Keeping track of concentrations

When doing kinetics and equilibrium problems, especially when you're doing an actual hands-on experiment, there are a lot of different concentrations to keep track of. The key to keeping them straight is mostly careful reading and organization, but there are a couple common definitions or descriptions that can help.
Stock Concentration
This is the one that comes up most common in a lab experiment. "Stock" refers to the large samples of reagent from which smaller amounts are taken for individual experiments. When you come to lab, the big bottles or carboys of solution that are on the side benches or in the dispensing hood are "stock" solutions and should have a "stock concentration" listed on the bottle. Most data analysis in lab begins with stock concentrations.
Initial Concentration
In either kinetics or equilibrium problems or experiments, we will often come upon something called an initial concentration. "Initial concentration" is (to me at least) quite fascinating because it's one of those things that we can do on paper that's just not physically possible to do in the real world. The "initial concentration" in a problem or experiment is the concentration of reactants after mixing everything together but before any reaction is allowed to take place. It's as if there was a little "start reaction" button on the side of the beaker, and nothing reacted until we pushed that button. In the real world, as soon as reactant solutions come in contact with each other, they begin to react, so the "initial concentration" is never the actual concentration we might observe in a reaction mixture. The initial concentration is most commonly calculated as a dilution of the stock concentration.
{In some specific reactions, we can probably observe an initial concentration because either the reaction is SO slow that we can mix the reactants before any measurable reaction has occurred, or because there's some external stimulus (like light or heat) required to make the reaction start. These reactions aren't that rare, but they're not reactions that we're likely to use very often in Gen Chem.}
Partial Pressures
When we're working with gases, we can often use Molarity to express the concentration of reactants, but we can also use partial pressures of the gaseous components. Partial pressures are a way to measure the number of moles of a specific gas in a mixture of gases. {Dalton's Law of Partial Pressures} That sounds an awful lot like a concentration... As with the vast majority of data analysis in chemistry, counting moles is the key to figuring out relationships between reacting species. If we know the concentration and volume of a liquid solution, we'll probably be calculating moles at some point. If we know the partial pressure of a gas, we'll probably also be calculating moles at some point. Chemistry is all about the mole!

The most important thing to do when approaching these problems is organization. This is especially true of equilibrium problems; if we can organize the information given in the problem, we'll be much more successful.

Did you hear about the chemist who was thinking very hard about removing excess solvent from a solution? She was concentrating.

Suggested Problems posted

A few people have asked about some suggested end-of-chapter problems from your book. I've picked some from our current chapters (http://www.drbodwin.com/teaching/genchem.php), this should be a good list for you to start with. Use these to identify the areas that you need to study/learn/review a little more, work through additional problems as needed. And always ask questions if you're unsure.

Exam keys all posted

I had a little technical "oops", all of this semester's Gen Chem II exams and keys are now posted. http://www.drbodwin.com/teaching/jbgenchem.html

A few people have asked about what equations and/or other information will be provided with the exam.  I cannot show you the exact information sheet that will be on the exam, but it is not all that different from the front page information you get with the exams you have taken (look at page 1 of Exam 4).  The ACS exam doesn't include everything I do, and it has a few things that I don't, but the big equations/formulas/constants are all there.

