What volume of 10 % (w/v) solution of `Na_(2)CO_(3)` will be required to neturalise 100 mL of HCl solution containing 3.65 g of HCl ?

Explanation

To solve the problem of determining the volume of a 10% (w/v) solution of Na₂CO₃ required to neutralize 100 mL of HCl solution containing 3.65 g of HCl, we can follow these steps:

### Step 1: Write the Neutralization Reaction
The balanced chemical reaction for the neutralization of Na₂CO₃ with HCl is:
\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]

### Step 2: Calculate Moles of HCl
First, we need to calculate the number of moles of HCl present in the solution. The molar mass of HCl is 36.5 g/mol.

\[ \text{Moles of HCl} = \frac{\text{mass of HCl}}{\text{molar mass of HCl}} = \frac{3.65 \text{ g}}{36.5 \text{ g/mol}} = 0.1 \text{ moles} \]

### Step 3: Determine Moles of Na₂CO₃ Required
From the balanced equation, we see that 2 moles of HCl react with 1 mole of Na₂CO₃. Therefore, the moles of Na₂CO₃ needed to neutralize 0.1 moles of HCl can be calculated as follows:

\[ \text{Moles of Na}_2\text{CO}_3 = \frac{0.1 \text{ moles of HCl}}{2} = 0.05 \text{ moles} \]

### Step 4: Calculate Molarity of Na₂CO₃ Solution
We know that the 10% (w/v) solution of Na₂CO₃ means there are 10 grams of Na₂CO₃ in 100 mL of solution. The molar mass of Na₂CO₃ is 106 g/mol.

\[ \text{Moles of Na}_2\text{CO}_3 = \frac{10 \text{ g}}{106 \text{ g/mol}} \approx 0.0943 \text{ moles} \]

The volume of the solution in liters is 0.1 L (since 100 mL = 0.1 L).

\[ \text{Molarity of Na}_2\text{CO}_3 = \frac{\text{moles of Na}_2\text{CO}_3}{\text{volume of solution in L}} = \frac{0.0943 \text{ moles}}{0.1 \text{ L}} = 0.943 \text{ M} \]

### Step 5: Calculate Volume of Na₂CO₃ Required
Using the molarity and the number of moles of Na₂CO₃ required, we can find the volume needed:

\[ \text{Volume} = \frac{\text{moles of Na}_2\text{CO}_3}{\text{molarity}} = \frac{0.05 \text{ moles}}{0.943 \text{ M}} \approx 0.053 \text{ L} \]

To convert this to milliliters:

\[ \text{Volume in mL} = 0.053 \text{ L} \times 1000 = 53.19 \text{ mL} \]

### Final Answer
The volume of 10% (w/v) solution of Na₂CO₃ required to neutralize 100 mL of HCl solution containing 3.65 g of HCl is approximately **53.19 mL**.

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