How much limestone in kilograms is required to completely neutralize a 3.8 x 109 L lake with a pH of 5.5?

Answer 1

pH 5.5 is what acidity in mol/L?
pH = -log(H^+)
How many mols H^+ in the lake? That's ?mol/L x 3.8E9L = ?mols H^+.

2H^+ + CaCO3 ==> Ca^2+ + CO2 + H2O
? mols H^+ x 1/2 = mols CaCO3.
Convert mols CaCO3 to kg.