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Answer each of the following: If the balance you have access l0 is only sensitive to mg measurements could you make 5 mlof 5 AM CaClz using this same approach (star...

Question

Answer each of the following: If the balance you have access l0 is only sensitive to mg measurements could you make 5 mlof 5 AM CaClz using this same approach (starting with the solid)? Explain. If you have 50 mM solution of CaChz how could YOu use it t0 make 5 ml of 5 AM CaCl;?In one of your labs later in the semester; you "Il be running an SDS-PAGE: This requires lot of running buffer; the composition of which is: 25 mM Tris-HCI; [92 mM glyeine; and 0.1% (vlv) SDS_ Wrile protocol t0 prepa

Answer each of the following: If the balance you have access l0 is only sensitive to mg measurements could you make 5 mlof 5 AM CaClz using this same approach (starting with the solid)? Explain. If you have 50 mM solution of CaChz how could YOu use it t0 make 5 ml of 5 AM CaCl;? In one of your labs later in the semester; you "Il be running an SDS-PAGE: This requires lot of running buffer; the composition of which is: 25 mM Tris-HCI; [92 mM glyeine; and 0.1% (vlv) SDS_ Wrile protocol t0 prepare I0x stock solution of this running buffer: Describe how You would treat this stock solution to generate the Ix running buffer: That is, what do yOu need t0 do t0 Iake 1Ox stock solution and creale a Ix solution?



Answers

A $1.0 \mathrm{~L}$ buffer solution is $0.125 \mathrm{M}$ in $\mathrm{HNO}_{2}$ and $0.145 \mathrm{M}$ in $\mathrm{NaNO}_{2}$. Determine the concentrations of $\mathrm{HNO}_{2}$ and $\mathrm{NaNO}_{2}$ after the addition of each substance: a. $1.5 \mathrm{~g} \mathrm{HCl}$ b. $1.5 \mathrm{~g} \mathrm{NaOH}$ c. $1.5 \mathrm{~g} \mathrm{HI}$

In this next question, you have 100 ml of a .1 Moeller. But for solution created with sodium hydrogen phosphate and sorry, sodium di hydrogen phosphate and sodium hydrogen phosphate. And it's ph is 6.8 If you add 10 ml Of one moller hcl to the solution, do you still have a buffer solution? Well, let's first determine the total moles of buffering species. If we have 100 mL at 1000.1 Mueller would have point to one moles. Now the real question is, is this point no one moles of each of them or both of them total? I'm not quite sure. But let's assume the What, what is usually assumed? And that is a .1 moller buffer solution is a .1 moller total. So that means would have .5 moles of each of them. So if we add 10 mL of a one moller hcl solution, we will be adding 10.1 moles of hcl. The .1 moles of Hcl will react with the base that is in the buffer that being sodium hydrogen phosphate. In the process, it will convert the sodium hydrogen phosphate into sodium di hydrogen phosphate. And then any extra will convert the sodium di hydrogen phosphate. Well just the hydro di hydrogen phosphate into um phosphoric acid. So the first.005 moles will create more di hydrogen phosphate. To the point where we have all of the of the buffering species as di hydrogen phosphate. Then the other .005. The other half of the .01 will convert half of the di hydrogen phosphate into phosphoric acid. So we will then have equal moller amounts of hydrogen phosphate or die hydrogen phosphate and phosphoric acid which is a buffer. So we will actually destroy the first buffer and create a new buffer. That new buffer being the weak acid phosphoric acid and it's weak. Conjugate base di hydrogen phosphate.