Standardized Final Exam

Your final exam will be a standardized American Chemical Society exam.  I know a number of you are a little anxious about this, but there are a few things to remember:
1. "Standardized" does not mean "impossible unless you're a genius".  "Standardized" just means that there are national averages to which your score can be compared.  This allows your performance to be scaled to Gen Chem students at any other school.
2. If I've done my job, I have helped you learn all the fundamental material in General Chemistry.  I certainly hope that I'm giving you as good an education in General Chemistry as you would get at any other university, in fact, I would hope that I'm giving you a better Gen Chem education than you would get at most other universities.  By occasionally using a standardized exam, I can see what things I'm doing well and what areas I need to improve to give future Gen Chem students the best possible education.
3. The format and length might vary slightly, but expect 70 questions and a 110 minute time limit.  That's almost 2 hours, and 1.5 minutes per question.  Some questions will be quicker than others, but at 1.5 minutes per question, none of them can be huge.  Many of the questions on my exams are probably 10-20 minute questions by the time you work through all the parts, none of the questions on the ACS exam will be anywhere near that complex.
4. 70 questions means that each question is worth XX% of the total score. {I'll leave the calculation of "XX" for you to do as a practice math problem.}  Scaling that to 200pts (the value of your final exam), means that each question would only be worth YYpts if I plugged your scores directly into my grade sheet.  {Again, "YY" is a practice math problem...}  On my exams, if you totally miss one of the big questions, you lose ~10%, sometimes more.  Yikes.
5. It is quite unlikely that I will plug your scores directly into my grade sheet, I will almost certainly be using some sort of scaling formula to calculate a score out of 200 points.  That doesn't mean you shouldn't prepare well for the exam, a good score can really help and a poor score can definitely hurt your final score/grade in the class.
6. There are resources available for you to use as practice.  The actual ACS Gen Chem exams are not published, BUT every year the ACS prepares exams for the Chemistry Olympiad for high school students.  These are not identical to the ACS Gen Chem exam, but the questions are similar and the style/format is almost the same.  The ACS posts their old Chemistry Olympiad exams online, so you can look them over to help you prepare.  Check out: http://portal.acs.org/portal/acs/corg/content?_nfpb=true&_pageLabel=PP_SUPERARTICLE&node_id=1508&use_sec=false&sec_url_var=region1&__uuid=1c147f5a-ee17-44e2-af0c-87ca446c65a8
 for 10+ years of previous Chemistry Olympiad exams from both the local and national level competition.

Good luck in your preparation.

Redox lab question

Quite a few people have questions about the redox lab, so let me give a hint/some guidance to everyone...

In the first part, you looked at reactivity and found Zn to be the most active metal, followed by Pb, then Cu.  I'll just use those three as an example, you will also need to include Ni and Ag in your assignment.  You measured the potential for a Zn/Pb cell, a Pb/Cu cell, and a Zn/Cu cell.  Is there a relationship between those measured potentials?  There's a relationship between those reactions, but how are cell potentials related to one another?

Zn(s) + Pb²⁺(aq)  ⇄  Pb(s) + Zn²⁺(aq)

Pb(s) + Cu²⁺(aq)  ⇄  Cu(s) + Pb²⁺(aq)

Zn(s) + Cu²⁺(aq)  ⇄  Cu(s) + Zn²⁺(aq)

Are cell potentials like kinetics (the cell potential for the overall process is determined by the lowest potential)?  Are cell potentials like equilibrium (the cell potential for the overall process is the product of the step-wise potentials)?  Or is there another relationship between the step-wise potentials and the overall potential?  When you think you see (observe) a relationship with the Zn/Pb/Cu system (hypothesis), check to see if the same relationship is true with some of the other cell combinations you measured.(test/experiment)

{Hmm, it looks like we could use the scientific method to analyze the data and results from this experiment.  Who would have guessed?!}

On your hand-in assignment for lab, the "calculated" cell potentials for pairs that are not next to each other refers to the treatment you see above.  You have measured all of the potentials for cell constructed from metals that are adjacent to each other in your activity series (step-wise potentials), so if there is a relationship between step-wise potentials and the potentials for cells constructed from metals that are not adjacent to each other in your activity series, you should be able to calculate the expected cell potential for those non-adjacent cells.

Voltaic Cells

Not a lot of information right now, but I just finished drawing a voltaic cell diagram.  Not perfect, but I'm pretty satisfied with it.

