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Calculate the potential at the equivalence point in the titration of 100 0.100 M Fe?+ in 0.500 M H,SO4 with 100 mL of 0. mL 0200 M MnOs Keq = 50 * 1062 E Fe3+IFe2...

Question

Calculate the potential at the equivalence point in the titration of 100 0.100 M Fe?+ in 0.500 M H,SO4 with 100 mL of 0. mL 0200 M MnOs Keq = 50 * 1062 E Fe3+IFe2+=0.771V

Calculate the potential at the equivalence point in the titration of 100 0.100 M Fe?+ in 0.500 M H,SO4 with 100 mL of 0. mL 0200 M MnOs Keq = 50 * 1062 E Fe3+IFe2+=0.771V



Answers

Calculate the pH at the halfway point and at the equivalence point for each of the following titrations. a. $100.0 \mathrm{mL}$ of $0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)$ titrated with $0.10 \mathrm{M} \mathrm{NaOH}$ b. $100.0 \mathrm{mL}$ of $0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)$ titrated with $0.20 \mathrm{M} \mathrm{HNO}_{3}$ c. $100.0 \mathrm{mL}$ of $0.50 M \mathrm{HCl}$ titrated with $0.25 \mathrm{M} \mathrm{NaOH}$

Let's calculate the pH at the halfway point and the equivalence point for each of the following. Titrate Asians for a we have ah, 100 mL. Uh, yeah, 0.10 Moeller H C seven h 502 and that's tight, treated with 0.10 Moeller sodium hydroxide. So let's calculate, um, the amount needed to reach the equivalence point. So starting with 0.10 Moeller of 100 milliliters of kassid 0.10 Mueller and V two of the base, so V two will be equal to 100 mL. So this would be the volume to reach the equivalence points and 50 mL is of anyway. Chatted is the volume, uh, to reach the halfway point, So let's do the halfway point first. So at the halfway points, we can calculate the moles of the H C seven h 502 which is equal the 0.10 Moeller and 0.1000 leaders in this war co 2.10 moles and we can calculate the moles of sodium hydroxide, which is equal the 0.10 Moeller and 0.5000 leaders and this is equal. That 0.50 moles. We can set up a Okay, these table here si seven h five to minus and each too old. This will be done in moles. And we have initially 0.10 and 0.500 minus 0.50 minus 0.50 plus 0.50 This will give me zero here and 0.50 and 0.50 To calculate, the pH will use the Henderson Hasselbach equation, which are P H is equal to the peak a plus the log of the C seven h five to minus over the H C seven h 52 and so p k s o negative log of K 6.4 times 10 to the negative five is the a given for the acid plus the log of 0.50 Moeller over 0.50 Moeller and this was going to work out to 4.19 plus zero. So the pH at the halfway point works out to 4.19 at the equivalence points. We've already calculated our moles of H C seven h 502 from up above. We found that this was 5020.10 moles and we can calculate the moles of the O H minus, which is 00.10 Moeller. And at the equivalence point, we have 100 mL or 1000.1000 leaders and this workout 2010 moles. A total volume here is 200 mL 100 mL of the Assad and 100 millimeters of the base. So we'll set up again. Nice table in moles. Yeah, and we have initially 0.10 moles, 0.10 moles zero minus 00.10 minus 0.10 plus 0.1000 and +010 No, we can calculate the molar ity of the C seven h 50 to minus which is equal to 2.10 moles over the total volume of 200 mL or 2000.200 leaders in this workout. 2.50 Moeller how we can set up a nice table. Si seven h five 02 minus at H 20 well heeled HC seven h five 02 plus O H minus Initial change. Librium Initial is 0.50 Complete Rights table here 1050 x x and x. This will be represented by a K B equation, HC seven h 502 which minus over si seven h 50 to minus the cabe, would be kw divided by the A 6.4 times 10 to the negative five sequel the X X and 0.50 minus X. Solving this quadratic equation for X, we find that it will be equal to 2.8 times 10 to the negative six going back to our base table hydroxide and morality is equal to 2.8 times 10 to the negative. Six. Mohler The P O. H can be calculated by taking the negative log of the hydroxide and malaria t just negative log 28 times 10 to the negative six and this would be equal to 5.55 and the pH at the equivalence point will be 14 minus 5.55 and this would be equal to 8.45 And there is our pH at the equivalence point, the next hydration. We have 100 milliliters of 0.10 Moeller C. Two, h five and H two, and this is tight treated with 0.20 Moeller a Channel three. So let's work out the volume two equivalence points. So solving for me to hear, we'll find that this is equal to 50 0 mL. So this is the volume to reach the equivalence points and UM, 25 mL would be the volume to reach the half equivalence point. So let's, um, first calculate the pH at the halfway points or the half of coolants. Point on will calculate the moles of C two, H five and each to sequel to 0.10 times 100 mL, or 1000.1000 leaders in this workout, 2.0 10 moles, and we'll calculate the moles of the H N 03 at the halfway point on the polarity is 30.0.20 Moeller and 25 mL of 250.0 to 50 Leaders would yield 500.50 moles H plus C two, h five and h two. Two c two h five an age three plus all three in Moore's. And initially, this is 0.50 0.10 and zero minus 0.50 minus 10050 plus 0.50 To leave, it was zero 0.50 point 0050 Let's calculate our pH. We first need to get a k a from the correspondent KB Value, which is given to us. So the K W one times 10 to the negative 14 divided by the baby value 5.6 times 10 to the negative four gives me the K a value of 1.8 times 10 to the negative. 