1.0 L solution is prepared by mixing 61 g benzoic acid `(pK_(a)=4.2)` with 72 g of sodium benzoate and then 300 mL 1.0 M HBr solution was added. The pH of final solution is :

Options

A. `3.6`
B. `3.8`
C. `4.2`
D. `4.8`

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the moles of benzoic acid
The formula to calculate the number of moles is:
\[ \text{Number of moles} = \frac{\text{Given weight}}{\text{Molecular weight}} \]
Given weight of benzoic acid = 61 g
Molecular weight of benzoic acid (C6H5COOH) = 122 g/mol

Calculating moles:
\[ \text{Moles of benzoic acid} = \frac{61 \text{ g}}{122 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 2: Calculate the moles of sodium benzoate
The molecular weight of sodium benzoate (C6H5COONa) is 144 g/mol.
Given weight of sodium benzoate = 72 g

Calculating moles:
\[ \text{Moles of sodium benzoate} = \frac{72 \text{ g}}{144 \text{ g/mol}} = 0.5 \text{ moles} \]

### Step 3: Calculate the moles of HBr added
The concentration of HBr = 1.0 M and volume = 300 mL = 0.3 L.

Calculating moles:
\[ \text{Moles of HBr} = \text{Concentration} \times \text{Volume} = 1.0 \text{ M} \times 0.3 \text{ L} = 0.3 \text{ moles} \]

### Step 4: Determine the reaction between HBr and sodium benzoate
HBr is a strong acid and will dissociate completely:
\[ \text{HBr} \rightarrow \text{H}^+ + \text{Br}^- \]
The sodium benzoate will dissociate into:
\[ \text{C}_6\text{H}_5\text{COONa} \rightarrow \text{C}_6\text{H}_5\text{COO}^- + \text{Na}^+ \]

### Step 5: Calculate the remaining moles after the reaction
Initially, we have:
- Moles of sodium benzoate = 0.5
- Moles of HBr = 0.3

The HBr will react with the benzoate ions:
\[ \text{C}_6\text{H}_5\text{COO}^- + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{COOH} \]
After the reaction:
- Moles of sodium benzoate remaining = \(0.5 - 0.3 = 0.2\) moles
- Moles of benzoic acid formed = \(0.3\) moles (from the reaction)

### Step 6: Calculate the total moles of benzoic acid
Total moles of benzoic acid after reaction:
\[ \text{Total moles of benzoic acid} = 0.5 + 0.3 = 0.8 \text{ moles} \]

### Step 7: Use the Henderson-Hasselbalch equation to find pH
The Henderson-Hasselbalch equation is:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Where:
- \(\text{pK}_a = 4.2\)
- \([\text{Salt}] = \text{Sodium benzoate} = 0.2\)
- \([\text{Acid}] = \text{Benzoic acid} = 0.8\)

Calculating pH:
\[ \text{pH} = 4.2 + \log\left(\frac{0.2}{0.8}\right) = 4.2 + \log\left(\frac{1}{4}\right) = 4.2 - 0.6 = 3.6 \]

### Final Answer
The pH of the final solution is **3.6**.

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