Question

How much limestone (in kg) is required to completely neutralize a 4.4 billion liter lake with a pH of 5.7?

Answer 1

from the pH find the concentration of H3O+

[H3O+] = 10^-pH = 10^-5.7 = 2 x 10^-6 mol /L

4.4 billion liters means 4.4 x 10⁹ liters

moles of H3O+ = MOlarity v volume = 2 x 10^-6 mol /L x 4.4 x 10⁹ L = 8800 mol

let see the balanced eqation

CaCO3 + 2H3O(+) ------> Ca(2+) + H2CO3 +2H2O

from the 1 mole of CaCO3 required 2 mole of H3O + or

1 mole of H3O+ required 1/2 mole of CaCO3 accordingly

8800 mol of H3O+ required 8800 / 2 = 4400 mol of CaCO3

mass of CaCO3 = moles x molar mass = 4400 x 100 g/mol

= 440000 grams convert in to kg

= 440000/1000

= 440 kg