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

Dissociation then hydration dissolving

Afraidsugar lol it’s the third one