Exam 3 keys

The keys for Exam #3 are fixed and should work now, let me know if there are still problems. Also, I think the blank exam #3's that were posted were old copies that still had a typo in the big titration problem (asking for the concentration of selenic acid instead of potassium sulfite...). I posted the corrected blank exam #3's.

Let me know if there are other questions, I'll be in my office for at least part of the day today...

A couple process questions...

Questions:

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First off, for some reason, and this might be something just happening at this time, exam 3a and 3b answer keys are not available online.

Also, I was wondering if we are going to be given any formulas on the final in addition to what was listed on exam 4a with the periodic table.

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Yep, there's something not right with the keys for 3a and 3b, I'll get them fixed first thing tomorrow morning when I get to my office.

The cover page for your final exam will contain the same information as the cover page for Exam 4. If there are any other constants or equations you would like to see on the cover page, let me know and I'll consider it.

OK, everybody concentrate...

Email question...

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I am studying for the exams, and I just had a random question. Why are there Molarity and Molality? Aren't they almost always going to be the same number since 1 g H2O = 1 mL H2O? I guess I'm just curious of when you would choose one measurement over the other.

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In practice, yes, they are almost always the same (to our ability to measure them...), but the reasons we need them both are:

1. Molality is mols of solute per kilogram of solvent. Since the mass of a substance does not depend upon its temperature, molality is a concentration unit that is independent of temperature.

2. The density of pure water at (somewhere in the neighborhood of) 17degC is exactly 1g/mL, with many zeros of sig figs after the decimal point. For most dilute aqueous solutions at moderate temperatures, it's pretty safe to assume that the density of the solution is pretty close to 1g/mL, but there aren't nearly as many sig fig zeros after the decimal point. If you're working with a solution that's not dilute or not relatively close to that 15-20degC temperature, the density will start to vary significantly. That means the molarity (M) will change, but the molality (m) should not, therefore the conversion between these two units will not be 1:1. For an example you can try at home (or Kise...), take a little bit of water and add sugar or table salt to it until you get a very concentrated solution. Carefully (and slowly) pour that solution down the inside wall of a glass of "pure" water... the solution is more dense than the pure water and will sink to the bottom without mixing (if you're very careful).

3. What if your solvent is not water? In your freezing point depression experiment, your solvent was cyclohexane, so there was absolutely NOT a 1:1 relationship between the M and m of the solutions you were making.

Again, in many "real" situations the difference between M and m in an aqueous solution you're using will be so small that it's ignored. Similarly for other concentration unit conversions, we tend to simplify them in practice because the solutions we are most often working with in a lab (and especially a biology or biochemistry lab) are probably aqueous and probably relatively dilute. As long as you understand the assumptions you're using to simplify your in-lab calculations, you should be able to adjust to different situations or solvents...

Redox...

A question from email:

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Hey Dr. Bodwin I had a question on balancing equations. I can't tell if this equation is a redox or not

2 C2H2(g) + 5 O2(g) 􀀧 4 CO2(g) + 2 H2O(g)

I was looking for some guidelines on how to tell if an equation is redox or not, and if it is I am also looking for help on how to write the half reactions.

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OK, I'll answer the easier part of this question first. If a reaction is redox, then the oxidation numbers of some (or all) of the elements have to change. Let's look at ox#'s here...

In C2H2(g), let's use the "rules"... if each of the H's is +1 and the molecule is neutral, then the two C's must total -2. (The sum of all the oxidation numbers has to equal the charge of the molecule or polyatomic ion.) That means each C must have ox# = -1. {I'm assuming that the two C's are identical.}

Since O2(g) is an uncombined element in it's natural/standard state, the ox# = 0.

For CO2(g) it's back to the rules... if each oxygen is -2, then the carbon must be +4.

For H2O(g), the rules pretty much do all the work for us, H is +1, O is -2.

So in this reaction, carbon is going from -1 to +4 and oxygen is going from 0 to -2, there are changes in ox#'s, therefore this is a redox reaction. {Which one is reduction and which one is oxidation? Hmm...}

Now for the harder part of this specific reaction... writing half reactions for a combustion reaction like this is a little more involved than writing half reactions for metal/metal redox reactions because the oxygen is incorporated into both products. That means that our stepwise procedure for balancing redox reactions doesn't work all that great here... fortunately, hydrocarbon combustion reaction can usually be balanced by inspection/trial-and-error. If you have to balance a redox reaction (and write out half reactions) on the final exam, it will not be a hydrocarbon combustion. You should still be able to assign oxidation numbers and identify which element is being oxidized or reduced, but these are probably easier to balance by inspection.

Other questions? Let me know...