In the lime soda process once used in large scale munici-pal water softening, calcium hydroxide prepared from lime and sodium carbonate are added to precipitate Ca2+ as CaCO3(s) and Mg2+ as Mg(OH)2(s): Ca2+(aq) + CO32-(aq) → CaCO3(s) Mg2+(aq) + 2 OH-(aq) → MgOH2(aq) How many moles of Ca(OH)2 and Na2CO3 should be added to soften (remove the Ca2+ and Mg2+) 1200 L of water in which [Ca2+] = 5.0x10-4 M and [Mg2+] = 7.0x10-4 M?

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Accepted Answer

Hello everyone today we are being asked to calculate the moles of sodium carbonate and calcium hydroxide needed to precipitate two different reactions. And so we have these chemical reactions here and here and we are given a 1700 liter sample of water that has four times 10 to the negative five molar of calcium and eight times 10 to the negative seven molar of iron. And so the very first thing we wanna do is we want to calculate moles of our calcium hydroxide And start doing that. We can take our volume are 1700 l and we can multiply this By the conversion factor that for every one mole of iron two plus we have eight times 10 to the negative seven moles. This is also equal to the malaria T. We can then multiply by our multiple ratio between iron and calcium hydroxide. And saying that for every one mole of iron We have one mole of calcium hydroxide. The multiple ratio came from our chemical equation here. The second equation had one as a coefficient in front of the iron and a one in front of the calcium hydroxide. It just wasn't drawn in. But that's how we derived our multiple ratio. And so we calculate this. Our units are going to cancel out this one mole should be one leaders that cancels out and we're left with 1.36 times 10 to the negative third moles of calcium hydroxide. The next thing you wanna do is you want to calculate the moles of calcium. See a two plus from that Previous answer. And so by doing that we take our 1. times 10 to the negative third moles of our calcium hydroxide. And we're gonna multiply by our multiple ratio of calcium hydroxide to regular calcium two plus ions. And so according to our equation we have for every one mole of calcium hydroxide we have one mole of C. A. Two plus in this calcium hydroxide here we only have one calcium, that's where the one mole comes from. When we simplify this, we get 1.36 times 10 to the negative third moles of C. A two plus or calcium two plus. Next we must calculate the moles of R. C. A two plus from the water sample provided. So from an Ht. 20 sample. and by doing that we can take our 1700 L once again and we can multiply by our other polarity which is going to be that one leader of that calcium two plus is going to equal four times 10 to the negative fifth Mueller of calcium two plus ions. In doing that, we receive our answer of 0. moles Of CA two plus. Next we must calculate our total calcium ions that we have. And so we can just add the two concert the two moles that we found for. So we can take our 1.36 times 10 to the negative third. We can go and add that to the 0.068 moles that we just sold for. And this should give us 0. moles of calcium two plus ions. Lastly, Will you step five over here, we're going to take the molds that we just calculated for. So the 0. Moles of CA two Plus. And we're going to add that to our multiple ratio of calcium two plus two. Our moles of sodium carbonate. And so we say that we have one mole of our sodium carbonate in a two C 03 for every one mole of C A two plus. Once again, this is derived from our clinical equation over here. And so we do that. We end up with an answer of 0. moles of sodium carbonate. And so our final answers. I'm going to be 0.96936 moles of sodium carbonate, as well as our 1.36 times 10 to the negative third moles of calcium two plus ions, calcium hydroxide islands. I hope this helped. And until next time

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Key Concepts

Stoichiometry

Molarity

Precipitation Reactions