Answer to Question #266243 in Chemistry for Royad

Question

50% pure 1 g CaCO3 is dissolved in 25 ml H2SO4 solution. A further 30 ml of 0.5 mol NaOH is required for complete dissolution of the mixture. So what was the concentration of H2SO4?

Answer 1

CaCO₃ + H₂SO4 = H₂CO₃ + CaSO₄

H₂SO₄ + 2NaOH = Na₂SO₄ + 2H₂O

1) Active mass of CaCO3 is: 1 g x 0.5 = 0.5 g.

n (CaCO₃) = m / M

M (CaCO₃) = 100.1 g/mol

n (CaCO₃) = 0.5 / 100.1 = 0.005 mol

According to the equation, n (H₂SO₄) = n (CaCO₃).

2) C_M = n / V

n (NaOH) = C_M x V =0.5 x 0.03 = 0.015 mol

According to the equation, n (H₂SO₄) = 1/2 x n (NaOH) = 1/2 x 0.015 = 0.0075 mol.

As H₂SO₄ reacted with both CaCO₃ and NaOH, the amount of used H₂SO₄ is:

n (H₂SO₄) = 0.005 + 0.0075 = 0.0125 mol

C_M (H₂SO₄) = n / V = 0.0125 / 0.025 = 0.5 M