Balancing Redox Reactions

Balancing redox reactions can be pretty simple for some system, but some redox reactions can be exceptionally challenging. To balance any redox reaction, we can follow a systematic set of steps, and every Gen Chem book happily provides a set of steps. In my experience, most books use rules that require a little bit of faith and function like a black-box. There are a LOT of places that mistakes can be made when balancing redox reactions, so I prefer to use rules that have built-in places to check my answer before I go through the whole process. Here they are:

Balancing redox rules (in acidic or neutral aqueous solutions):

1. Assign oxidation numbers to all atoms in the equation

2. Identify the oxidation and reduction half reaction

3. In each half reaction, balance all atoms except hydrogen and oxygen

4. In each half reaction, add electrons to the reactant or product side to balance the change in oxidation state. Note: you are not adding electrons to balance charge

5. In each half reaction, add water molecules to balance any oxygen atoms

6. In each half reaction, add H⁺(aq) to balance any hydrogen atoms

7. At this point, the half reactions should be balanced, check the charge balance to confirm

8. Multiply each half reaction by an appropriate integer to balance the electrons involved in the oxidation and reduction processes

9. Add the half reactions together

10. Again, the resulting reaction should be balanced, check the charge balance to confirm

11. Cancel out any spectator species, sit back and pat yourself on the back for writing such a lovely balanced redox equation.

These rules work well, BUT rely upon some assumptions. First, since we're using water and H⁺(aq), the reaction must be taking place in aqueous solution that is neutral or acidic. That's OK for most redox reactions in Gen Chem, but once in a while we'll run into a rxn that takes place in basic aqueous solution. How do we handle that? Think about it... basic solutions have excess (relatively speaking) hydroxide ions. Hydroxide ions react with H⁺(aq) ions to form water. If a reaction is taking place in basic aqueous solution, balance it according to the above rules and then add:

12. Add enough OH^-1(aq) to each side to react with all the H⁺(aq) that is present

13. Check charge balance

14. Cancel any excess water

As with any process, practice is the key, so practice balancing redox rxns, then practice a little more, and when you think you have it all figured out, practice a couple more times. We'll do some of that in class...

Oxidation Numbers

Oxidation numbers describe the balance between electrons and protons on an atom, whether that atom is happily floating around all by itself or part of a massive molecule. Oxidation numbers can be determined two different ways: by using rules, or by looking at structure. Let's look at the rules first.

Oxidation Numbers by the Rules:
1. For neutral, uncombined elements, Ox# = 0. Examples: Fe(s), H₂(g), Hg(l), Ne(g)
2. For monoatomic ions, Ox# = charge. Examples: Fe²⁺(aq) {Ox# = +2}, P^3-(g) {Ox# = -3}
3. Oxygen is almost always Ox# = -2, except in O₂ {Ox# = 0, Rule #1} and peroxides {Ox# = -1}
4. Hydrogen is almost always Ox# = +1, except in H₂ {Ox# = 0, Rule #1} and hydrides {Ox# = -1}
5. The sum of the Ox#s on all the atoms in a polyatomic molecule or ion is equal to the charge on the whole polyatomic molecule or ion.

Let's look at a redox reaction and assign Ox#s by the rules:

Cl₂(g) + 2 O₂(g) ↔ 2 ClO₂(g)

Cl₂(g) : Rule #1, Ox# = 0

O₂(g) : Rule #1, Ox# = 0

ClO₂(g) : Rule #3, oxygen is Ox# = -2.

ClO₂(g) : Rule #5, (Ox# Cl) + 2(Ox# O) = 0 (the charge on a neutral molecule)

(Ox# Cl) + 2(-2) = 0

(Ox# Cl) = +4

So in this redox reaction, each Cl is going from 0 to +4, losing 4 electrons, Losing Electrons is Oxidation; and each O is going from 0 to -2, gaining 2 electrons, Gaining Electrons is Reduction.

