31.26 ml of 0.165 M solution of `Ca(OH)_(2)` is required to just neutralise 25 ml of citric acid `H_(3)C_(6)H_(5)O_(7)` . Then correct regarding this is/are

Options

A. n-factor of citric acid is 3
B. Molarity of citric acid is 0.138 M
C. Molarity of citric acid is 0.029 M
D. n-factor of citric acid is 2

Answer

To solve the problem, we need to determine the n-factor of citric acid and its molarity based on the information provided about the neutralization reaction with calcium hydroxide, \(Ca(OH)_2\).

### Step 1: Calculate the moles of \(Ca(OH)_2\)

Given:
- Volume of \(Ca(OH)_2\) solution = 31.26 mL = 31.26 × \(10^{-3}\) L
- Molarity of \(Ca(OH)_2\) = 0.165 M

Using the formula for moles:
\[ \text{Moles of } Ca(OH)_2 = \text{Molarity} \times \text{Volume} = 0.165 \, \text{mol/L} \times 31.26 \times 10^{-3} \, \text{L} \]

Calculating:
\[ \text{Moles of } Ca(OH)_2 = 0.165 \times 31.26 \times 10^{-3} = 0.00516 \, \text{mol} \]

### Step 2: Calculate the moles of \(OH^-\) produced

From the dissociation of \(Ca(OH)_2\):
\[ Ca(OH)_2 \rightarrow Ca^{2+} + 2OH^- \]
This means 1 mole of \(Ca(OH)_2\) produces 2 moles of \(OH^-\).

Thus, the moles of \(OH^-\) produced:
\[ \text{Moles of } OH^- = 2 \times \text{Moles of } Ca(OH)_2 = 2 \times 0.00516 = 0.01032 \, \text{mol} \]

### Step 3: Determine the moles of \(H^+\) in citric acid

For neutralization:
\[ \text{Moles of } H^+ = \text{Moles of } OH^- = 0.01032 \, \text{mol} \]

### Step 4: Calculate the n-factor of citric acid

Citric acid \(H_3C_6H_5O_7\) can donate 3 protons (\(H^+\)) in total, thus:
\[ \text{n-factor of citric acid} = 3 \]

### Step 5: Calculate the molarity of citric acid

Given:
- Volume of citric acid = 25 mL = 25 × \(10^{-3}\) L

Using the formula for molarity:
\[ \text{Molarity of citric acid} = \frac{\text{Moles of } H^+}{\text{Volume of citric acid in L}} = \frac{0.01032 \, \text{mol}}{25 \times 10^{-3} \, \text{L}} = 0.4128 \, \text{mol/L} \]

However, since citric acid has an n-factor of 3, we need to adjust the moles of citric acid:
\[ \text{Moles of citric acid} = \frac{\text{Moles of } H^+}{\text{n-factor}} = \frac{0.01032}{3} = 0.00344 \, \text{mol} \]

Now, calculate the molarity:
\[ \text{Molarity of citric acid} = \frac{0.00344 \, \text{mol}}{25 \times 10^{-3} \, \text{L}} = 0.1376 \, \text{mol/L} \approx 0.138 \, \text{mol/L} \]

### Conclusion

Based on the calculations:
1. The n-factor of citric acid is 3.
2. The molarity of citric acid is approximately 0.138 M.

Thus, the correct statements are:
- n-factor of citric acid is 3 (Correct)
- Molarity of citric acid is 0.138 M (Correct)
- Molarity of citric acid is not 0.029 M (Incorrect)
- n-factor of citric acid is not 2 (Incorrect)