Question

A contains 0.25 mole of HCl per 250 cm³ of solution. B is a solution of Na2CO3. In a titration experiment, 25 cm³ of B required 23.50 cm³ of A for a complete reaction. Calculate:

i. Concentration of B in mol/dm³
ii. Number of sodium ions present in 1 dm³ of solution B
iii. Mass of Na2CO3 required to prepare 200 cm³ of solution B
iv. Name the suitable indicator for the titration and the color at the end of the titration
v. Describe briefly how water can be obtained from the titration mixture.

Answer 1

i. The concentration of solution B is 5 mol/dm³. ii. There are 0.25 moles of sodium ions present in 1 dm³ of solution B. iii. The mass of Na2CO3 required to prepare 200 cm³ of solution B is 13.25 g. iv. A suitable indicator for the titration is Phenolphthalein, which turns pink at the end. v. Water can be obtained from the titration mixture by evaporation.

Step-by-step explanation:

i. Concentration of B in mol/dm³:

To find the concentration of solution B, we need to use the equation:

• Concentration (mol/dm³) = Moles of solute / Volume of solution (dm³)

In this case, 23.50 cm³ of solution A (HCl) is required to react with 25 cm³ of solution B (Na2CO3). Since the molar ratio between HCl and Na2CO3 is 2:1, we can calculate:

• Moles of Na2CO3 = 0.25 mol HCl × 0.5 = 0.125 mol
• Volume of solution B = 0.025 dm³ (since 25 cm³ = 0.025 dm³)
• Concentration of B = Moles of Na2CO3 / Volume of solution B = 0.125 mol / 0.025 dm³ = 5 mol/dm³

ii. Number of sodium ions present in 1 dm³ of solution B:

The balanced chemical equation between Na2CO3 and HCl is:

• Na2CO3 + 2HCl → 2NaCl + H2O + CO2

From the equation, we can see that 1 mole of Na2CO3 produces 2 moles of NaCl, which means that the number of sodium ions in 1 dm³ of solution B is twice the number of moles of Na2CO3:

• Number of sodium ions = 2 × 0.125 mol = 0.25 mol

iii. Mass of Na2CO3 required to prepare 200 cm³ of solution B:

We can use the same equation as in part i to find the mass of Na2CO3:

• Mass of Na2CO3 = Moles of Na2CO3 × Molar mass of Na2CO3

The molar mass of Na2CO3 is:

• Na = 22.99 g/mol, C = 12.01 g/mol, O = 16.00 g/mol
• Molar mass of Na2CO3 = (2 × Na) + C + (3 × O) = (2 × 22.99 g/mol) + 12.01 g/mol + (3 × 16.00 g/mol)
• Molar mass of Na2CO3 = 105.99 g/mol
• Mass of Na2CO3 = Moles of Na2CO3 × Molar mass of Na2CO3 = 0.125 mol × 105.99 g/mol = 13.25 g

iv. Name the suitable indicator for the titration and the color at the end of the titration:

• A suitable indicator for the titration of HCl with Na2CO3 is Phenolphthalein. It changes color from colorless to pink at the end of the titration when the solution becomes slightly basic.

v. Describe briefly how water can be obtained from the titration mixture:

• Water can be obtained from the titration mixture by evaporating the solution to remove the excess reactants and the indicator. The remaining water can then be collected and used as desired.