Question

Another question...

I just had one more thing to add- I'm having trouble understanding, conceptually, why a strong acid-strong base combination makes a poor buffer system. When you compare titration curves of a strong HA/strong A- with strong HA/weak A- (or weak HA/strong A-), they look the similar to me...they both have areas before (and after, actually) the equivalence point where pH does not change rapidly, but only in the weak/strong combination is this considered a buffer range. Why is this? I think the answer has something to do with the purpose of a buffer (to neutralize added acid or base), but I'm having a hard time understanding how a strong acid-strong base combination fails to do this... Also, I found this flash web-page, and personally I find it helpful to visualize these sorts of things, thought you might want to pass it on to the rest of the class: http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/buffer12.swf

This really gets to the root of why and how a buffer works. Buffers control pH because there is a significant quantity of both the conjugate acid and conjugate base present in solution to react with added acid or base. Let's put some numbers on it to convince ourselves. Let's say we have a solution that contains 10,000 molecules of HA and 10,000 A^- ions. If I throw 50 H⁺ ions into this solution, the ratio will be 10,050:9,050. It has changed, but it hasn't changed much so the pH of the solution remains nearly constant.

Now let's think about a strong acid solution. In that strong acid solution, the concentration of conjugate acid is effectively zero because strong acids (essentially) completely dissociate, so the ratio of HA to A^- is more like 1:1,000,000 (or more). If you add a little OH^- to this mixture, it's not going to react with HA because there's so little of it present. It will react with H₃O⁺ that's floating around free in the water and the resulting solution will be influenced not by the HA/A^- equilibrium, but by the H₃O⁺/H₂O/OH^- equilibria present in water. This is the pH leveling effect of aqueous solution. It is possible to make solutions that have negative "pH" or "pH" above 14, but these are not regular aqueous solutions and the definition of "pH" has to be stretched a little bit to understand the acid/base character of these solutions. Convinced? Pull out your calculator and calculate the pH of a 6M aqueous solution of a strong acid. If we assume that 6M strong HA results in 6M H⁺ ions (or H₃O⁺ ions if you prefer), the pH of that solution would be -0.78. What about a 1M solution? It would have a calculated pH of zero. In practice, these are not the measured pH's of these solutions because the ionization/autoionization of water kicks in and limits the observed pH.

Thanks for the web site, it's got a few other flash animations that might be helpful.
http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/flash.mhtml

Question...

An email question...

Qu: Is the pH of the first equivalence point of a polyprotic acid necessarily basic? And vice versa, is the pH of the first equivalence point of a polyprotic base necessarily basic? For example, in the winter 2006-exam 3 practice exam (#5), we are drawing a titration curve for the 1M potassium sulfite (kb=1.6*10^-7) with 1M HClO4. The answer key seems to show the first equivalence point being slightly acidic, but when I actually go through the calculations I find the pH at the first equivalence point to be 7.204 (I got that using an ICE table, with the assumption that at the 1st equiv point [k2SO3]=[KHSO3], if that is not correct could you go through the calculations?).

There may be a labelling problem here, so let me work through this. If the concentration of sulfite (conjugate base) and hydrogen sulfite (conjugate acid) are equal, this would be a buffer, it would not be at an equivalence point. {The potassium ion is a neutral spectator, so I'll leave it out...} When this titration reaches the equivalence point, all of the sulfite ion has been converted to hydrogen sulfite ion, so at that first equivalence point, this can be considered to be a solution of hydrogen sulfite ions. If you want to calculate a pH for this, you could use a table (not exactly an equilibrium table, but similar...), but let's try to talk through it instead.

Let's say that to start our titration we have V mL of 1M sulfite solution. To reach the first equivalence point, we have to add V mL of 1M perchloric acid, so the total volume of the mixture is 2V mL. (This is probably not strictly true, but it's a close enough assumption for our purposes here.) So if the initial concentration of sulfite ions was 1M, the concentration of hydrogen sulfite ions at the first equivalence point will be 0.5M. This means that at the first equivalence point, the pH of this mixture would be the same as an authentically-prepared 0.5M solution of HSO₃^-(aq). From the K_b given in the problem, we can calculate that K_a = 6.25x10^-8 for HSO₃^-(aq). Now we should set up an equilibrium table for the reaction:
HSO₃^-(aq) + H₂O(l) <=> H₃O⁺(aq) + SO₃^-2(aq)
Go ahead, set it up on paper right now before you read the rest of this.

OK, what assumptions can we make to simplify this problem? Based upon its K_a, HSO₃^-(aq) is a weak acid, so it's probably safe to assume that "x" will be small compared to 0.5M. That means your math simplfies to:
6.25x10^-8 = x² / 0.5
Solving, x = 1.77x10^-4 = [H₃O⁺], so pH = 3.75. This may seem kind of low, but sulfurous acid is one of the strongest weak acids, so its titration curve should tend toward the acidic side. If you really wanted to analyze this titration and draw a super-accurate titration curve, you could also calculate the pH of the "half equivalence points". These are the points where [conj acid] = [conj base], so when you plug them into the Henderson-Hasselbalch equation they are the points where pH = pK_a of the conjugate acid. For sulfite/hydrogen sulfite/sulfurous acid, these fall at pH = 7.2 and 1.9.