11. Let's use this in our Henderson household back equation to calculate the pH at the halfway point c two, h five and H three plus over c two, h five h two and the P K is negative. Log 1.8 times 10 to the minus 11, plus the log of 0.0 050 Moeller over 500.50 Moeller and this would be equal to 10.74 plus zero and the pH at the halfway point works out to 10.74 Now let's calculate the pH at the equivalence points. So at the equivalence points, we already have worked out the moles of the C two h five NH two from above. This is 0.10 moles, and we can calculate the moles of a Channel three at the equivalence point, which is 30.0.20 Moeller. And we're adding 50 mL, or 500.500 leaders, and this will be 0.10 moles. The total volume at the equivalence point would be, uh, the 100 mL of the C two, h five n h two and the 50 mL of a Channel three. So the total volume of the equivalence point is 150 mL, so H plus add C two, h five and H +22 C two, h five and h three plus on. We'll do this. In malls I see e and 0.100100 minus 0.10 manage 0.10 plus 0.10 Calculate the malaria C two, h five and h three plus would be 30.10 Moles, divided by the volume 150 mL 4.150 liters Works out 2.67 Moeller. Now we can set up a nice table to calculate the P H. C two h five NH three plus plus H +20 for you h 30 plus at C two, h five and H two. Initial change. Equilibrium Initially 0.67 theater a stable 0.67 minus x x x que a here would be equal T h 30 plus C six, h five and H two over c six, h five and h three plus e k a. We calculated at the halfway point we can look back and see that it's equal to 1.8 times 10 The negative 11 sequel the x x 0.67 minus X and solving for X. You will find that it's equal to 1.1 times 10 to the negative. Six. Going back to our ice table H 30 plus is to find his ex, which is 1.1 times 10 to the negative. Six. Mohler and Therefore, the pH at the equivalence point is negative. Log of the Hydro Nieminen concentration Negative log 1.1 times 10 to the negative six and the pH at the halfway or at the equivalence point is equal to 5.96 For Part C, we've got 100 milliliters of 0.50 molar hydrochloric acid. This is tight. Traded with 25 Moeller Yes, 0.25 Mueller sodium hydroxide. It's calculate the volume required to get to the equivalence point c one B one that's equal to see to to enough 0.50 Moeller 100 middle leaders 0.25 Mueller C. Two. That's very V two. Solving for me to hear, we'll find that this is equal to 200 0 mL. So this is the volume required to reach the equivalence points and 100 mL. Half of this would be the volume required to reach the halfway point eso halfway to the equivalence point. So first of all, let's find the pH at the halfway point, we congratulate the malls of AH H plus from HCL similarities 0.50 Moeller and 100 mL, or 1000 leaders, which is equal to 05000 moles and the moles of the H minus from the sodium hydroxide 0.25 Moeller 100 mL to the halfway 1000.2 point 10 Israel's leaders and this is equal the 0.0 to 500 moles on. Do you concede that the moles We have a strong as a strong base. So the moles of H plus are present in excess. And this would be the 0.5000 bulls are You should do the right Sig figs here, I'm gonna go back. This should be to the correct number of Sig figs here on board 0.50 moles and 0 to 5 moles to sig figs. Each 0.50 moles minus 0.25 Moles would be equal the 0.25 moles. The volume the total volume of the halfway point would be 200 mL for the morality of the H plus that's present in excess. At the halfway point, eyes equal 2.25 moles and 200 mL, 2.200 leaders and we get 1 to 5 Moeller and Therefore, the pH would be equal the negative log the beach plus, which is equal to negative log of 0.1 to 5, and this is equal to a pH of 50.90 and this is the pH at the half way point on. Now let's offer the pH at the equivalence points. We know that the MOL of H plus from the first part of Part C here is equal to +050 moles. Let's calculate the moles of O H minus. We know the mole aren t is 0.25 Mueller. At the equivalence point, there will be a volume of 200 mL, or point to users or leaders. This will yield 0.50 moles H plus plus so each minus U. C H 20 I see moves rules moles 0.50050 minus 0.50 minus 0.50 00 So, in this case here, the H plus similarity is equal to the O. H. Minus mill arat E. Therefore, the solution is neutral at the equivalence point, and the morality of each plus is due solely to the self finalization of water. Their form similarity of the H plus is equal to 1.0 times 10 to the negative. Seven. Mueller In the self organization of water, the pH is equal the negative log plus negative log 1.0 times, 10 to the negative. Seven Fuller believe units here, and we'll get a pH of 7.0 This would be the pH at the equivalence point.

