400 g Na2CO3 is dissolved in 7500 mL H2O and titrated with H

4.00 g Na2CO3 is dissolved in 75.00 mL H2O and titrated with HCl. 66.0 mL of HCl solution was required. What was the molarity of the HCl solution? Na2CO3(aq) + 2HCl(aq) 2NaCl(aq) + H2O(l) + CO2(g)

Solution

Mass of Na₂CO₃ = 4.00 g.

Molar mass of Na₂CO₃ = 106 g/mol

Moles of Na₂CO₃ = mass / molar mass = 4.00 / 106. = 0.0377 mol

Initial molarity of Na₂CO₃ = moles of Na₂CO₃ / volume of solution in L = 0.0377 / 0.0750 = 0.503 M

Balanced equation,

Na₂CO₃ (aq.) + 2 HCl (aq.) ---------> 2 NaCl (aq.) + H₂O (l) + CO₂ (g)

Using neutralisation formula,

M₁ * V₁ / n₁ = M₂ * V₂ / n₂

0.503 * 75.00 / 1 = M₂ * 66.0 / 2

M₂ = 1.14 M

Molarity of HCl solution = 1.14 M