This is MCAT General Chemistry 2, we'll be doing a few sample problems. This is another passage based question. We want to remember that a lot of the information contained in passages is totally unnecessary to actually answer the questions. So let's read through, and we'll try and pick up a few things. But we'll try not to spend too much time learning anything from the passage. Read full transcript
Passage IV, questions 18 to 22. Vinegar owes its distinct odor and taste to the acetic acid AA it contains. AA is a weak acid with an acid dissociation constant Ka of 1.8 times 10 to the -5. In the vapor phase, AA exhibits strong deviations from ideality. This is revealed by examining the compressibility factor, z, of AA and various gases.
Compressibility is defined by the equation z equals PV over nRT. Now before we get too far, let's make sure we underline here our acid dissociation constant and this equation, z equals PV over nRT. Continuing on, the condition z = 1 represents the ideal gas. Plots of z at 300K for AA and various gases are shown in the figure below. So we've got a figure, but we don't want to spend too much time looking at it.
We just note that it exists, and we're looking at z versus pressure. Deviation from ideality can have many causes. But in the case of acetic acid, vapor dimerism is the primary agent. Some trimerism and tetramerism also occur, but they are not nearly as large scale and systemic. This dimerization leads to two molecules of AA behaving as one combined molecule.
The dimer is shown below. And here we have our dimer. So that's it for the passage, now we can look at the questions. 18, under ideal isobaric conditions, how does the occupied volume of 3 moles hydrogen gas at -70 degrees C compare to the volume of 2 moles of xenon gas at 180 degrees C and 2,000 psi?
Now, a couple of things to notice in this problem. The first thing is that we need to know definitions. If we don't know what isobaric means, we're gonna be in trouble. Isobaric is a constant pressure. And the next thing here is, you might notice that this problem is entirely unrelated to the passage.
This is a discrete question embedded in the passage. Well, we want to know how the volume of hydrogen compares to the volume of xenon. And if it's ideal, it means we can use the ideal gas law, or PV equals nRT. We could rearrange this and say that V equals nRT over P. And so let's do this, we'll compare the volume of hydrogen to xenon. And we'll say volume of hydrogen to the volume of xenon, equals n hydrogen times R times T hydrogen over P hydrogen, over n xenon times R times T xenon over P xenon.
Now at this point, we'll see if we can cancel anything. The ideal gas law is a constant, so those Rs drop out. We also don't know the pressure of the hydrogen. They didn't say what the pounds of hydrogen were. We know that xenon was at 2,000 psi. Well, indirectly, they actually did tell us.
Because they said it was isobaric conditions, meaning that the pressures are the same. So we can get rid of the pressure. What we'll be left with, V hydrogen to V xenon equals n hydrogen times the temperature of hydrogen over n of xenon times the temperature of xenon. Now at this point, we need to remember that we need to use absolute temperatures in the ideal gas equations.
-70 degrees C is not an absolute temperature. So if we add 273 to this, we would say that hydrogen is at approximately 200 Kelvin. And the xenon is at 180 plus 270, or approximately 450 Kelvin. Now we can plug those in. Our volume of hydrogen to our volume of xenon, equals 3 moles of hydrogen gas times 200K, over 2 moles of xenon gas times 450 Kelvin.
3 times 200 is 600, over 2 times 450 is 900. So 600 to 900, or 600 divided by 900 is two-thirds. So this is going to be a 2 to 3 ratio, or B. Question 19, in industrial processes where AA dimerization must be avoided, the best operating conditions are?
Now we don't know much about dimerization. But we do know that the passage told us that it was one of the reasons that AA deviates from ideality. So how do we make a gas more ideal? What conditions would favor ideal gasses? Anything that makes it so those gasses don't see each other, so the molecules don't have interactions.
Well that happens when we have a high temperature and a low pressure. Choice A is high temperature, high pressure, that's incorrect. B is high temperature, low pressure, so that's looking good. C is low temperature and high pressure, that's the exact opposite. And D is low temperature, low pressure. So B looks like it's the right answer.
A high temperature and a low pressure would keep acetic acid molecules away from each other because they would be moving fast. And since it's low pressure, they would be in a larger volume, meaning that we wouldn't have dimerism occurring. Question 20, a solution of acetic acid and its conjugate salt would make an effective buffer at what pH?
What do we know about buffers? We know that a buffer has a pH that equals the weak acid's pKa. Do we know the pKa of acetic acid? Not off the top of our head, so it must have been given to us somewhere. First paragraph, we have a Ka of 1.8 times 10 to the -5. pH equals pKa, so we want to know pKa of 1.8 times 10 to the -5.