For many substances, it's actually easier to assign Ox#s by looking at the structure. The process is very similar to finding Formal Charge, the electrons are just assigned a little differently. Formal Charge assigns electrons as if all bonds are purely covalent, meaning that all bonding pairs of electrons are split with one electron given to each atom in the bond. Oxidation Number assigns electrons as if all bonds are purely ionic, meaning that all of the bonding electrons go to the more electronegative element in the bond.

Oxidation Number by the Structure:
1. Draw a good Lewis Structure
2. Assign all bonding electrons to the more electronegative element in the bond
3. Compare the electrons assigned to each atom to the valence electrons of the neutral element

Again, let's look at an example, in fact, let's look at the same example as above, ClO₂(g). Drawing a good Lewis Structure:

This is a very interesting molecule, it violates the octet rule and it has an unpaired electron. Looks like it would be pretty reactive. Looking at the electronegativities, oxygen is more electronegative than chlorine, so all of the bonding electrons will be assigned to oxygen, giving each oxygen 8 assigned electrons and the chlorine 3 assigned electrons.

Neutral oxygen has 6 valence electrons, we've assigned 8 electrons to oxygen, so the oxidation number for oxygen in this molecule is -2, just like we predicted using the rules. Neutral chlorine has 7 valence electrons, we've assigned 3 electrons to chlorine, so the oxidation number for chlorine in this molecule is +4, again, just like we predicted using the rules. Both methods work. Why would you ever use structures when the rules work? Try looking at hydrogen peroxide, H-O-O-H, or a more complex molecule like glucose. The rules don't always give the best picture of what's happening in a molecule, and this can ultimately make it harder to predict reactivity or other behaviors.

Redox - Definitions

If you want to understand chemistry, you have to follow the electrons. If electrons are transferred during a chemical reaction (as opposed to just being rearranged...), then a reduction-oxidation process is taking place. To help keeping track of the electron transfer in redox processes, we can use a couple acronyms/mnemonics related to the definitions of reduction and oxidation.

OIL / RIG - “Oxidation Is Losing electrons” / “Reduction Is Gaining electrons”

LEO / GER - “Losing Electrons is Oxidation” / “Gaining Electrons is Reduction”

Hmm, why is “reduction” associated with gaining electrons? Remember, electrons are negatively charged, so gaining electrons increases the number of negatively charged particles associated with an atom which reduces its net charge. But are we really looking at “charge” to determine redox chemistry? Sometimes it seems like it, but other times the charge doesn't seem to line up with the processes. We really have to look at oxidation number, which is related to charge, but a with some subtle differences. One way to distinguish charge and oxidation number is that “charge” can be used to describe the net overall balance between electrons and protons in a system that might contain multiple atoms, but “oxidation number” describes the balance between electrons and protons for each individual atom in a structure regardless of its size. Oxidation numbers sound important enough for their own post, look for it soon.

Another thing that can cause some mix-ups is the term “oxidation” or {in verb form} “oxidize”. {Or for the British English spellers in the crowd, “oxidise”.} The element oxygen is very often involved in redox reactions. Is oxygen usually undergoing oxidation or reduction? You know you want to say oxidation, the words look so similar... But if we start with molecular oxygen, O₂(g), it's almost always going to gain electrons. Gaining Electrons is Reduction. GER, indeed! The important thing to remember here is that reduction and oxidation are ALWAYS coupled processes. You can't have one without the other. This leads to some other terminology...

The process of one substance undergoing oxidation causes something else in the system to be reduced. The substance that is being oxidized is the reducing agent or reductant because it is causing reduction to take place.
The process of one substance undergoing reduction causes something else in the system to be oxidized. The substance that is being reduced is the oxidizing agent or oxidant because it is causing oxidation to take place.

So if oxygen is usually being reduced, it is a good oxidizing agent.

Qualitative Analysis of Metal Cations - Week 2

This week in lab you're going to be analyzing an unknown mixture of the metal cations you studied last week.  To do this, you need to look over the tests you did last week and find a way to sequentially use some or all of those tests to separate the cations.  This is NOT a "run all the tests and figure it out later" experiment, you have to have a plan.
To develop your plan, assume you are starting with a mixture of all 5 cations, and you want to separate them into 5 different containers.  The key to developing a good flow chart is the ability to separate solids from each other when multiple things precipitate.  For example, if the first step of your flow chart is "add chromate", you will precipitate all 5 cations as their chromate salts.  You have no way to separate these solids from each other, so this would be a VERY bad first step.  Look over some of the multi-step tests you did last week.  If adding some reagent causes 2 or 3 of the cations to form precipitate, and you have a way to separate those precipitates from each other (with another step in the multi-step test you performed last week), then that might be a good place to start.
As an example, what if you had a mixture of NaCl, NaNO₃, sand, and sawdust.  How could you separate them?  If you added water, the NaCl and NaNO₃ would dissolve, the sand would sink, and the sawdust would float.  That accomplishes some of the separation, but what about the dissolved salts?  If there was a reagent you could add to make chloride ions form a precipitate, like maybe Pb²⁺(aq), you would be able to separate the chloride from the nitrate.
{OK, picky people in the crowd, that doesn't exactly separate "NaCl" from "NaNO₃", but it illustrates the point!}
Given the tests you performed, there are a few different flow charts that will work to separate the 5 cations you're working with, so if you have something that's a little different from someone else, that's OK.  It would be great if you compared your flowchart to someone else's and had a discussion about the similarities and differences, it might lead you BOTH to make better flowcharts/plans.

Titrations I - Monoprotic/Monoprotic

When we're dealing with acid-base chemistry, we have to be able to determine the concentration of the acids and bases that we're using. This most often means performing a titration where one of the components is known. The absolutely 100% most important thing about titrations is to remember that titrations are stoichiometry problems, and they should be treated as such. There's nothing new and magical about titrations, they're just stoichiometry problems applied to a specific situation. The 4 steps to solve every stoichiometry problem are:

1) Write a balanced chemical equation

2) Calculate moles of one component (“moles of known”)

3) Convert moles of known to moles of interest using the ratios in the balanced chemical equation

4) Convert moles of interest to whatever you want to find

With acid-base titrations, it is most often convenient to monitor pH of the solution. To explore the pH-dependence of a titration, we can look at a system where we know all the concentrations. Let's say we are titrating 25.00mL of 0.85M nitrous acid, HNO₂(aq), with 0.85M KOH(aq). At the beginning, we have a solution of a weak acid, so we can calculate the initial pH using a K_a expression. Looking up the K_a of nitrous acid online gives a couple different values, let's say that it's around 5x10^-4 . K_a is just like every other equilibrium, so let's set up a table...

	HNO2(aq) +	H2O(l) ↔	H3O+(aq) +	NO2-1(aq)
[ ]initial	0.85M	XXXX	10-7 M	0 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.85 – x) M	XXXX	(10-7 + x) M	x M

Plugging in values:

That expression is solvable by the quadratic formula (and I encourage you to solve it for practice and to convince yourself that the things we're about to do are valid), but we might be able to simplify it if we make some assumptions about the size of “x”. Since this is a weak acid (K_a < 1), it's probably reasonable to assume that most of the HNO₂ molecules will still be intact when this system reaches equilibrium, so although we will lose some (Δ [ ] = -x), the value of “x” will probably be quite small compared to 0.85, so we can assume that (0.85 – x) ≈ 0.85. At the same time, although “x” is probably small, 10^-7 is really small, so there's a pretty good chance that (10^-7 + x) ≈ x. With these two assumptions (which we have to check later) in place, the K_a expression simplifies to:

x = 0.0206

Before we do anything else, we need to pause to check that the assumptions we made are indeed reasonable. 0.0206 is truly massive compared to 10^-7 so the second assumption is great. For the first assumption, it's a little closer; 0.0206 is certainly smaller than 0.85, but is it smaller enough?

