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# Conservation of Charge - Acid/Base Systems

Take Home Test Two
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on 9 May 2011

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#### Transcript of Conservation of Charge - Acid/Base Systems

Ascorbic acid, ............ , commonly known as Vitamin C is an essential nutrient for humans and certain other animal species, in which it functions as a vitamin. Some people believe that taking large amounts of vitamin C each day can help prevent colds, cancer, and other maladics. You have discovered that the 1st pKa of ascorbic acid is 4.1 + A/100 after long hours of investigating in your bat cave. If this is the case, make a recommendation on the maximum dose of vitamin C someone would want to take so that their pH would not drop below 7.0. Assume the only buffer in the blood is the bicarbonate system with an initial concentration of and . Likely the pH of your blood would not remain at 7.0 because the kidney would produce ammonia to remove hydrogen ions from the blood. How much ammonia (NH3) would be required in the blood to counteract the added protons from the ascorbic acid? First Thoughts There are three vital steps to beginning any conservative principles problem:
identify what it is exactly that you are looking for, the "find"
based on your find begin to build the most feasible system possible to utilize
based on your system and system boundaries begin to make proper assumptions to simplify the problem Assemble: Find:
max dose of vitamin C to keep pH above 7.0
amount of ammonia required to counteract the protons given off by the dissociation of ascorbic acid Diagram: Blood System The entirety of the blood circulating through the body is the system. By doing making this our system we will greatly simplify the amount of work we have to do. If you don't see it now don't worry, we'll get to it in a bit. Assume:
closed, dynamic, reactive
only buffers in the body are the bicarbonate and ammonium systems.
.
.
.
1 tablet of vitamin C = 1 gram
complete dissociation of vitamin C
presence of buffer does not shift dissociation equilibirum of vitamin C
V (blood) = 5L but.. why? Well I'll tell ya! This relates back when we made our system just the volume of blood in the body. By doing so we made a closed system which will greatly simplify the equations we will utilize in the calculate. The closed system means no charge moves across the system boundary and we are left with only the generation, consumption, and accumulation terms which is intuitively all we want when dealing with charge conservation problems. Other assumptions will be explained later as we get to them. So now let's start thinking about the data we're given and the extra data we will need look up or calculate. Extra Data:
for bicarbonate:

for ammonium:

Ka bicarbonte:

Ka ammonium: Given:
Initially have: and
.. *Ka' denotes a literature value Vitamin C Overdose This one I don't think I could overdose on her.. That will do for extra data, as you can see we had to extropolate and research for them, which you should almost always always have to do for this section if necessary. Equations: 0 0 Based on our assumptions we can reduce the accounting equation down to what is above, our governing equation for the problem specialized to our situation with 's' denoting a specific species. Here are a few extra equations that will surely be used later in our calculate. Now, the basis. For this problem a simple basis doesn't really stick out. That's because we essentially have three mini systems (the buffers and vitamin c) inside our larger system of the blood. So we can go ahead and choose The one basis that will keep them all together. Thus Basis: Volume of Blood = 5 L Calculate:
For this problem our calculate will be large, as we're going to begin by splitting our system up into three smaller systems, manipulating our governing equation for each, and then applying our numbers to solve the problem. The reason for this is because we're dealing with two weak acids, vitamin C and bicarbonate, so we need to make simplifying manipulations to our charge balance equation with respect to each molecule such that the presence of the buffer doesn't shift the dissociation equilibirum of vitamin C (as we assumed). SO we'll begin the vitamin C system. This equation will come in very useful later with slight manipulation. As we've established the vitamin C will completely dissolve into the system, so we need to establish specific governing equations for the base and proton that are generated from this dissocation. For the base: There is no initial value for the anion, and no consumption, therefore we know this equation holds. Also, by a wonderful turn of events we'll find that this same equation can be established for the protons released into the system by vitamin C (based on the same theory). The hydrogen ions will have their own equation not based on our governing equation, but on one of the extra equations we added into our equation section. For this type of problem we're only interested in the final number of mols of hydrogen in the system at a specific pH, there for we can use this equation to calculate the mols of hydrogen:

beacuse we know that: For carbonate we know there will only be consumption, thus our equation will be: For ammonium:
we know there will only be generation based on the number of ammonia molecules generated: ammonia molecules will be consumed, and there will be an initial amount based on the number of hydogren ions present, therefore:
We'll calculate the number of hydrogen ions present the same way we did for bicarbonate: For Vitamin C:
We know immediately that there is no generation of vitamin C in the system, as it is being consumed by the individual. Which ironically is also all that's being done by the system. Thus we can go ahead and reduce our equation like so: Now the equations for Bicarbonate:
we know that there is no consumption of this molecule in the system, however because it is a buffer it is already in the system so there will be an initial value. Thus we can reduce our governing equation to: Now using all these new equations, the rest of our problem should be fairly simple. The problem is set up such that vitamin C as already completely dissociated, and the hydrogen ions given off from this are left to bind with the bicarbonate buffer system. Therefore we'll begin our first calculations with this system. Such as for any system, we'll take all the values we know and then choose the most suiting equation to solve for what we want. In this case, we know the Ka, initial amount of bicarbonate, and the amount of hydrogen ions in the system at the pH of 7. So to calculate the amount of carbonate consumed, what should we do? Right! (I hope) we'll use the Ka equation and manipulate our charge equations for bicarbonate and end up with an equation like this: We know that in principle, the number of carbonate ions formed will be equal to the number of bicarbonate molecules formed, so we can reduce our equation even a little further. No we have only one unknown in our equation, and it's a simple algebra problem from here. One thing we did not do however, was convert our initial concentration of bicarbonate to moles, so do that (multiply by the volume of blood), then plug and chug and you should get an answer along the lines of: n' = .0087 mols This is the number of moles of carbonate produced when then pH is at 7, so what does that tell us? Take a moment to think it over. Maybe grab a snack, some orange juice (I find with cranberry juice makes an excellent combination), try something with fish oil, it's great for your health and thought processes. Alright then, it means that that is also the number of hydrogen ions that will be in the blood at a specific time! And because, if you recall, we're looking for the number of hydrogen ions required to have a blood pH of 7.0 it means that's how many ions of hydrogen were given off by the dissociation of vitamin C. Thus we can rewrite the Ka equation for vitamin C: *I'll leave it to you to prove this expression, keep in mind which molecules are generated and consumed in a 1:1 ratio and be careful with manipulation of the governing equations. The chapter in (what is hopefully still) yourbook 5.9.2 does an has a few examples working through this. Pluggin' and chuggin' you should get an answer along the lines of: .227 mols of H generated (about 40 grams of Vitamin C) If you don't get this number exactly it's ok, you probably had a different pKa of vitamin C So now in order to calculate the number of mols of ammonia to we need first calculate the exact number hydrogen ions needed to drop the blood pH from 7.41 to 7.0, which is done simply: Recall the equation for hydrogen ions we determined for NH4 The number of mols required is 3.00 x 10^-7

Now analyze the equations we wrote for ammonia and ammonium, then apply them to Ka equation, and you'll get: Remember, consumption and generation terms are equal here, and the H+ final is at a specific pH. Pluggin' and chuggin' here you'll get an answer along the lines of: 3.03 x 10 ^-7 mols (about 5.15 x 10^-6 g of NH3) Finalize: