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Titrated with 0.008 MEDTA and required7.6mLto reachtheendpoint Whatis thetotalhardhess(RPm A50.0 mL of drinking water was 00o Jof the titrated- water? (MW CaCO3-IOO...

Question

Titrated with 0.008 MEDTA and required7.6mLto reachtheendpoint Whatis thetotalhardhess(RPm A50.0 mL of drinking water was 00o Jof the titrated- water? (MW CaCO3-IOO g/mol)6.08 mg/L130 mg/L0.06 mg/L121.6 mgfL

titrated with 0.008 MEDTA and required7.6mLto reachtheendpoint Whatis thetotalhardhess(RPm A50.0 mL of drinking water was 00o Jof the titrated- water? (MW CaCO3-IOO g/mol) 6.08 mg/L 130 mg/L 0.06 mg/L 121.6 mgfL



Answers

If $100.0 \mathrm{~mL}$ of a water sample is titrated with $0.01000 \mathrm{M}$ EDTA, what is the titer of the EDTA in terms of water hardness/mL?

Considering this mixture, let's determine the volume of 0.1 Mueller nitric acid that must be added to achieve a final pH of 7.21 We have to consider all the neutralization is that take place? So let's start with our strong base. Strong base is K O H. So from here, which we can calculate the moles of K o. H. It's a strong base which is equal to the moles of H minus. So we get zero 500 Mueller times 100 mL 0.1000 Leaders will find that this is equal to a 0.500 moles of oh H minus and will then look at our strong acid from HCL. The malls of HCL Zico, TV moles of H Plus and Armel Arat E Here is 0.750 Mueller volume is 200 mL or 2000.2000 leaders and we'll find that the moles are equal 2.150 moles of H plus. Now we'll have some neutralization here We see that the H plus is greater than the O. H minus. Therefore the we can calculate the malls of H plus in excess should be equal 2.150 moles of H plus minus 3.500 moles of the H minus the minus would be completely neutralized, and we'll be left with 0100 moles of each plus left over in excess. Now let's look at our next two bases. They're both weak bases on DWI air told that we have p 043 minus, uh, ignoring the sodium spectator from any three p 04 and we have C n minus. Ignoring these again the sodium spectator so r k A of the contra get acid here, H p 04 to minus is equal to 4.8 times 10 to the negative 13 and R K for the conjugal H C N would be equal to 6.2 times 10 to the minus 10. Comparing these values here, we see that the K A for H P 04 to minus is, uh, much less so. This would be the weaker acid or the stronger base so the acid would first react. Eso the H plus well, first react the excess H plus well. First react with the P 043 minus. So the malls of N A three Peel four is equal to the moles of Peel 43 minus. And we have 0.100 Mueller and 50 mL Their 500.500 leaders in this workout. 2.500 moles p 043 minus. And so now we compare this to our H plus and uh H plus is still greater. So moles of H plus in excess is equal. 2.0 100 moles of the H plus minus 3.500 moles of the peel for three minus, and we're left with 30.500 moles of the H plus in excess. And now second, the H plus in excess will react with the C N minus. And so the moles of the in a CNN sodium spectator so moles CN minus and this is equal to 0.150 Moeller times 0.500 leaders, and this is equal to a 0.750 moles of CN minus and now this time we can see that the sea end minus will not be the same here. So h plus plus c n minus will produce H c N Andi. We'll set up a nice table and moles here. The H Plus is 0.500 CN minus is 0.750 This is zero minus 0.500 minus 0.500 plus 0.500 This would work out to zero 0.250 point 00500 Using the results of this, let's go ahead and way have excess CN minus so the C and minus will now react with be added nitric acid. And we use our Henderson Hasselbach equation. We have all of our values here on D P H desire to 7 to 1 p. K A for H CNS negative log 6.2 times 10 to the minus 10. We don't know how much of the nitric acid so plus log 0.0 to 50 Now I'm gonna add some nitric acid that would shift this to the right so minus X plus x so this will be minus X and 00500 plus x by adding the nitric acid solving this for X, you can solve this and plugging this into my equation solver. I find that X is equal 2.243 So that tells me that the moles of H N 03 added Z equals 2.243 Moles of a Channel three. And so the volume of a Channel three added a Z we told were given the majority 0.243 moles. And we know that as one moles per leaders that similarity and we consol this and it'll yield 0.243 leaders, or 24.3 mil leaders of H and 03 that needs to be added to achieve a Ph of 7.21 to this mixture.

