3 gram of acetic acid is mixed in 250 mL of 0.1 M HCl. This mixture is now diluted to 500 mL. 20 mL of this solution is now taken is another container `1/2` mL of 5M NaOH is added to this. Find the pH of this solution. (`pK_a = 4.74`)

Options

A. (a) 5.22
B. (b) 4.74
C. (c) 6.23
D. (d) 3.47

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Calculate moles of acetic acid (CH₃COOH)
Given:
- Mass of acetic acid = 3 g
- Molar mass of acetic acid (CH₃COOH) = 60 g/mol

**Moles of acetic acid:**
\[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{3 \text{ g}}{60 \text{ g/mol}} = 0.05 \text{ mol} = 50 \text{ mmol} \]

### Step 2: Calculate moles of HCl
Given:
- Volume of HCl = 250 mL
- Molarity of HCl = 0.1 M

**Moles of HCl:**
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} = 0.1 \text{ mol/L} \times 0.250 \text{ L} = 0.025 \text{ mol} = 25 \text{ mmol} \]

### Step 3: Total moles in the diluted solution
The mixture is diluted to 500 mL. We need to find the concentrations of acetic acid and HCl in this diluted solution.

**Concentration of acetic acid in 500 mL:**
\[ \text{Concentration} = \frac{50 \text{ mmol}}{500 \text{ mL}} = 0.1 \text{ mmol/mL} \]

**Concentration of HCl in 500 mL:**
\[ \text{Concentration} = \frac{25 \text{ mmol}}{500 \text{ mL}} = 0.05 \text{ mmol/mL} \]

### Step 4: Calculate moles in 20 mL of the diluted solution
**Moles of acetic acid in 20 mL:**
\[ \text{Moles} = 0.1 \text{ mmol/mL} \times 20 \text{ mL} = 2 \text{ mmol} \]

**Moles of HCl in 20 mL:**
\[ \text{Moles} = 0.05 \text{ mmol/mL} \times 20 \text{ mL} = 1 \text{ mmol} \]

### Step 5: Calculate moles of NaOH added
Given:
- Volume of NaOH = 0.5 mL
- Molarity of NaOH = 5 M

**Moles of NaOH:**
\[ \text{Moles} = \text{Molarity} \times \text{Volume (L)} = 5 \text{ mol/L} \times 0.0005 \text{ L} = 0.0025 \text{ mol} = 2.5 \text{ mmol} \]

### Step 6: Determine the limiting reagent
**Reaction:**
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
- Moles of HCl = 1 mmol
- Moles of NaOH = 2.5 mmol

Since HCl is the limiting reagent, it will completely react with NaOH.

### Step 7: Calculate remaining moles after reaction
After the reaction:
- Moles of HCl left = 1 - 1 = 0 mmol
- Moles of NaOH left = 2.5 - 1 = 1.5 mmol

### Step 8: Remaining moles of acetic acid
The acetic acid does not react with NaOH, so it remains unchanged:
- Moles of acetic acid = 2 mmol

### Step 9: Calculate concentrations for Henderson-Hasselbalch equation
**Concentration of salt (sodium acetate) formed:**
\[ \text{Moles of salt} = 1.5 \text{ mmol} \]

**Concentration of acetic acid:**
\[ \text{Concentration} = \frac{2 \text{ mmol}}{20 \text{ mL}} = 0.1 \text{ mmol/mL} \]

### Step 10: Use the Henderson-Hasselbalch equation
Given:
- \( pK_a = 4.74 \)

**Henderson-Hasselbalch equation:**
\[ \text{pH} = pK_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Substituting the values:
\[ \text{pH} = 4.74 + \log\left(\frac{1.5}{2}\right) \]
Calculating:
\[ \log\left(\frac{1.5}{2}\right) = \log(0.75) \approx -0.1249 \]
Thus,
\[ \text{pH} = 4.74 - 0.1249 \approx 4.6151 \]

### Final Answer
The pH of the solution is approximately **4.62**.

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