Answer to Question #342173 in General Chemistry for Kian

Question

If 1.25 g of sodium carbonate was dissolved in water and the solution titrated with 22.7 ml of hydrochloric acid to a medal orange end point, calculate (i) the molarity and (ii) the percent by mass (density of sution = 1.02 g/ml) of the hydrochloric acid solution

Answer 1

Na₂CO₃ + 2 HCl => 2 NaCl + CO₂ + H₂O

1. According to the above reaction, 1 mole (106 g) of Na₂CO₃ reacts with 2 moles of HCl. we have 1.25 g of Na₂CO₃ and we can find how many moles of HCl react with it:

106 gr —> 2 mol

1.25 gr —> x mol

x = 1.25 * 2 / 106 = 0.0236 mol.

2. We can now find the molar concentration.To do this, divide the amount of solute (HCl) by the volume of solution (L):

C_M = n / V = 0.0236 mol / 0.0227 L ≈ 1.04 mol/l.

3. We can now find the percentage concentration of the solution. To do this, we need to divide the mass of solute (HCl) by the mass of the solution and multiply by 100:

Percent by mass= Mass of Solute/ Mass of Solution * 100

Massa of Solution = V (volume) * p (density)=22.7 * 1.02 = 23.154 gr .

Percent by mass = 0.86 gr / 23.154 gr * 100= 3.71 %

ANSWER: Molarity = 1.04 mol/l

Percent by mass = 3.71% .