In this problem, we're going to calculate the concentration of calcium in a blood sample and units of milligrams of calcium per mil, A leader of blood. We are told that we have a 10 millimeter sample of blood, so we need to calculate the mass of calcium in units of milligrams and then divided by that volume of blood which is already in millet leaders. We use the equation from the previous problem and from that problem, once we balance it, this is the overall reaction that is taking police. We are told that we're told the volume and the polarity of a came in a for a solution that is required for Ty Trish in of the ox elite Ion. And we know that since the compound C a C 204 has see a with a charge of two plus cancelling out with CTO for the charges three minus that, if we calculate the number of moles of C two for minus that were used in this tie Trish in and that would relate to you and be equal to the number of moles of calcium which we can then convert through its more mass into milligrams of calcium. So that is how we're going to approach this problem. And we know that polarity is equal to moles per leader. And so that means that the moles of K m n O. For that we have is equal to its malaria t times, its volume and leaders and were given both of those. So we can start by calculating in the number of moles of K m N 04 that were required for the Thai tradition and somewhere to see a sea. 204 came in. 04 has equal and opposite charges cancelling out. So the number of moles of Canada four is equal to the number of moles of M N 04 minus ions and that is equal to nine point 56 times 10 to the negative Fourth Moeller times when we convert the volume into leader 0.0 242 leaders and when we want to play, there's together that comes out to about 2.31 times 10. Do you the power of negative five moles of M N 04 minus Now, based on the overall balanced reaction, we can relate this to the number of moles of the Oxlade ion CTO for Tu minus, using the ratio of two moles of M. N 04 minus. Reacting with five moles of C 24 Tu minus Again, we said that one more of C 204 to minus responds to one mole of calcium, since it has a charge of two plus, and they're in a 1 to 1 ratio in that formula for that compound. And now we can convert this into a mass of see a two plus. So one more of calcium is equal to 40 0.8 grams of calcium from the periodic table. Then we convert grams two milligrams. We know that one gram is equal to 1000 milligrams. So now when we want to play all those ratios together, we get the mass of calcium in milligrams, and that comes out to you about 2.32 milligrams of calcium. And remember that we were told this is in a 10.0 Miller liter sample of blood. And so these are the units that we want. So when we take 2.32 divided by 10 that comes out to zero point 232 milligrams of calcium. Her Miller liter of blood

Let's look at some tight rations data for a vinegar and sodium hydroxide, tight rations. And here is the information we're given. We have a 10.0 zero mill leader sample of vinegar, and that's in a quick solution of acetic acid. And its trade tie traded with zero point 50 six to Moller sodium hydroxide. And it requires 16 point 58 mill leaders of sodium hydroxide. And there's two parts for this problem. Um, I've got one more piece of data, all right, it down here, even though we're not going to use it later. And we're also told that the density of the vinegar is 1.0 six grams per centimeter cubed. So those are the four. I can't get my cubed in there. Centimeter cubed. There we go. One point 006 grams per centimeter cubed. So those are the four pizza of data that were given one, 23 and four. We have to calculations to dio what is the molar ity of the acetic acid? That's the vinegar and herself. Second calculation is what is the mass percent of acetic acid in the vinegar? Okay, let's begin. I'm gonna go to the next page and start check my camera and it's good. So let's do density first whips Now I gotta check it again, because that was very weird. Um, so let's first find the polarity. And to find the mill Araji, we need to know that there's a 1 to 1 ratio of 1 to 1 ratio of hydrogen ions in vinegar to the O H ions, so we don't have to worry about any changes to any concentrations. The concentrations that were given, um, will be the concentrations of the ions that were interested in. So we're going to figure out how many moles of h of the O age minus. We have using the data for, um the that any wage. And then we're going to take moles divided by leaders to get the mill. Araji, Let's do it. So to find the Moores that's coming cross strange moles of any wage, we're going to take our polarity, which we were given as zero point. I wrote it down here somewhere, 5062 We're gonna multiply that by the middle leaders, which was zero 0.16 58 leaders and that will equal 0.8 39 three Moller. And that's the polarity of our acetic acid. Next, let's find our, um, mass percent. And to do that, we're gonna assume that we have a 1000 mill leader sample. Then we're going to use density to get the mass of our sample. Will we use malls to get the mass of acid, and then we'll do our percent killed. I wonder if I have enough to do that here. I probably don't Let's go to the next page. So I had. If I assume I have a one 1000 mill leader sample and I have one 0.6 grams per milliliter, I have 1006 grams. That's my That's my whole solution. That's the whole thing. Whole kit and caboodle. And then to find the mass of my acid, I'm going to take my mole Arat E. What am I gonna do? I'm gonna take my polarity, which is moles per leader, and I have one leader so that I can just assume that that's my moles. And I want to multiply that by the molar mass of a CD gas it, and that will give me the mass of the asset. Let's do that. So I have 0.8 was that three 93 molds of asset and the older man's of a CD. Gas that I looked up was 60.0 five grams, and that gives me 50 point 40 grams of acid, and my mass percent will simply be 50.40 grams, divided by 1006 grams times 100 and my magic number is five 0.10%. Problem solved.

In this problem were asked to determine the polarity when varying amounts of Salyut are dissolved in various amounts of solution, Our first prompt involves 2.92 moles of methanol being dissolved in 7.16 liters of solution. So we know the polarity is in units of moles per leader. So we can just go ahead and divide our moles by the amount of solution that we have and we will find our polarity. By doing this calculation, we find that arm Olara T is equal 2.4 08 Moeller. Our second problem involves 7.69 miller of ethanol being dissolved in 50 milliliters of solution. We don't necessarily have to do this step, but we can do a few unit conversions just to confirm that we are performing this calculation in the right units. So we see. But by doing these two unit conversions, we actually cancel out are 21 thousands that were either multiplying by or dividing by to achieve this unit conversion. So we could have just divided our mill Immel by our milliliters and gotten the same answer. So when we do this math, we find that arm Olara T is equal 2.154 Moeller in this last scenario were given a massive Yuria and were given a certain volume of solution. We confined our moles of Yuria by taking the mass that we were given in the problem statement and dividing by the molar mass. So doing this gives us our moles of our Salyut which comes out to be 0.4 20 moles of Yuria and we were also given a volume in milliliters and our units of similarity is usually done in moles per leader So we can do another unit conversion here and we have a volume of 0.275 leaders. So to find arm polarity will just Duthie molds that we calculated above divided by the volume of our solution and we find that are resulting polarity is 1.53 Moeller


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