After all of that discussion, let's take a step back. For the exam problem mentioned, I was not expecting that anyone would calculate exact pH's of the equilivalence and half-equivalence points, I was more interested in qualitatively reasonable equivalence points. All of those calculations would have taken a lot of time and wouldn't have really resulted in "better" picture of the titration curve. In this case, sulfite ion is a weak base being titrated with a strong acid, so the solution should initially start out basic. The first equivalence point in this titration should be on the acidic side so as long as the first equivalence point was below 7, I was satisfied. Obviously, the second equivalence point should be at an even lower pH than the first, and it should take twice as much titrant to reach the second equivalence point. Some of the more common (and frustrating) errors in this question are that people don't always label their axes, or don't label them correctly, or titrate in the wrong direction.

Good luck with your preparation, and again, the exam will take place on the second class after we return. If you have questions before then, email me and I'll answer them to the blog.

Be safe.

Exam and Mastering Chemistry Update

I'm glad to hear (and see) that so many of you are contributing to the sandbagging efforts. Because of the flood-related delays, Exam 3 will take place on the second day after classes resume, with the first day devoted to review. At this point, it is expected that classes will resume on Monday, so we will review on Monday and the exam will be Wednesday. If classes get pushed back again, we'll adjust similarly. I have also pushed the deadlines on Master Chemistry Assignments 12 & 13 back to March 31st. Work on them as early as you can.

Be safe.

One of these days it'll crest...

OK, see you all on Monday for the exam. I'll be back in town on Friday and will at least stop in my office. Email any questions and I'll answer to the blog.

Flood, flood and more flood...

With all the flood issues and canceled classes, I haven't bothered putting any updates on the blog. The only issue is with Chem 210L this week. Since the University has canceled Tuesday and Wednesday labs, we'll also cancel Thursday labs.

Help out with flood control in whatever way you can, I'll see you all on Friday.

New Mastering Chemistry...

There are two new Mastering Chemistry assignments posted:
Assignment 12 - Due 3/27 - mostly titrations and buffers
Assignment 13 - Due 3/29 - mostly solubility equilibria

Remember, we do not have class on Monday, 3/23, or Wednesday, 3/25, and Exam 3 has been pushed back to Monday, 3/30. I will be out of town until Thursday night, but I will be in email contact, send me any questions you have.

It looks like Fargo and Moorhead will be looking for volunteers to fill sandbags this week, if you have the time and ability to help out, here are a couple links to websites that might have info:

https://extranet.casscountynd.gov/Flooding/Pages/Volunteer.aspx
http://www.cityofmoorhead.com/flood/
http://www.inforum.com/

Have a good week, I'll see you all on Friday in class.

Solubility product constants

On Friday we looked at solubility product constants and how we can use K_sp to understand how salts dissolve in water. Don't forget, just like K_a and K_b and K_w, K_sp is just an equilibrium constant that refers to a specific reaction type. It follows all the same rules and trends as any other equilibrium constant.

I hope you all have a great break. There are a couple announcements regarding the next couple weeks:
1. I will be gone to the American Chemical Society's National Meeting and Exposition in Salt Lake City, UT, March 21-26, so my section of Chem 210 will not meet on Monday or Wednesday of the week after break. My Chem 210L section will meet on Tuesday at its regularly scheduled time.
2. Our next exam is scheduled for March 27th, the Friday after break. Since I will be out of town until Thursday evening, a few people expressed concern that they would have a difficult time preparing for the exam if I am out of town. Although I have the utmost faith that you would all be able to do a splendid job preparing in my absence, I can see that this might give rise to a little extra anxiety, so Exam 3 will move back a day to Monday, March 30th.
3. Although I will be out of town, I will be in email contact and I will be posting to the blog. Those posts will include answers to any questions I get via email as well as other info and sample problems. Check the blog for new content. If you have questions to email me, please include specific information. I won't have my textbook with me, so asking "How do I do #37 in Chapter 15?" will make it kind of difficult for me to give you a meaningful answer.
4. I will be setting up at least one (probably 2) Mastering Chemistry assignments for the week after break. I will have those assignments go "live" as soon as I organize them and will post a notice to the blog.
5. For those of you in my lab, I will post some info for you all as well. I'll probably try to post it on my web page with a notice on the blog, but it may end up here. I will have a few more graded assignments (including your last lab report) to be handed back in lab. At this point, Dr. Edvenson will be your lab instructor.

I'm sure I'll think of a few other things over the next couple days, I'll post any info that comes up.

Additional aqueous equilibria....

Oops, looks like I missed a couple days. Sorry 'bout that.

So far this week, we have looked more closely at buffers. The key to most buffer problems is to try and wrestle the information you are given into the Henderson-Hasselbalch equation and solve for the bit of info that you're missing. The nitrite/nitrous acid buffer that we looked at in class also (sort of) demonstrated something called the common ion effect. We'll get more into common ion effects tomorrow.

