{M} \]

- Concentration of CH₃COONa:
\[ \text{Concentration of CH₃COONa} = \frac{1.2 \, \text{mmol}}{18 \, \text{mL}} = \frac{1.2}{18} \, \text{M} = 0.0667 \, \text{M} \]

### Step 5: Use the Henderson-Hasselbalch equation to find the pH
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Where:
- pKₐ of acetic acid = 4.75
- \([\text{Salt}] = [\text{CH}_3\text{COONa}] = 0.0667 \, \text{M}\)
- \([\text{Acid}] = [\text{CH}_3\text{COOH}] = 0.0333 \, \text{M}\)

Substituting the values:
\[ \text{pH} = 4.75 + \log\left(\frac{0.0667}{0.0333}\right) \]
\[ \text{pH} = 4.75 + \log(2) \]
Using \(\log(2) \approx 0.301\):
\[ \text{pH} = 4.75 + 0.301 = 5.051 \approx 5.05 \]

### Final Answer:
The pH of the mixture is **5.05**.

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