How much limestone (in kg) is required to completely neutralize a 4.5 billion liter lake with a pH of 5.3

1105.7kg of limestone must be added

Explanation:

A solution is neutralized when pH = 7.0. That is [H+] = 1x10⁻⁷M.

The initial [H+] is 10^-5.3 = 5.01x10⁻⁶M.

That means we need to neutralize:

5.01x10⁻⁶M - 1x10⁻⁷M = 4.91x10⁻⁶M

The moles of H+ in 4.5 billion liters = 4.5x10⁹L are:

4.91x10⁻⁶M * (4.5x10⁹L) = 22095 moles of H+

That reacts with limestone (CaCO3), as follows:

2H+ + CaCO3 → H2O + CO2 + Ca²⁺

Moles of limestone needed to neutralize the lake are:

22095 moles of H⁺ * (1 mole Limestone / 2 moles H⁺) = 11047.5 moles of limestone.

Molar mass of limestone is 100.09g/mol. Mass of limestone we need to add are:

11047.5 moles of limestone * (100.09g / mol) = 1.106x10⁶g of limestone. In kg:

1.106x10⁶g of limestone * (1kg / 1000g) =

1105.7kg of limestone must be added