Lakes that have been acidified by acid rain can be neutralized by...

Question

Lakes that have been acidified by acid rain can be neutralized by 
liming, the addition of limestone (CaCO3). How much limestone 
(in kg) is required to completely neutralize a 4.3 billion 
liter lake with a pH of 5.5?

Accepted Answer

Hello everyone today, we're giving the following scenario and asked to find the mass of a statue that's lost. So the question says acid rain is a serious environmental problem that damages the ecosystem, destroys materials and has a negative has negative effects on human health. A limestone made buddhist statue in India has endured damaged caused by acid rain. If 1.546 times 10 to the seventh liter of rain water with a ph of 4.5 completely reacts with the statues. What mass of the statue is lost? So the first thing we have to do is calculate the concentration of hcl from its ph and so we can say that the concentration of our H. Plus is equal to 10 to the negative ph and this situation that ph is 4. And this is going to give us 3.16 times 10 to the -5 molar. Next we want to determine what the reaction equation is. And so we have our limestone which is C. A. C. 03 solid. It's going to react with two equivalence is of H. Plus and the acquis form that's gonna give us calcium two plus in the acquis form plus water in the liquid form plus C. +02 in the gas form. And so before we determine the mass of C. C. A. C. +03 lost, we have to include the molar mass of limestone which of C. A C. +03 which is 100.9 g per mole. And now we can solve for the mass of C. A. C. 03 that is lost or that has reacted. And so first we're going to start off with our leaders. So we have 1.546 times 10 to the seventh leaders. And then we're going to use a conversion factor that says that for every one liter of rainwater we have 3.16 times 10 to the negative five moles of H plus. We're then going to use a multiple ratio of H plus ions and calcium carbonate. So we can say that for every one mole or for every two moles of H plus ions, we have one mole of C A C +03. And then finally we want to determine the moles and grams. So therefore we have to use the molar mass. We say one mole of this, C a C +03 is equal to 100. g. And that's why we use the molar mass earlier. And so we need this answer in kilograms. And so to go from grams to kilograms, we can say that for every one kg that's equal to 10 to the third grams. And so this is going to give us our final answer of 24.4 kg lost. I hope this helped. And until next time

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