Neutral salts

The conjugate of a strong acid or strong base is neutral. That's just something we tend to accept, memorize, and move on. But why? Those two words are a big part of the reason I'm a chemist.

Let's take a look at a strong acid and see if we can make sense of this. Of the typical strong acids, nitric is usually the weakest, and nitric is also the only one that might have a K_a listed in standard tables. The stronger strong acids have really useful K_a values listed in the tables like "large" or "strong"... I don't have a "large" button on my calculator, so let's just use nitric acid and we'll hopefully see why the other strong acids would follow the same trend if we had a value for their K_a.

The K_a for nitric acid is usually listed at around 25. That means the K_b for nitrate ions is:

That's a REALLY weak K_b, but we can go ahead and calculate the pH of a solution just like in any other situation. How about the problem: What is the expected pH of a 0.500M solution of sodium nitrate? {By the way, we could make similar arguments to show that the sodium ions don't affect the pH, but we'll save those for another day...} As with all good equilibrium problems, it's probably not a bad idea to start with a table:

Now we can set up the K_b expression and plug in the numbers we have:

We should be able to simplify that with some assumptions... Let's assume that "x" is much smaller than 0.500 and much larger than 10^-7. That gets us the simplified expression:

Solving this expression, we get x = 1.41x10^-8, which gives us a pOH = -log(1.41x10^-8) = 7.85, and pH = 6.15. Hmm, that's not neutral, that's acidic. The whole point of this was to prove that nitrate was a neutral ion. This is a disaster.

BUT WAIT!

We made some assumptions. We didn't check our assumption after we solved for "x". This is why I always tell you to check assumptions... We assumed that "x" would be much smaller than 0.500, which it is, but we also assumed that "x" was much larger than 10^-7, which it absolutely is not! So the assumptions we made were an oversimplification of the problem and that's where we entered the danger zone. Looking back at our K_b expression, we can only simplify it to:

That's still going to require the quadratic formula to solve. I'll let you work out the details, but the result should be that x = 7.96x10^-9. That means:

[OH^-1]eq = 10^-7 + (7.96x10^-9) = 1.08x10^-7 M

pOH = -log(1.08x10^-7) = 6.9667

pH = 14 - 6.9667 = 7.0333

That's not exactly 7.0000000000 neutral, but it's pretty darn close, especially if we're thinking about this in terms of selecting a visual acid-base indicator for a titration. 

Lab Reports

The exact requirements for a lab report will vary from field to field, class to class, even instructor to instructor. If you're looking for an example of a "good" lab report for my Gen Chem class, try this one:
http://www.drbodwin.com/teaching/genchemlab/iodinationlabreport12a.pdf
The most common problem I see in lab reports is that students don't always explain the experiment and its results in a way that makes it (somewhat) clear that the concepts behind the experiment are understood. The purpose of a Gen Chem experiment is almost never "We collected a bunch of numerical data, made some observations, and calculated/determined this result". What does that result mean? How is that result related to the concepts we talked about in class? How can that result help inform the exercises and exam questions you'll see in the classroom?
One of the harder things for students to do is get a feel for "reasonable" answers because Gen Chem level students don't have a lot of experience looking at these answers. Activation energy is a great example of this. If you have no feeling for how activation energy relates to the observed rate of a reaction, you might calculate an activation energy of 25 J/mol for some problem. Is that a fast reaction or a slow reaction? If you've only every done on-paper activation energy problems, that might be a hard question to answer. The advantage of doing experiments is that you have personally observed what happened, you've gained experience that will help you make some of these judgement calls. For the iodination of acetone experiment, the reaction is fast enough to easily observe, but it's not so fast that it blows up in your hand. The activation energy for the iodination of acetone is somewhere around 80-100 kJ/mol. If 80-100kJ/mol is the activation energy for a reaction that's "kinda fast, but not super fast", what do you think about that 25 J/mol activation energy reaction? {Pay attention to units.}
Lab experiments are a great way to build your knowledge base. When you're writing a lab report, think about the bigger picture and show the reader that you've recognized the link between classroom exercises and the first-hand experience you've had in the lab.