Lakes that have been acidified by acid rain HNO3 and H2SO4 c

Lakes that have been acidified by acid rain (HNO3 and H2SO4) can be neutralized by a process called liming, in which limestone (CaCO3) is added to the acidified water.

What mass of limestone (in kg) would be required to completely neutralize a 15.3 billion-liter lake that is 1.9×10⁵ M in H2SO4 and 8.6×10⁶ M in HNO3?

Solution

Write the balanced chemical equations for the reactions of CaCO₃ with H₂SO₄ and HNO₃.

CaCO₃ + H₂SO₄ ---------> CaSO₄ + CO₂ + H₂O …..(1)

CaCO₃ + 2 HNO₃ --------> Ca(NO₃)₂ + CO₂ + H₂O ……(2)

As per the stoichiometric equations,

1 mole CaCO₃ = 1 mole H₂SO₄ = 2 moles HNO₃.

Moles of H₂SO₄ in 15.3 billion-liter lake water = (15.3 billion-liters)*(1.0*10⁹ L/1 billion0liters)*(1.9*10^-5 M) = 290700 moles.

Moles of CaCO₃ corresponding to 290700 moleS H₂SO₄ = (290700 mole H₂SO₄)*(1 mole CaCO₃/1 mole H₂SO₄) = 290700 moles.

Moles of HNO₃ in 15.3 billion-liter lake water = (15.3 billion-liters)*(1.0*10⁹ L/1 billion0liters)*(8.6*10^-6 M) = 131580 moles.

Moles of CaCO₃ corresponding to 131580 moleS HNO₃ = (131580 mole HNO₃)*(1 mole CaCO₃/2 mole HNO₃) = 65790 moles.

Total moles of CaCO₃ required = (290700 + 65790) moles = 356490 moles.

Molar mass of CaCO₃ = (1*40.078 + 1*12.01 + 3*15.9994) g/mol = 100.0862 g/mol.

Mass of CaCO₃ required in kilograms = (356490 moles)*(100.0862 g/mol)*(1 kg/1000 g) = 35679.73 kg (ans).