So in this video, we're going to the answering questions. 60 from Chapter 17 which is a one litre buffer solution, is 60 17 1 10.1 to 5 Mueller in H and 02 and 20.145 Moeller in Indiana to determine the concentrations of agent to an entertainer, too, after the addition and each substance. So before we even dive in, I wanted to point out that things are made a little bit easier for us by our total volume equaling one leader. That means that our amounts and moles are just gonna be the same number as our concentrations and polarity because we're just dividing by one meter. Um, so in part A, we have 1.5 grounds of each seal. So we calculate our number of moles of h steal from that mass divided by the molar mass of HCL, which is 36.46 and we end up with 0.411 for moles. Um, now we want to put that into an ice box where initially we have that point. 0411 formals of HCL. We have our 4110.145 moles of energy minus and our 0.1241245 Moles are agent know too. We know that all of our HCL is going to react because of the strong acid so well. So we decrease HCL and energy minus by the 0.4114 moles and we increase our products by the same amount. We end up with no HCL equilibrium and 0.104 moles of Eno to minus in 0.166 miles of H and 02 And we don't really care about the chloride ions, but we have point out for 11 for miles of them. Um and I are concentrations are going to be the exact same is our amounts and Mel's because we're dividing by just want leaders So we have 0.104 Moller and I and our 2.166 Moeller agent or two. So let's go ahead and do this same thing for part B where we're adding a strong base, your solution so this time are strong base is going to be neutralized by our acid by the H no too. So let's go ahead and figure out how many moles of sodium hydroxide we're starting with by dividing the last by the molar Mass. That gives this 0.37375 moles of sodium hydroxide or just hydroxide. Um, and we put those values into the initial line in our icebox. We know that all of our sodium hydroxide is going to react because it's a strong base, so we decrease each of our reactions by point or 37 by then. In cleats are products by 370.375 and we end up with 0.875 miles of agent or two. Which means age No. Two is 20.75 Moller, and we have 0.83 moles of any, you know to, which means that Aaron and Artoo is 0.8183 Moeller, in Pricey, were again adding a strong acid to our buffer solution. So are strong acid is going to be neutralized by the weak base. That's present. So to get are a number of moles of H I. We divide the masked by the more last minute with 0.117 moves. Um, we fill in the initial Leininger Ice Thoughts with that point on 117 and then the 1170.145 point 117 1170.145 1 to 5 with inner tube minus an agent. No, too. We know that all of our inch I is going to react because it's a strong babe. Sorry. A strong acid. Um So we decrease each of the reactions by 0.117 and increase each of the products by 0.117 and we end up with 0.133 Moles of inner to minus, which means R N A N 02 is 20.133 moller and 0.137 Moles of H and know, too, which means our virginity was 0.137 Moeller.

All right. So for this problem, we're trying to figure out how to make a one leader solution of 10.5 moles per liter buffer at a ph of 7.5. What we're given is in a h to peel for and a witch of one mole solution of in a wage. So first we figure out what identify our acid and base or in this problem, So we look at her too ingredients, so to speak. So basically, we have h to appeal for minus and any which what we get from that we get H p a four to minus and some water and some sodium and a soda sodium ion. So we know the H two p o for minus is gonna be our asset because it loses a hydrogen and each peel for two minus is gonna be our contact base. So when looking at the handles Henderson Hasselbach equation, we know that our ratio is gonna be H P 04 to minus over h two p 04 minus and we fill in the rest of it. We know that Ah h two p 04 minus and each p 04 to minus mixture has a PK equal to 7.2. So filling in the rest of this equation PK being 7.2, we want a pH of 7.5. Let's attract some went to from each side and we'll get 0.3 is equal to log of our ratio. You can get rid of this by 10 of the power of each side. Scared of the log. It's a 10 to power of 100.3 is equal to three shoe of a concentration of our country basis with concentration of our acid, which is equal to tend to the power of 0.3. Just one. It's too sorry. 2 to 1. So we have three parts in our buffer solution, and two is gonna have to be congregate base. Someone's gonna have be the acid. So we know that we're gonna have to make these parts using our in age to p o for and are in a witch. We have our acid from the in a age to peel for, but we're gonna have to convert some of it. Teoh H p 04 to minus using the n a o. H. Just a matter of how much So we know we need Ah, one leader of the 10.5 more per leader. Buffer solution. So how many moles are involved in that? Oh, yes. There's gonna be points your five moles and that 0.5 moles. And that one leader is gonna be of our acid plus our congregate base. So we're gonna need 0.5 moles of N A. H two p 04 to start. Some of that will remain the way it is in solution and then some of it, we're going to convert to our conjugal base with in a O. H. So 0.5 moles in a H two p 04 one mole of it has a mass of about 100 and 20 grams. So to start out with, we're gonna need six grams in a H two p o. For we also know we have we're gonna need 0.5 moles went 05 moles of our acid and conduct based together, we know two parts are good, only going to be the acid, and one part is gonna be the country basis. We're going to buy this by three. It's gonna be 0.1 point zero 1667 Bulls Maria, multiply this by two because we need to convert two parts to our contact base. So we need 0.33333 moles of H P 04 tu minus so into a little 0.5 moles of both our acid and count. You get base two parts of that are gonna be our contact base. In one part of it is gonna be our congregate s is gonna be our acid, so we know we need 0.33 three and so on. Moles at age. Pude 04 to minus. But how do we get that? Well, we're gonna add an equivalent Mueller amount. Mowlam, Moles of Noh because we know with our equation H two p 04 to minus plus O H minus. We get water in our contact base, so let's just want minus right there. So this is a 1 to 1 ratio one one. So we know we're gonna need an equal amount of moles of O nao age to convert 0.33333 moles of in a age to p 04 Teoh h p 04 to minus. We know the amount of Moore's of any wage we need. Uh, all we have for enemy in a oh, age. Is that one more per litre solution? So we're gonna plug that in? We know we need 0.3333 bowls. I don't know the leaders yet. We know it's one mole. So when we saw for this, we know we need When. 03 33 Leaders video age to convert 0.33333 moles of N a H two p 42.33333 moles of our contact base age P O Ford's who minus Teoh account for those two parts of our country at base. This is also equal to 33.3333 milliliters. I'm in a village, so we know we're going to start out with six grams of an age to peel for then to convert to parts of that two h p 04 to minus, we're gonna add 33.333 milliliters of in a ohh. And then we're gonna add dilute with water upto one leader to get our one leader 10.5 moles per liter buffer with a pH of 7.5.