That's going to be negative log of 1.8 times 10 to the -5. Well, this is a pretty complex calculation, so let's try and approximate it. Negative log of 1.8 times 10 to the -5, is going to be in between negative log of 1 times 10 to the -5 and negative log of 10 times 10 to the -5, or 1 times 10 to the -4.
So it'll be in between negative log of 1 times 10 to the -5, or 5, and negative log of 1 times 10 to the -4, or 4. So our pH is going to be in between 5 and 4. A is 3.18, that is not between them. B, 4.74, that's between the two. C, 6.65, no.
D, no. Notice that D was also a pH that was greater than seven. This is basic, and most likely not going to be the pH for an acid. We can check and make sure that 4.74 is reasonable. If we look, 1.8 times 10 to the -5 should be closer to the 5 side, because it's a really low number.
As it gets larger, it'll shift towards the 4. But since this is a really low number, we would expect it to be closer to 5, and it is. Question 21, Gases Y, W, and X could represent? Now at this at this point, we actually need to look at the figure and compare these.
So we've got Y, W, and X, what could they be? Well, the passage told us that z =1 is the ideal gas. So the further away we get from that is the least ideal. Y, W, X means we want to see most ideal. W is going to be the least ideal, and then X will be in the middle. This is one of those things where they're trying to trip us up by switching the order around.
It doesn't go most to least, it goes most, least, middle. Make sure you keep track of that when they start throwing variables in. You don't want to accidentally mess up the order when you can easily pair the most and least ones correctly. It'd be an easy way to lose points, so keep track of the variables and make sure you order them in the order that they give you.
Don't just assume that it's going to be from the most to the least, or the least to the most. So let's look here, we expect our first gas to be the most ideal. Argon, nitrogen, neon, hydrogen fluoride. Well, of all these, argon, nitrogen, neon, you would expect these ones to behave fairly ideally.
Noble gases, a diatomic. Hydrogen fluoride, however, is one of the few compounds that has hydrogen bonding. So already D is looking suspect. We go down here, xenon, that's a noble gas. Looks like D is not gonna be our answer. Next, let's look again, argon, oxygen, chloromethane.
This one right here is polar. Oxygen is a diatomic. We would expect oxygen to be more ideal than this one. A is gone. B, nitrogen, krypton, with a flourosilane. This one is also polar.
These ones are fairly ideal. It looks like B might be the wrong answer as well. C then, noble gas, silicon bromide. This is probably gonna have a dipole moment. CO2, this is a linear molecule that would most likely have its dipoles cancel. Based on the fact that this thing is ideal, and this guy which we'd expect to be the least ideal of these, C looks like it's the answer.
So again, the big point from this question is that we can't assume that things are ordered most to least. So we need to make sure we read the question and answer them according to the way the question wants them to be answered. It's an easy way to miss points if we just assume that it's ordered most to least. Question 22, according to the figure, what is the concentration of Gas W at 40 atmospheres?
Now, anytime we see an according to the figure, we know we need to look at the figure. So we go back here, 40 atmospheres, we'll draw this up to W. At that point, we can then draw this over, and we say that z is 0.6. So what is the concentration of Gas W? Well, concentration is in molarities or moles per liter.
How do we get moles per liter using this z equation? Well, we could rearrange this. If z equals PV over nRT, n over V or moles over liter would give us that. So we can multiply n here, and divide by V. We move the z to the other side, we'll have n over V equals P over zRT.
Now at this point, we just need to plug things in. What was P? Well, that was 40 atmospheres. z, we found to be a 0.6. R is the ideal gas constant, or 0.08. And T, what's T?
Do we see T anywhere? It's not here in this problem, it's not here in this plot. However, the label for the plot says it was at 300 Kelvin. So we can multiply 300 Kelvin here. So we've got 300 times 0.6 times 0.08. 6 times 0.08, we've got 40 over 300 times 0.048.
We can move a few decimals. This one that way, this one this way. 3 times 0.48, almost 3 times 5. So we have 40 over 15. 40 goes into 15, we can divide 5 out of the top and bottom and get eight-thirds.
Eight-thirds reduces to 2 and two-thirds, or 2.67 molar. We look through our choices, A is 2.78, B is 2.03, C is 0.27, D is 6.32. So because we rounded, most likely A is the right answer. It's pretty close to what we have. You can cancel B, C, and D.
And if we were to do the exact numbers we would get 2.78. So in a problem like this, if you had rounded a little bit and you came to the answer choices, and they were too close to call, at that point you'd have to go back and use exact numbers. But most of the answers you'll get will be several magnitudes away, so you can round here and there, and still be in the right vicinity.
Most problems, however, will have very clean divisions and multiplications. And often, you'll see answers left as mixed numbers. And that concludes this problem walk-through.