Lakes that have been acidified by acid rain can be neutralized by limiting the addition of limestone how much limestone is required to completely neutralize a 4.3 billion liter lake with a ph of 5.5

From the given pH, we calculate the concentration of H+:

[H+] = 10^-pH = 10^-5.5

We then use the volume to solve for the number of moles of H+:

moles H + = 10^-5.5M * 4.3x10^9 L = 13598 moles

From the balanced equation of the neutralization of hydrogen ion by limestone written as

CaCO3 (s) + 2H + (aq) → Ca2 + (aq) + H2CO3 (aq)

we use the mole ratio of limestone CaCO3 and H + from their coefficients, which is 1 mole of CaCO3 is to react with 2 moles of H+, to compute for the mass of the limestone:

mass CaCO3 = 13598mol H + (1mol CaCO3/2mol H+)

(100.0869g CaCO3/1mol CaCO3) (1kg/1000g)

= 680 kg