You are provided with anhydrous Na2Co2 of 0.52g which was dissolved in 100ml of distilled water 25ml of the stock solution of Na2Co3 prepared was titrated against HCl to arrive at average titre value of 20cm calculate

Question

The concentration of the HCL in mol\dm3
The concentration of the Na2Co3 in mol\dm3
The number of miles of Na2Co3 in the stock solution
The number of miles of Na2Co3

Answer 1

To solve the problem, we need to follow a series of steps:

1. Calculate the molar mass of Na2CO3.
2. Determine the concentration of Na2CO3 in the stock solution.
3. Calculate the number of moles of Na2CO3 in the stock solution.
4. Find the concentration of HCl solution based on the titration data.

Step 1: Calculate the Molar Mass of Na2CO3

The molar mass of Na2CO3 can be calculated using the atomic masses of its constituent elements:

• Sodium (Na): approx. 23 g/mol
• Carbon (C): approx. 12 g/mol
• Oxygen (O): approx. 16 g/mol

\[ \text{Molar mass of Na}_2\text{CO}_3 = 2 \times 23 + 12 + 3 \times 16 = 46 + 12 + 48 = 106 \text{ g/mol} \]

Step 2: Calculate the Number of Moles of Na2CO3 in 0.52 g

Using the molar mass, we will calculate the moles of Na2CO3 in 0.52 g.

\[ \text{Number of moles of Na}_2\text{CO}_3 = \frac{\text{Mass}}{\text{Molar mass}} = \frac{0.52 \text{ g}}{106 \text{ g/mol}} \approx 0.00491 \text{ mol} \]

Step 3: Calculate the Concentration of Na2CO3 in the Stock Solution

The stock solution of Na2CO3 was prepared by dissolving the above moles in 100 mL (0.1 L) of distilled water.

\[ \text{Concentration of Na}_2\text{CO}_3 = \frac{\text{Number of moles}}{\text{Volume (in dm}^3\text{)}} = \frac{0.00491 \text{ mol}}{0.1 \text{ dm}^3} \approx 0.0491 \text{ mol/dm}^3 \]

Step 4: Calculate the Number of Moles of Na2CO3 in the 25 mL Used for Titration

Since only 25 mL of the stock solution was used for titration, we calculate the number of moles in that volume.

\[ \text{Volume of aliquot} = 25 \text{ mL} = 0.025 \text{ dm}^3 \] \[ \text{Number of moles of Na}_2\text{CO}_3 \text{ in 25 mL} = 0.0491 \text{ mol/dm}^3 \times 0.025 \text{ dm}^3 \approx 0.0012275 \text{ mol} \]

Step 5: Calculate Concentration of HCl Based on Titration Data

The reaction between Na2CO3 and HCl can be represented as:

\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]

From the equation, we see that 1 mole of Na2CO3 reacts with 2 moles of HCl, which means for every mole of Na2CO3, we require 2 moles of HCl.

For the moles of Na2CO3 titrated (0.0012275 mol):

\[ \text{Moles of HCl used} = 2 \times 0.0012275 = 0.002455 \text{ mol} \]

Step 6: Calculate the Concentration of HCl

The average titre value of HCl used in the titration is 20 cm³, which must be converted to dm³:

\[ \text{Volume of HCl} = 20 \text{ cm}^3 = 0.020 \text{ dm}^3 \]

Now we can find the concentration of HCl:

\[ \text{Concentration of HCl} = \frac{\text{Number of moles of HCl}}{\text{Volume of HCl}} = \frac{0.002455 \text{ mol}}{0.020 \text{ dm}^3} \approx 0.12275 \text{ mol/dm}^3 \]

Summary of Results:

• The concentration of Na2CO3 in the stock solution is approximately 0.0491 mol/dm³.
• The number of moles of Na2CO3 in the stock solution is approximately 0.00491 mol.
• The number of moles of Na2CO3 in the 25 mL aliquot is approximately 0.0012275 mol.
• The concentration of HCl is approximately 0.12275 mol/dm³.