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Simply determine the concentration of each:these were added to 500 milliliters of dehydrated H2O to make a IX PBS Component Molar mass milligrams Concentration in (...

Question

Simply determine the concentration of each:these were added to 500 milliliters of dehydrated H2O to make a IX PBS Component Molar mass milligrams Concentration in (g[mol) measured M NaCl 58 43.5NazHPO412011.6KH2PO413621.8KCl3916.3

Simply determine the concentration of each: these were added to 500 milliliters of dehydrated H2O to make a IX PBS Component Molar mass milligrams Concentration in (g[mol) measured M NaCl 58 43.5 NazHPO4 120 11.6 KH2PO4 136 21.8 KCl 39 16.3



Answers

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes.
a. 0.0200 mole of sodium phosphate in 10.0 mL of solution
b. 0.300 mole of barium nitrate in 600.0 mL of solution
c. 1.00 g of potassium chloride in 0.500 L of solution
d. 132 g of ammonium sulfate in 1.50 L of solution

After six Problem 32 asks us to calculate the concentration of all ions present in each of the following solutions was important to note. These are all strong electrolyte solutions. So we're not gonna have any of our initial molecule in there. It's all going to be ions. So in order to solve these will first need to start of the balanced chemical equation, showing the dissociation of each of these molecules and water. And then we can look at the specific number of moles of each of the products that we end up having. So let's go with a We're starting with sodium phosphate. That's an a three and a three p o. For So we're going to do first is we're going to write out what that associates into in a balanced form. Okay, so that has three an A plus molecules and one p o for three minus. Okay, So now what we can do is we can calculate the molar ity of our starting material. Remember that polarity is molds for leader and were given the number of moles here. And we can calculate leaders from milliliters by dividing by 1000 right? So that 0.2 Moles over and 10 mills is 100.1 leader when we divide by 1000. So 0.2 divided by 0.1, gives us a polarity of to Mueller. Okay, so now what we can do is we can convert that to prod the product side. So if we have for every one mole of sodium phosphate, we have three moles of sodium and one wall of phosphate. So if we have two moles of sodium phosphate, that's gonna be six moles of sodium and two moles of phosphate. Right? So let's try again with problem with be here. So we have 0.3 moles of barium nitrate. So we're going to start again by reading out are balanced chemical equation. So we'll have very, um, nitrates. And that's B A n 03 to that associates into one. Very, um, and to know three minus molecules. Okay, what's the polarity of our barium nitrate here? So we have 0.3 moles over and 600 mils divide that by 1000 is 10000.6 leaders. Okay, so we have 0.3 divided by 0.6. That gives us a mill aridity of 0.5 again, we can do the same thing. So if we have 0.5 Moeller of berry of nitrate One mole of barium nitrate is one mole of barium, two moles of nitrate. So we have the same, um, moles of burials, barium nitrate and twice as much of metric quote. Okay, let's go up to see now. So now we have potassium chloride again. We'll start by writing outer equation So potassium chloride k c L associates into one K plus and one C l minus. Now we aren't given the number of moles here, so we'll have to calculate that out first. So we have one gram of K c l. And if we tell healing up Moeller mass of K C l adding the massive potassium plus the mass of chlorine, we find out that 74 grams per mole Okay, so calculating that through one divided by 74 we find that we have 0.135 molds. All right now we can divide that by our leaders, and we're already given that in leaders. So 0.5 leaders. So we find that the polarity of K. C. L itself is 0.135 invited but 0.5 0.27 So that's similarity of R K C L. That is also going to the similarity of r K and R C L. Because we noticed we have a 1 to 1 to one ratio of all of them. So everything has the same polarity there. Okay, let's look at the last one here. We have ammonium sulfate, so we're gonna write that out again as a balanced equation. So ammonium sulphate is at H four two s so far. Right? So we get to pneumonia for everyone s so far. Okay, so now let's continue and calculate the polarity again. We're not given the number of moles, so have to convert here. So it's 1 32 grams of ammonium sulfate, and one mole of that is 132 Gramps. So that tells us we have one mole. Now we're gonna find polarity by dividing by leaders. So one mull over 1.5 leaders and we find that that is 0.67 Mueller. Okay, so now looking are ratios here. So that tells us that we have 0.67 Moeller of sulphate. We multiply that by two for ammonium, which gives us 1.34 more eso in total. To be able to calculate these kinds of problems, you need to balance out a chemical equation. Find out how many moles of your initial molecule that you had and it just used.

