What's the best way to prepare for a cumulative Final Exam? That can be a little ominous, and the "best" way to study can vary considerably from person to person. At this point, you've (hopefully) found something that works for you, whether that means a quiet corner in the newly remodeled library or a lounge with a little more "white noise". Go with the method that works for you!
With that as a foundation, what should you study? It's probably not a good idea to just sit down with the textbook and start reading on page 1. Use the tools you have available to focus and prioritize your studying time and energy!
1. Start with the exams you've already taken!
Look back at your exams. There are things that you knew very well when you took Exam 1 that have gotten a little foggy over the past few months. The good thing is that you knew that material fairly recently, so it'll probably just take a little review to freshen up those concepts and problems.
What about things you didn't do well on earlier exams? Chemistry (and many other fields...) is a cumulative subject. We tend to look at things from a bunch of different directions, and we often approach a concept or problem 2 or 3 different ways over the course of the semester and year. In the past few months, we might have looked at something differently in a way that suddenly makes perfect sense to you.
Use your exams to jot down and prioritize things to study. It won't be a perfect list, but it will give you a good starting point.
2. Review notes from class
Once you've identified the things you need to review, look back at your notes from class.(Sometimes this serves as a reminder to take better notes in the future!) For some topics, a brief reminder from your notes will be all that you need to bring these concepts and problems up from the cold, dark storage room of your memory. For others, it's a good way to once again identify and prioritize your study topics.
3. Use the book
For topics that are still fuzzy, look up key words in the index to help you find a good place to start. Not sure you remember integrated rate laws? Look it up in the index and hit those pages!
It's also not a bad idea to skim over sections that you're pretty sure you understand. Once you're fairly comfortable with a topic, it becomes easier to pick up some of the more subtle points.
End-of-chapter problems can also be helpful because they present information a little bit differently than I do. I'm not talking "better" or "worse" here, the book just uses different wording to ask the same questions I ask. At the end of my class, I don't just want you to be good at answering questions that I write, I want you to understand chemistry, even if someone else is asking the questions.
If you get through all of those steps and still have time, energy, and a thirst for more knowledge, use your favorite tool to search the World Wide Interwebs Net or stop in and chat with me. This isn't a perfect list of study tips, but it's a good place to start. Identify topics, prioritize your studying time and energy, look for connections and common themes in the material, and good luck. You can do it!
Showing posts with label solutions. Show all posts
Showing posts with label solutions. Show all posts
2014-05-02
2014-03-30
Pre-Exam 3 email questions...
A few questions have come in by email, here they are:
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I was going through some of the old exams and the problems that use the Henderson - Hasselbalch. In those examples when it asked for the concentration of the conjugate base over the concentration of conjugate acid, but in the key only the moles of both are put in those places. Why?
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Keep an eye on the weather... Unless MSUM officially closes campus tomorrow, we will have class and the exam as planned. I'll be there at 7:30.
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I have a question regarding problem 9 on exam 3a from spring 2013. I understand that the calculated x value doesn't fit under the assumptions. I see where the first two values come from. I was wondering where the -1.238 x 10^-3 came from?
{(x)(x)} / (0.516 – x) = 2.40 x 10^-3
0.516 is the initial concentration and 2.40 x 10^-3 is Ka.
x^2 + (2.40 x 10^-3)x + (-1.238 x 10^-3 ) = 0
In order to use the quadratic formula, we have to solve the equation to the form:
ax2 + bx + c = 0
The (-1.238x10-3) term comes from (2.40x10-3)(0.516).
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I was going through some of the old exams and the problems that use the Henderson - Hasselbalch. In those examples when it asked for the concentration of the conjugate base over the concentration of conjugate acid, but in the key only the moles of both are put in those places. Why?
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This is a little mathematical shortcut. Since both the conjugate acid and conjugate base are in the same total volume of solution, the volumes mathematically cancel so I left them out. For example, if we had a buffer made from 0.65mols of HA and 0.55mols of A-1 in 800.0mL of buffer solution, that last part of the Henderson-Hasselbalch would look like:
You can always keep the volume in there and calculate the actual concentrations of each component, you should get exactly the same answer either way.
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I dont understand how you can derive pH values from pka's given in the question. I also dont understand how pH can be calculated at each eq point in the titration curve as in number 11 on spring 2013 where it asks what indicator to use.. it says use the 2 pkas...how does this help?
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I dont understand how you can derive pH values from pka's given in the question. I also dont understand how pH can be calculated at each eq point in the titration curve as in number 11 on spring 2013 where it asks what indicator to use.. it says use the 2 pkas...how does this help?
