Asked by amy
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
How much limestone in kilograms is required to completely neutralize a 5.8×109 L lake with a pH of 5.6?
thanks
Answers
Answered by
amy
I meant 5.8×10^9 L
Answered by
DrBob222
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.
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