This question asks us to calculate each of the following quantities for a It wants the polarity of the solution prepared by diluting 37 mil milliliters of 370.250 Moeller potassium chloride to 150 milliliters. This uses the simple equation. See one of the one equal see to be, too, Um and so we begin with the initial concentration tells us was 0.25 Mueller the initial volume. We're taking his 37 milliliters and does the fall you. The final volume will be 150 milliliters. So you plug in those three parts, divide by 1 50 on both sides to get the second concentration by itself. And that simply gives this 0.617 Miller. So this is the concentration. After you dilute this initial solution to 150 milliliters and B, it wants the polarity of the solution prepared by diluting 25.71 milliliters of 0.706 Mueller ammonium sulfate 2 500 millilitres. It's a very similar set up the initial concentration 5000.706 Moeller the initial volume. 25.71 milliliters the final volume. 500 milliliters. Sulfur seat to, and you get 0.363 Bowler for See It's a little bit more complicated. It tells us that it wants the polarity of sodium ion in a solution made by mixing 3.58 milliliters of 0.348 Moeller sodium chloride with 500 milliliters of 6.81 times, 10 to the negative second Mueller sodium sulfate. And to assume the volumes or additive so we have to do is figure out how many moles sodium ions we get from each of those two solutions. Then divide that by the total volume of the solution to get the polarity of sodium miles. So for the sodium chloride solution, it tells us it was 0.348 bowler or 0.348 miles per leader. We multiply that by the volume, but we need volume units and leaders to cancel out. So we divide this by 1000 so middle leaders and leaders will cancel and then we're left with moles of sodium chloride. But we want moles of sodium ions. So we do this unit conversion for every one mole of sodium chloride. There is one mole of sodium ions because there's one sodium in this formula and then that will cancel. Moles and sodium chloride were left with moles of sodium ions. So from the sodium chloride solution, we will get 0.125 Moles of studying minds now for the sodium sulphate solution and tells us the concentration with 6.81 times 10 to the negative second molar or most per liter. Similarly, we multiply by the volume, but again, we need that leader. So divide by 1000 so the mill leaders and leaders cancel. We're left with moles of sodium sulphate again do a unit conversion, but this time sodium sulfate has to sodium ions per one sodium sulfate, so it's 2 to 1. Moles of sodium sulfate cancel. We're left with moles of sodium ions. So from the sodium sulphate Klute solution, we get 0.681 moles of sodium ions. So to find the concentration of Saudi Iam's sodium ions, we add these two moles together. 0.125 plus points here 681 moles of sodium and we divide by the volume We got 3.58 milliliters from the first solution and 500 from the second. So this is most per mil leaders. We need moles per leader. So we multiplied by 1000 milliliters. Cancel and we're left with moles per leader. And if you do this bath, you would find that the concentration of sodium ions by mixing these 22 solutions is 0.138 Moeller.