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There are a couple ways that pKa (or pKb) can lead to a pH. One possibility is in a question like "What is the expected pH of a 0.618M solution of ammonium nitrate solution?" In this question, you can set up a Ka-type equilibrium for ammonium ions and use the Ka of ammonium to calculate [H3O+] and pH. This is similar to the problem I posted yesterday (http://chemistryingeneral.blogspot.com/2014/03/neutral-salts.html). This method can be used to calculate the initial pH for a titration. It would also work to approximate the pH of an equivalence point. Let's think about that...
For the titration of phosphite ions with hydrochloric acid, we can calculate the initial pH by setting up a Kb-type equililbrium and using the Kb of phosphite ion to calculate [OH-1] and pOH and pH. At the first equivalence point in this titration, we have a solution that we can think of as HPO3-2(aq) because we have added just enough acid to complete the following equation exactly once:
PO3-3(aq) + H+(aq) <=> HPO3-2(aq)
Between equivalence points, we have buffering regions of the titration curve... at the mid-point of this buffering region, the pH is equal to the pKa of the weak acid of the mixture. If we know the pKa (and therefore the pH) on either side of the equivalence point we're interested in, we can get a pretty reliable estimate of the pH of that equivalence point. On the titration curve below, if the pKas that bound the equivalence point are 6 and 9, the pH of the equivalence point should be right between them at pH = 7.5.Keep an eye on the weather... Unless MSUM officially closes campus tomorrow, we will have class and the exam as planned. I'll be there at 7:30.
2014-01-19
Quiz 1 questions...
I've gotten a few questions about Quiz 1. Here are a few hints...
Question 1: How many moles of nitrogen atoms are in a given mass of iron(III) nitrate?
Start with a balanced chemical formula. How many moles of N are in each mole of iron(III) nitrate? Use the mole ratio to get moles of N from the moles of iron(III) nitrate.
Question 2: How many moles of chlorine atoms are in a given volume of a given concentration vanadium(IV) chlorate solution?
This is almost the same as question 1, but you need to use concentration and volume to determine the moles of vanadium(IV) chlorate instead of mass and molar mass.
Two other helpful hints:
1. Don't wait until the day before a quiz is due to work on it. This is especially true because...
2. When you have a question, show me your work. If you have the problem all set up, it's much easier for me to look at what you've done and help you work through the problem correctly. It's always easier to help you when you can give me an idea of where you're getting confused.
Question 1: How many moles of nitrogen atoms are in a given mass of iron(III) nitrate?
Start with a balanced chemical formula. How many moles of N are in each mole of iron(III) nitrate? Use the mole ratio to get moles of N from the moles of iron(III) nitrate.
Question 2: How many moles of chlorine atoms are in a given volume of a given concentration vanadium(IV) chlorate solution?
This is almost the same as question 1, but you need to use concentration and volume to determine the moles of vanadium(IV) chlorate instead of mass and molar mass.
Two other helpful hints:
1. Don't wait until the day before a quiz is due to work on it. This is especially true because...
2. When you have a question, show me your work. If you have the problem all set up, it's much easier for me to look at what you've done and help you work through the problem correctly. It's always easier to help you when you can give me an idea of where you're getting confused.
2013-07-18
In-Class problems 2013-07-18
Today was buffers day in class so we looked at 294 different ways to use the Henderson-Hasselbalch equation. Remember, the Henderson-Hasselbalch equation is just a rearrangement of the Ka expression we use for understanding any acid... You can always plug directly into the Ka expression and get the same result you will get using Henderson-Hasselbalch. So the problems...
1. Combine 12.642g of chlorous acid
with 15.372g of lithium chlorite, dilute to 500.0mL. What is the
expected pH of this buffer? (Chlorous acid Ka = 1.12x10-2)
Plug into the Henderson-Hasselbalch
equation...
And solve...
2. Prepare 500.0mL of a 0.650M HN3/N3-1
buffer at pH = 5.10 from HN3(s) and NaN3(s).
(Hydrazoic acid Ka = 1.93x10-5)
Start off by solving for the ratio of
conjugate acid to conjugate base using either the
Henderson-Hasselbalch equation or just an unmodified Ka
expression. I'll use H-H...
[N3-1]
/ [HN3] = 2.4297
[N3-1]
= 2.4297[HN3]
Now that we know the ratio
of these concentrations, we can solve for the actual concentrations
by using the relationship...
