Working with ammonia

From my perspective, aqueous ammonia is a fascinating reagent to use in the lab. As a type of matter, it is a gas dissolved in a liquid, which seems pretty wild. In many cases, aqueous ammonia is just a fairly typical weak base that's nothing all that special as long as you're using the Bronsted-Lowry definition of a base. Those are great features of aqueous ammonia, but they really pale in comparison to what we see when we start combining aqueous ammonia with metal ions, especially transition metal ions. The key to thinking about aqueous ammonia in these situations is to remember that aqueous ammonia is always involved in a K_b-type equilibrium:

NH₃(aq) + H₂O(l) <=> NH₄⁺¹(aq) + OH^-1(aq)

This means that in any solution of aqueous ammonia, there are both ammonia molecules and hydroxide ions. If we think about the Lewis definitions of acids and bases, this means that floating around in every solution of aqueous ammonia, there are nitrogen-based lone pairs of electrons on ammonia molecules and oxygen-based lone pairs of electrons on hydroxide ions. Different metal ions have different affinities for different types of lone pairs, so sometimes when a metal ion is added to aqueous ammonia it forms complexes with ammonia while other times it forms complexes with hydroxide.
How do we tell which is which? Whenever possible, by comparison with know reactions. If the observed reaction between a metal ion and aqueous ammonia looks identical to the reaction of that same metal ion with a known hydroxide source {like NaOH(aq)}, then the metal ion is probably more attracted to oxygen lone pairs and is reacting with the hydroxide ions in the aqueous ammonia. If, however, the observed reaction between a metal ion and aqueous ammonia is different from the reaction of that same metal ion with NaOH(aq), then the metal ion is probably reacting with the ammonia molecules in the aqueous ammonia solution.
Differential affinities between metal ions (Lewis acids) and different Lewis base donors is a very diverse field and was a driving force in chemistry before newer instrumental methods were developed. It's still an important consideration in chemistry and physics and biology... Biology? That's right! Every biological system that contains metal ions (especially transition metal ions) relies heavily upon differential binding affinities to function correctly. And that's just one of the many reasons why biologist need to understand chemistry...

Qualitative analysis of metal cations

This week in lab, you'll be using the chemical tests you observed last week to separate (and identify) the metal cations in an unknown mixture of cations. It's probably better to think about this as a separation rather than just an identification because you will be given 1 sample and through a series of chemical (or physical) steps you will end up with up to 5 different metals in 5 different container. Some tips:
1. Flow chart - You have to organize your procedure to do well. The logical way to do this is with a flow chart for this type of a problem... a sequence of steps with decision points and branches along the way. You might not be a "flow chart person", but it makes it MUCH easier to follow through on a logical set of tests if you become a bit of a flow chart person for this experiment. Here's an example of a flow chart for separating and identifying anions if you're looking for an example: http://chemlab.truman.edu/chemlab_backup/CHEM131Labs/QualFiles/Figure2.gif
2. Assume you have all 5 metals - OK, there's not much chance that you'll get the sample that has all 5 metals, but you might. Even if you don't, design your flow chart and approach so that it will work for any possible unknown in this experiment.
3. The first step is the key! - Any step in which you are potentially making 2 or more precipitates represents a potential problem because you need to be able to separate those solids from one another. If you can't separate the solids from one another, it's not a useful test. For example, chromate makes a BUNCH of precipitates with the metals we are using, but there's no way (in our list of chemical tests) to separate those solids from one another. Adding chromate would be a horrible first step in your flow chart, but it might be handy later on when you only have to test for the presence of a single metal cation.
4. Positive tests - You need to have a positive confirmation test for all the metals, do not try to infer the presence or absence of a specific metal cation based upon negative results.
5. Chemical equations - If you come to lab prepared with a good plan, the "wet" part of this lab will not take three hours. This is a lab report experiment and as part of your lab report, you will have to write out balanced chemical equations for ALL of the tests you performed in the first week of the experiment. {quit rolling your eyes, it's not that hard and it's good practice!} Take advantage of your time in lab to make sure you understand the chemical equations and ask your instructor and lab assistant for feedback.

Good luck on your unknowns. One final note on your flow chart: this is one of those experiments where there is not a single correct answer. There are probably half a dozen or more variations on the "correct" flow chart that all work well.

Class this week

As was mentioned in class, I am at a conference for part of this week so we willnot be having class today (Monday, 2013-April-08) or Wednesday (2013-April-10). We will also not have lab this week. I will be back on Friday, see you then.