Welcome. We're going to be discussing and doing some calculations for ion concentrations. In each of these problems, we're going to be given, um, how much volume we're gonna be given volume. And we're going to also be given either grams, which we will then convert moles. So I'm gonna put that here or moles. Either way, we're gonna have to get two moles we're going to be asked to solve. For how? Maney. But I shouldn't say how many. What is the concentration? So we'll be doing this similarity and the way that, um, this problem puts it is concentration of all ions present all ions. So I will report that appropriately, we're told that each one of the substances were going to be dealing with is a strong electrolyte, which means it dissociates completely. So these are going to be simple calculations. Okay. For our first problem, we're given the following were given that we are dealing with calcium nitrate, which has one calcium and two nitrate ions. Um, so that's our substance. And we have 0.100 moles in 100.0, not 1100 point zero mill leaders of solution Okay, so the first thing we're gonna dio and I should have finished up here were given these. Well, I'll do it down here. We're gonna find the polarity polarity of the compound. The mill arat e of the calcium nitrate and mill Araji is moles per liter. So for our first problem, we're gonna take our mold 0.100 moles and divide that by 0.1000 Leaders don't forget to turn this into leaders. And as you can see, quite simply, the concentration of this is one 0.0 Moeller and that see a and all three to Okay, Now we know if we have 1.0 Moeller see a and 03 to I could multiply this by a factor of one C A n 032 contains one. See, a two plus equals 1.0 Moeller. See a two Plus I could do the same calculation, but I'm just going to this time put two nitrates for one calcium nitrate equals two Moeller and 03 plus So now we were asked to find the concentration of all I am is present. So the concentration of calcium was 1.0 The concentration of the nitrate was, too. Let's do our second problem and check my camera and it's good. Our second problem were given 2.5 bowls of sodium self eight and this is in 1.25 Leaders of solution. So our mill arat e is moles per leader is 2.5 moles divided by 1.25 leaders. And that will be 2.0 polarity. Any to s 04 Now, in any to s 04 we can see that we have to. So Diem's and one cell fate. So all we have to do is multiply as we did in that last problem. Multiply the number of each eye on times Armel, Arat e To get the concentration of that I on the concentration of sodium is going to be 2.0 times two 4.0 Moeller and a to or any plus sorry about that. And for sulphate, which is s a four to minus, we're gonna have our 2.0 times one equals 2.0 Moeller and a sulphate s 04 tu minus. And that's our answer for the second problem. The third problem were given Mass were given 5.0 zero grams of ammonium chloride. And I looked up the molar mass of ammonium chloride. And it is 53.49. I'm just gonna put that in parentheses. So remember, a first order of business will be to convert grams two bulls, and we can do that by taking 5.0 grams and multiplying that by one mole is 53 0.49 grams. The answer for this. Oh, I didn't do the answer for this. I put this whole thing into my equation. Remember, our equation from polarity is bowls per leader. So what I did here is I did by this gonna me quit my moles. So I did 5.0 I put that in my numerator. And then I put my, um volume, which I forgot to write up here. My given volume. Let me go back up. Here is five 100 0.0 mill leaders, which is 0.5000 leaders, 0.5000 Leaders, do your mouth on this and you get to two Sig figs. Zero 0.19 Moeller in a church for C l Well, this is a substance that has one ammonium and one chloride. So the concentration of each of these will be the same. Easy. And one more. This one. We have 1.0 grams of potassium phosphate and potassium phosphate is to 12.27 from all our mass, and we are given 250 milliliters solution. So again, we're going Teoh polarity is a good equal balls and my moles. I'm going to do my conversion right here. 1.0 grams in 212.27 grams per mole and then my volume and leaders right here. Did I do this one? I believe I did, but it looks like I got the same answer. Let me just verify that real quick. Yes, I did. Um, I got 0.19 019 Moeller. So this means for K. Plus I have three ions. And for the fast fate, I'm gonna move that down the line p 043 minus. I have one. So I'm gonna multiply my quantity of each eye in within, um, one formula unit, times similarity. And I get 0.57 polarity for my K plus and 0.19 mil Araji for my peel form three minus. Okay, thanks.

This question asks us to calculate the concentration of all Lions President each of the following solutions 1st 1 is 10.1 mole of C A, N 032 or calcium nitrate. So what's first convert that, um actually, for this one, it's easy enough to start with. Gives a small or we need to do is divide by the A number of leaders. Same thing is multiplying by one over. The volume tells us 100 milliliters. That means you for that leaders 1000 no leaders, our leader, new leaders, cancels out. We're left with moles per leader, which is the use. It's for Milady. That's perfect. And we get one more C A and oh three two. That's the construction of the whole solution. They want each of the individual ions concentrations, so calcium in this compound there's only one calcium for molecule, so they will have the same concentration. But for nitrate, there are two nitrate molecules are ions per, uh, molecule. So it will be double to Mueller and a three minus great next one 2.5 moles of any two s a four or sodium sulphate in 1.25 liters solution. This is real straight forward to start divide by the volume, which is already in leaders. That gives us dancer of two more where sodium sulphate and, uh, now for each molecule of sodium sulfate there to sodium ions so we'll double this concentration to four. Moeller that a plus. And there's one sulfate per molecules that concentration. We'll stay the same. Two more sulphate, sort of the 3rd 1 It tells us that five grams of N. H four CLR ammonium chloride in 500 militar solutions, I first need to convert this two moles of ammonium chloride. One more of an H four c o is equal to 53.49 grams. Then we're gonna divide by the volume. This is 500 mils 100 milliliters convert this toe leaders no leaders her leader all units to cancel grams moles, ups No moles do not cancel but milliliters cancel like with most earlier, which is what we want. This answer is 0.187 those of an H for C l and each of these ions is one eye on per molecule. So the concentration is the same free to the ions 1.87 more where and h for plus and wait. One, 87 Moeller c minus and finally one gram of K. Threepio four. Try potassium phosphate in 250 milliliters. Again convert this two moles. Ah, one more obsessed tract house and possibly eight is 212.27 grams. As above. Divide by the volumes. Turn two milliliters and convert to leaders 1000 milliliters per leader. Grams cancel and middle leaders cancel left with most per litre, which is what we want. This gives us an answer of 0.18 Eat more Okay, three p Go four now for each of the ions, they're three potassium items per molecules will multiply this times three. That gives us 30.0 565 more where que plus And there's one phosphate ion pro molecules. So that is the same 10.1 heat. Eight more lost speed Pio for remind us


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