Lakes that have been acidified by acid rain can be neutralized by liming, the addition of limestone (CaCO3).
How much limestone in kilograms is required to completely neutralize a 5.8×109 L lake with a pH of 5.6?
thanks
2 answers
I meant 5.8×10^9 L
CaCO3 + 2H^+ ==> Ca^+2 + H2O + CO2
What is the H^+?
pH = 5.6; therefore, H^+ = 2.51E-6M
How many mols H^+ do you have? That's M x L = 2.51E-6M x 5.8E9L = 14,569 moles.
How many moles CaCO3 will it take? It will take 1/2 of the H^+; therefore. 14,569/2 = 7,284 moles CaCO3.
How many grams is that?
g = moles x molar mass = ??
Convert that to kg.
Check my numbers.
What is the H^+?
pH = 5.6; therefore, H^+ = 2.51E-6M
How many mols H^+ do you have? That's M x L = 2.51E-6M x 5.8E9L = 14,569 moles.
How many moles CaCO3 will it take? It will take 1/2 of the H^+; therefore. 14,569/2 = 7,284 moles CaCO3.
How many grams is that?
g = moles x molar mass = ??
Convert that to kg.
Check my numbers.
2 answers