So this question we have ch three and it's still just methyl amine reacting with water. Rather, we're dealing with death looming wrapped in FH shell. But I'm writing out with the expression for the baby is, um whenever reacts with water becomes see a stream an edge three plus in o h minus. In other words, it's the base with a K B that is equal to 4.4. I'm sense a negative fourth now. So this is a Thai tradition that where the curve is gonna look something like this. Remember Ph here? Acid volume. We started a high pH. It goes down as we had acid. And so in this first heart, we're told we have 0.5 leaders of one Moeller c h three n h two and we're adding 05 leaders of five Mueller hcl And so I'm just by inspection, we can see that Will we have the same amount? But we have half the concentration of hcl. So in other words, we have enough H CEO to react halfway with all the amount of CHD. It is too. No, 0.5 is half of one. And so what? That tells us. Is that Yeah. Half of the C H three and H three to went and became C H three and H three plus their concentrations are gonna be equal now, the reason this is nice. I'm going about this later because there's a trick that whenever the concentration of ch three an interval of like, for instance, in this case, the the base that we started with and the cons you get acid of that base. So this is the base we started with conjured acid. When those two concentrations are equal, the PKB is equal to the P o age. And so when I take the negative log of this value I get with the P O. H. Is which is equal to 3.3 and send the pH is equal to 10.7. And that's exactly what we should expect because we're still having a mostly basic solution now at the Psychiatric Point, which is also known as the equivalence plan. What that means is that all of this C h three and h three to that we started with has become its contrasts. The a story and it's too close. And so our job is to figure out how much ch three initially we started. So we have 0.5 leaders if we must. By that by one mole per leader will see that the leaders cancel out and we had 10.5 moles of C H three and H two to start with. Now that means that when we do have v hcl reactive it all the way at equivalence point at the story key metric point. We're going to have 0.5 moles Oh ch three and h three plus because that's his con. Get acid now before we jump and say OK, 7.5 more. That's cool. But we have to watch out for is that we've added volume and so now we have to think about how much value has been added. So we are. We already had an initial volume of 0.5 years. All the state the volume total It goes 0.5 leaders from the beginning. Plus, however, much was added. So our other solution we'll do this in black was a point five Moeller HCL solution Front to get 0.5 Moeller Ah, a solution to the, uh it's a reach the equivalent point. We realized we'd have to put in twice as much as we put it over here. So we're gonna have to be doing 0.1 leader. So our total volume is 0.5 liters plus 0.1 leader. That means 0.15 liters. And so now the concentration of CH three and a street plus at equilibrium. Very clever department at the equivalents point is 0.0.5 moles over 0.15 Leaders. I don't I didn't actually go ahead and calculate that because our next step is so if this is all we really have out of the bullet point, we want to find the cave that governs. This is so remember K W, which is equal to tend to negative 14th is equal to the K B of a base Times the k of its contrast were given this KB that's that 4.4 times sense negative Fourth and we have that. So if we want to find the k of ch three industry plus, we could do that and that is equal to 2.27 Time sense in eight of 11 required me. So Huguely 2.27 times Centenary of love it. And so now we have a k A. And we have an initial concentration, it calculated. But we have that we confuse a classic. Like I asked Association I stable, which I'm not gonna give you should be really powerful of this by now. If you're not, go back and work on some simple lasted association questions. You'll need it in the long run. But we'll find that the k that is gonna be able to x squared over the initial concentration, which is that 0.5 over 0.15 which I guess I just looking at that I can see that that's 1/3 um and then we're X is equal to the concentration of H 20 plus. And so when I do that, I will find the concentration of a studio. Plus, I would. She got the axe, and I found that that's equal to 2.75 times tense and your six Mohler. And then I took the negative log of the constellation age three or plus, which is, by definition, the pH and I found the pH was eclipse of 5.5, which is what we should expect. It slightly is

Sketch of titrate incur for the tie tray Shin of 50 mL of 0.250 Moeller Pagenaud too, with one Moeller of any Ohh. So we make our or graph here. PH is on the side to seven wells. Margraff will look like this with our equivalence point. Got a pH of eight point 35

According to the given data. In the caution, we have concentration off its seal concentration off material. Amine on volume off me, Tyler Me, the baby off me Tyler means 4.4 into tender super minus four so we can determine the K by this given for this formula on bigger is negative. Log off. Okay, which turns out to be 10.64 for but a off the ocean 50 ml off 0.50 m it seal solution is added given is the chemical reaction after reaction moles We can use this formula to data mine the final pH off the solution which comes out to be 10.4 to solve the part be first we need the psychometric point on chemical reaction which is represented by the given form. Given the action after reaction moles 0.5 is the on so putting in the values we are able to get the polarity which comes out to be 0.333 Using the Henderson equation, we can have the final Piotr which comes out to be 5.56 And it is all explained by these given steps


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