(0.0206 / 0.85 ) *100 = 2.4%

Usually, the limit is around 5%, so we should be OK here as well. The pH of this solution at the beginning of the experiment should be:

pH = -log[H₃O⁺] = -log(0.0206) = 1.686

What happens to the pH when we start adding KOH(aq)? Well, KOH is a base, so the pH will go up, but how will it go up? We can calculate that. Starting with 25.00mL of 0.85M nitrous acid, let's add 5.00mL of 0.85M KOH(aq). To make it easier to keep track of everything, let's start by writing out a chemical equation and converting to moles:

HNO₂(aq) + KOH(aq) ↔ H₂O(l) + KNO₂(aq)

Mols HNO₂: (0.02500L)(0.85M) = 0.02125mols HNO₂

Mols KOH(aq): (0.00500L)(0.85M) = 0.00425mols KOH

We can assume that all of the OH^-1(aq) that is added will react with the H⁺/HNO₂ that is present in solution, so after this 5.00mL addition, we should have a solution that contains:

0.02125 – 0.00425 = 0.017mols HNO₂(aq)

This nitrous acid is now in (25.00 + 5.00 = 30.00mL) of solution {if we assume that the volumes are additive}, so the concentration of nitrous acid is:

0.017mols / 0.03000L = 0.5667M

And the concentration of nitrite ions is:

0.00425mols / 0.03000L = 0.1417M

We can calculate the pH by treating this like an equilibrium problem, just like above, setting up a table. The pH of the solution should be the same as if we had made a new solution by adding 0.017mols of HNO₂ and 0.00425mols of NO₂^-1 to enough water to make 30.00mL of solution:

	HNO2(aq) +	H2O(l) ↔	H3O+(aq) +	NO2-1(aq)
[ ]initial	0.5667M	XXXX	10-7 M	0.1417 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.5667 – x) M	XXXX	(10-7 + x) M	(0.1417 + x) M

Making similar assumptions, the expression simplifies to:

x = 0.002000

Checking assumptions again, they are all valid, so:

pH = -log[H₃O⁺] = -log(0.002000) = 2.699

We can continue adding 5.00mL portions of the KOH(aq) and calculating to get the values shown in the table:

mL KOH(aq) added	pH
0.00	1.686
5.00	2.699
10.00	3.125
15.00	3.477
20.00	3.903

But what happens when 25.00mL of the KOH(aq) solution is added? At that point, the mols of OH^-1 that have been added is exactly equal to the mols of acid that were present in the original solution. This is an equivalence point in the titration. If we look at the balanced chemical equation for the process,

HNO₂(aq) + KOH(aq) ↔ H₂O(l) + KNO₂(aq)

The solution we have produced by titration is the exact same solution that we would have produced if we had simply dissolved 0.02125mols of KNO₂ in enough water to make 50.00mL of solution. Since NO₂^-1(aq) is the conjugate base of nitrous acid, we can think about it being involved in a K_b equilibrium with water, K_b(NO₂^-1) = 2x10^-11. Set up a table...

	NO2-1(aq) +	H2O(l) ↔	OH-1(aq) +	HNO2(aq)
[ ]initial	0.425M	XXXX	10-7 M	0 M
Δ [ ]	- x M	XXXX	+ x M	+ x M
[ ]equilibrium	(0.425 – x) M	XXXX	(10-7 + x) M	x M

Again, we can make assumptions similar to the above examples to get x = 2.92x10^-6 = [OH^-1].

pOH = -log[OH^-1] = -log(2.92x10^-6) = 5.535

pH = 14 - pOH = 8.465

If we keep adding KOH(aq), the mixture will continue to get more basic, eventually leveling off as the concentration of excess hydroxide reaches a limit. {Sounds like a fascinating calculus problem, the concentration should asymptotically approach 0.85...} If we plot this pH data, we get a titration curve like the one shown below.