Hello, everyone. My name is Ali. Now we will answer the questions that we have mixed for here. It's consisting off H E and 02 glass key in tow. This make sure is acidic buffer existing in volume on going toe. Five litre because he said 500 elite or so it's open five liters. The concentration off each in orderto is open. One more clarity and the concentration off key and auto is a 0.15 polarity. And he told it, You know, too is weak acid as a B K is equal. Three boys 34 The question here is which Well, it happened when we and one off this compound tow these power formation is we exceed their buffer range or no, but the meaning off exceeded the proper range or not the buffer rangy. If we can squeeze a B age off these perform, it will be equal beak, a blast look, the concentration off K and on toe divided by each in or two and by substitue will be sweep on 34 plus. Look the concentration off K in order which is open on five and consideration off each end or two which is open it one and the result be h will be 3.516 this be H, which is 3.516 The buffer capacity is maximally to increase the bee etch by one. So from 3.516 to 4.516 or two 2.51 Sex wen will not be exceeded zuppa for rage but more than this by two unit. Or they listened this by one unit. That's mean me exceeds up off our past from the definition, a buffer capacity to increase the beach by one or decrease it by one. So now in the 1st 1 he said, we will add, So do you mind ropes Sign which it's amount. 200 and 50. Mellie Grant. Does this represent the helm in your morn? It represent 250 times pointing negative three grand, divided by its molecular weight, which is 40 grammar moon when she's representing him anymore, or point or 6 to 5. So what have been willingly exist? Amount is a buffer. Capacity will be exceeded or no Now nobody is Noh What effect on edge in or two it decrees its amount. So the new concentration off each and you will be the initial number off moon off H in order to which is all going to one time spy opening. Fine. Negative. Tow the number off more off in which added, which is all point or 6 to 5 divined it by then. You won't and its effect on the concentration off K and auto, which is a congregate based soccer basic, saw the intellectual increase its amount, so its initial amount is on 0.15 times by or going five litre, which is a present, the number of more plus the number of moon added, which is all point or 6 to 5 divided by the voice. So now when we calculated and you'll be agin you, it will be the beacon a which we steal it before it's just three points. 34 Blast Look, Zach, Concentration Off K and two, which she's a stated here, opine on five times by open front glossies. It's a value off. These ethical created by calculator is all point or fine. Six to fi and the value off these is O point for 37 fine, so by substituted here it will be over 0.56 Toe finder Volume in the year or point. All 4375 My violet on volume goes volume. So when we calculated we found a new B H is 345 by combining is a B H new after having disease amount off no edge with the initial value 3.516 So we found that the steel's about far no exceeded its a buffer capacity. That's me. We still in a bar for rage. So all these amount of you know EJ not exceeding the buffer capacity. But what about the other solution? We will substitute. You don't like these muscle for the other three solution and we will found that we steal in a buffer range not exceeded. Okay. Thank you.


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