What about that ratio in the Henderson-Hasselbalch equation?

A few people have asked about some buffer questions that use the Henderson-Hasselbalch equation. In the proper Henderson-Hasselbalch equation derived from Ka, the pH and pKa are related by "log of the concentration of conjugate base over concentration of conjugate acid". Great, that works. When we're actually using the H-H equation, we can get away with a little shortcut in the calculation BECAUSE we're talking about a single buffer solution that has some conjugate acid and some conjugate base. Here's an example from an old exam:

The set-up calls for concentration {implied by the square brackets}, but when I worked through the calculation, I used moles. THAT'S CHEATING!!!! Well, not really... If we wanted to convert both of those moles to molarity, we'd divide each of them by 0.5000L, and you can go ahead and do that if you like, BUT since we're dividing the numerator and the denominator by the same volume, those volumes cancel out (mathematically) and give the same ratio as when we use moles directly. Let's look at a simpler example of this using simple whole numbers so I don't have to find a calculator... Let's say we have 2 liters of buffer solution that contains 4 moles of conjugate base and 2 moles of conjugate acid. The concentration of conjugate base is (4moles/2L = 2M) and the concentration of conjugate acid is (2moles/2L = 1M), so the ratio of conjugate base concentration to conjugate acid concentration is (2M/1M = 2). If I just use moles, the ratio is (4moles/2moles = 2). Same ratio.
So when we're using the Henderson-Hasselbalch equation correctly for buffer calculations, we can often just use the ratio of moles rather than the ratio of concentrations because it gives the same answer.

Exam questions in my email...

A couple questions have trickled in to my email box...
--Question 1------------
Quick question, what does it mean to be diprotic or monoprotic? And how do you know if something is diprotic monoprotic?
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This refers to how many acidic protons (H+) there are on an acid, how many protons can be donated. Something like HCl(aq) only has 1 H+ to donate so it's monoprotic. Phosphoric acid has 3 acidic H+ so it's triprotic. Usually it's just a matter of "count the H's" in the formula, but for organic acids (and some others...) there are H's that are not acidic. Think of the acetic acid you used in lab... CH3COOH(aq)... the 3 H's that are connected to carbon are not acidic, only the H attached to oxygen is acidic, so this is a monoprotic acid.
I will often use shorthand terminology and refer to bases as being "monoprotic" to mean that the base can accept only 1 proton or "diprotic" if the base can accept 2 protons.


--Question 2------------
On one of the practice exams there is a question that asks for a detailed titration curve. What exactly would you be looking for in something like that as far as details to include?
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You caught me... I usually don't include those drawings in the keys because I type the keys and those titration curves are a little hard to draw electronically... I've drawn a couple and they look horrible. When drawing a titration curve like this, you should label any equivalence points, indicate the species present in solution, label any pH's you can reasonably estimate (or might know...), maybe something like this:

You don't have to label all of those pH values, but again, if you know the pH values for some of those points, you might as well use them to draw a titration curve that's a little more quantitatively accurate.

Other questions, let me know.

Titration question...

Email question:
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I am having trouble with this question from a previous exam: You find that 25.00mL of phosphorous acid requires 41.39 mL of Sodium hydroxide to reach the second equivalence point. What is the concentration of the phosphorous acid solution?

I understand the process of what to do in this question, but am really struggling understanding why we are supposed to use the equation that you used in the explanation: H3PO3(aq) +2 OH-1(aq)†(一)2O(l) +HPO3-2(aq)
Earlier in this problem, we broke down this into three different equations, and I don't understand why we don't use 
H2PO3-1 + OH-1 --> H2O + HPO3-2?
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The equation you list gets us from the first equivalence point to the second equivalence point. In the titration described in the equation, we're going from phosphoric acid all the way to the second equivalence point, so we're doing 2 steps of the potential 3 step process. To get to the second equivalence point, have to go through the following steps:
H₃PO₃(aq) + OH^-1(aq)(aq) <=> H₂O(l) + H₂PO₃^-1(aq)

H₂PO₃^-1(aq) + OH^-1(aq) <=> H₂O(l) + HPO₃^-2(aq)

So the whole 2-step process to get from the beginning of the titration to the second equivalence point has the net equation:

H₃PO₃(aq) + 2 OH^-1(aq) <=> 2 H₂O(l) + HPO₃^-2(aq)

That's where the 2:1 stoichiometry comes from.