0.650M = [HN3]
+ [N3-1]
0.650M = [HN3]
+ 2.4297[HN3] = 3.4297[HN3]
[HN3] = 0.1895M
[N3-1]
= 0.650 – [HN3] = 0.650 – 0.1895 = 0.4605M
To make a solution that's 0.1895M HN3
at 500.0mL, we need...
(0.5000L)(0.1895M) =
0.09475mols HN3
(0.09475mols
HN3)(43.029g/mol) = 4.077g HN3(s)
To make a solution that's 0.4605M N3-1
at 500.0mL, we need...
(0.5000L)(0.4605M) =
0.23025mols N3-1
(0.23025mols
N3-1)(65.011g/mol) = 14.969g NaN3(s)
So we should be able to make the target
buffer by combining 4.077grams of HN3 and 14.969g of NaN3
in enough water to make 500.0mL of solution.
NOTE: If you're actually making a
buffer, be very careful about the order of addition of the
components. Whenever you're combining a solid or concentrated
solution with a solvent, it's usually a good practice to add the
solid or concentrated stock to the larger volume of solvent slowly
with very good mixing. Dissolving and/or mixing can liberate a LOT of
heat in some cases that could be dangerous if the order of addition
is reversed.
NOTE2: Hydrazoic acid is not a solid at
room temperature, and the pure liquid is a non-trivial safety risk...
We can talk about this type of a buffer on paper, but there's very
little chance you (or I) will ever prepare or use a hydrazoic
acid-based buffer.
3. What is Ka of a weak
acid, “HA”, if a solution made by dissolving 0.316mol HA and
0.327mol A-1 in water and diluting to 750.0mL has a pH of
9.374?
Plug in to Henderson-Hasselbalch or the
generic Ka expression...
9.374 = pKa +
log (0.327 / 0.316)
pKa = 9.359
Ka =
4.374x10-10
4. What is Kb of a weak
base, “B”, if a solution made by dissolving 0.143mol B and
0.158mol HB+1 in water and diluting to 400.0mL has a pH of
5.975?
Similar to the previous problem, plug
in to Henderson-Hasselbalch or the generic Ka expression...
5.975 = pKa +
log (0.143 / 0.158)
pKa = 6.018
pKb = 14 –
6.018 = 7.892
Ka = 1.043x10-8
2013-05-14
Email question - van't Hoff factor
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Can you please explain how to determine the i factor. Now matter how much I read the book and look at examples I can't figure it out. Thanks,
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There are a couple things that could be "the i factor", but I'll assume this question is about the van't Hoff factor that's used when determining changes due to colligative properties. Remember, colligative properties are properties of a solution that depend only upon the number of solute particle, not the identity of those solute particles. It doesn't matter if those solute particles are molecules, or cations, or anions, or a big random mixture of them all. When an ionic compound dissolves in water, it dissociates (at least somewhat) into its component ions. The number of ions (particles) that each formula unit breaks into when it dissolves is the van't Hoff factor. For example, if NaCl(s) is dissolved in water, it forms 1 Na+1(aq) ion and 1 Cl-1(aq) ion, so each formula unit (NaCl) forms 2 particles (ions) in solution; the van't Hoff factor is "2". For polyatomic ions, the ions don't break down into their individual atom, they stay polyatomic ions, so if K2SO4(s) is dissolved in water, it forms 2 K+1(aq) ions and 1 SO4-2(aq) ion in solution; the van't Hoff factor is "3".
Can you please explain how to determine the i factor. Now matter how much I read the book and look at examples I can't figure it out. Thanks,
----------
There are a couple things that could be "the i factor", but I'll assume this question is about the van't Hoff factor that's used when determining changes due to colligative properties. Remember, colligative properties are properties of a solution that depend only upon the number of solute particle, not the identity of those solute particles. It doesn't matter if those solute particles are molecules, or cations, or anions, or a big random mixture of them all. When an ionic compound dissolves in water, it dissociates (at least somewhat) into its component ions. The number of ions (particles) that each formula unit breaks into when it dissolves is the van't Hoff factor. For example, if NaCl(s) is dissolved in water, it forms 1 Na+1(aq) ion and 1 Cl-1(aq) ion, so each formula unit (NaCl) forms 2 particles (ions) in solution; the van't Hoff factor is "2". For polyatomic ions, the ions don't break down into their individual atom, they stay polyatomic ions, so if K2SO4(s) is dissolved in water, it forms 2 K+1(aq) ions and 1 SO4-2(aq) ion in solution; the van't Hoff factor is "3".
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