10 mL citric acid `(H_3 C_6H_5O_7)` is neutralised completely by 35.6 mL of 0.312 M NaON solution. The molarity of the solution of citric acid is

Options

A. 1.11 M
B. 0.45 M
C. 0.11 M
D. 0.37 M

Answer

To find the molarity of the citric acid solution, we can follow these steps:

### Step 1: Identify the given information
- Volume of citric acid solution (V1) = 10 mL
- Volume of NaOH solution (V2) = 35.6 mL
- Molarity of NaOH solution (M2) = 0.312 M
- The formula for citric acid is H₃C₆H₅O₇.

### Step 2: Determine the n-factor for citric acid and NaOH
- The n-factor for NaOH (N2) = 1 (since it can donate one hydroxide ion).
- The n-factor for citric acid (N1) = 3 (since it can donate three protons).

### Step 3: Use the neutralization reaction relationship
According to the principle of neutralization:
\[ n_1 \times V_1 = n_2 \times V_2 \]
Where:
- \( n_1 \) = n-factor of citric acid
- \( V_1 \) = volume of citric acid
- \( n_2 \) = n-factor of NaOH
- \( V_2 \) = volume of NaOH

### Step 4: Rewrite the equation in terms of molarity
Since we need to find the molarity (M1) of citric acid, we can express it as:
\[ n_1 \times M_1 \times V_1 = n_2 \times M_2 \times V_2 \]
Substituting the known values:
\[ 3 \times M_1 \times 10 \, \text{mL} = 1 \times 0.312 \, \text{M} \times 35.6 \, \text{mL} \]

### Step 5: Solve for M1
Rearranging the equation to solve for M1:
\[ M_1 = \frac{1 \times 0.312 \times 35.6}{3 \times 10} \]
Calculating the right side:
\[ M_1 = \frac{0.312 \times 35.6}{30} \]
\[ M_1 = \frac{11.1392}{30} \]
\[ M_1 = 0.3713 \, \text{M} \]

### Step 6: Round to appropriate significant figures
Rounding to two decimal places, we get:
\[ M_1 \approx 0.37 \, \text{M} \]

### Final Answer
The molarity of the citric acid solution is approximately **0.37 M**.

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