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: Understand the Neutralization Reaction
Citric acid (H₃C₆H₅O₇) is a triprotic acid, meaning it can donate three protons (H⁺ ions). Sodium hydroxide (NaOH) is a strong base that can neutralize these protons. The balanced reaction can be written as:

\[ \text{H}_3\text{C}_6\text{H}_5\text{O}_7 + 3 \text{NaOH} \rightarrow \text{Na}_3\text{C}_6\text{H}_5\text{O}_7 + 3 \text{H}_2\text{O} \]

This indicates that one mole of citric acid reacts with three moles of NaOH.

### Step 2: Calculate the Milliequivalents of NaOH
To find the milliequivalents of NaOH, we can use the formula:

\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (mL)} \]

For NaOH:
- Normality (N) = Molarity (M) because NaOH has one hydroxide ion (OH⁻).
- Molarity = 0.312 M
- Volume = 35.6 mL

Calculating the milliequivalents of NaOH:

\[ \text{Milliequivalents of NaOH} = 0.312 \, \text{N} \times 35.6 \, \text{mL} = 11.1392 \, \text{meq} \]

### Step 3: Relate Milliequivalents of Citric Acid to NaOH
Since citric acid is triprotic, the milliequivalents of citric acid will equal the milliequivalents of NaOH used in the reaction:

\[ \text{Milliequivalents of Citric Acid} = \text{Milliequivalents of NaOH} = 11.1392 \, \text{meq} \]

### Step 4: Calculate the Molarity of Citric Acid
Using the formula for milliequivalents:

\[ \text{Milliequivalents} = \text{Normality} \times \text{Volume (mL)} \]

We can express the normality of citric acid in terms of its molarity (M) and its valency factor (3):

\[ 11.1392 \, \text{meq} = 3 \times M \times 10 \, \text{mL} \]

Now, solving for M:

\[ M = \frac{11.1392 \, \text{meq}}{3 \times 10 \, \text{mL}} = \frac{11.1392}{30} = 0.3713 \, \text{M} \]

### Step 5: Final Result
The molarity of the citric acid solution is approximately:

\[ \text{Molarity of Citric Acid} \approx 0.3713 \, \text{M} \]

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

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