Buffers let us control pH of a solution, but how do we determine concentration of an acid or base in solution? Titrations let us see how the pH of a solution changes as we add acid or base. We talked about titrations as stoichiometry problems in Gen Chem I, now we want to understand what's happening in a pH titration in a little more detail. This is easiest to do by looking at a pH titration curve. Using a titration curve, we can pick out regions where buffers are formed (that allows us to calculate the Ka of an acid), and equivalence points where the amount of acid and base added to the titration reflects the stoichiometry of a balanced reaction.

Don't forget about the Mastering Chemistry assignment, due tomorrow. For anyone who hasn't already heard, since labs were cancelled on Tuesday because of the lovely little snow storm, we have cancelled Chem 210L labs for the week. You will be doing the Titrations, Indicators and Buffers experiment the week after Spring Break.

Chapter 15 to Chapter 16

Today we pretty much finished up Chapter 15 and started Chapter 16. If you didn't notice, it's because those two chapters are very very closely related to one another. We looked at an acid equilibrium problem and moved into buffers, we'll look more closely at buffers on Monday....

There's a new Mastering Chemistry assignment, due Tuesday. Don't forget about the MC assignment due on Sunday.

Have a good weekend...

Predicting acid strength

Today we looked at how the molecular structure of some acids can give us clues about their relative strength. We also briefly talked about how to treat sig figs when dealing with pH calculations. I also handed out a problem set for practice, if you didn't get one I'll post in on the class web page.

Hmm, not much else to say. 1st round playoff game tonight for the women's basketball team, 6pm tip. See you all Friday....

Calculating pH of a solution

Today we combined the ideas of Ka (or Kb) with pH (or pOH) to predict the pH of an aqueous solution of acid or base. These are the same type of equilibrium problems we've been working with for the last couple weeks, they're just applied to a specific chemical situation. We can often simplify these problems by making some assumptions about the size of "x" compared to the initial or final concentrations.

There is a new Mastering Chemistry assignment posted, due Friday.

The MSUM women's basketball team is starting out the playoffs as the 3rd seed at home this Wednesday at 6pm against St. Cloud State.

Acids and bases!!

Today we jumped into acids and bases with both feet including Arrhenius' definition of acids and bases, the Bronsted-Lowry definition of acids and bases, and a couple "new" equilibrium constants (K_a, K_b, K_w) that let us assess the strength of an acid or a base. We just barely touched on pH, we'll get more into that on Monday.

I will also have your exams on Monday.

Tomorrow night (Saturday) is the last regular season home game for the Dragon basketball teams, come on out and watch our Dragons put a hurt on UM-Crookston. Women tip at 6pm, men at 8pm. Post season play starts next week....

Have a good weekend.

Questions...

Can you explain these two questions? Thanks!
9. Which of the following cannot be explained by LeChatelier’s Principle?
a. Removal of a gaseous product from a reaction drives the reaction to completion.
b. Any salt will dissolve in water, even though the amount might be extremely small.
c. The activation energy of a reverse chemical process is related to the activation energy and
ΔH of the forward process.
d. Less of a salt will dissolve in a solution that has a “common ion” than in pure water.
e. The equilibrium of an exothermic process will be more reactant-favored at higher
temperatures

"a" - removing a product will drive an equilibrium toward products, that's almost the definition of LeChatelier's Principle (I'll call it LCP from here on...)
"b" - Think about solubility in terms of a chemical reaction/equilibrium, salt + water yields ions in solution. Water acts as a reactant here, so adding more water will stress the equilibrium and shift it toward products.
"c" - This is purely kinetics, so it'd be tough to explain it with LCP.
"d" - Again, think about solubility as an equilibrium. NaCl(s) <-> Na⁺(aq) + Cl^-(aq). If you use a "solvent" that already has a bunch of chloride ions dissolved in it, it will push the equilibrium left.
"e" - for an exothermic process, "heat" is a product of the reaction. If you add a product, the equilibrium will shift toward reactants.

14. You have found the following value in a table of equilibrium constants at 25ºC:
3 O2(g) 􀀧 2 O3(g) Kc = 6.25x10-58
2 NO(g) + O2(g) 􀀧 2 NO2(g) Kc = 2.25x1012
What is the equilibrium constant for the reaction:
NO(g) + O3(g) 􀀧 NO2(g) + O2(g)
a. 3.60x1069
b. 1.41x10-45
c. 2.03x1053
d. 6.00x1034
e. 9.38x10-52

How do we have to rearrange the two given reactions so they add up to the final reaction? Looks like the first rxn has to be reversed and multiplied by 0.5, so the "adjusted" equilibrium constant for that step is (1/K_c1)^0.5. The second rxn has to be multiplied by 0.5, so the "adjusted" equilibrium constant is K_c2^0.5. Therefore, the equilibrium constant for the final rxn is: (1/K_c1)^0.5 K_c2^0.5 = 6x10³⁴. {Just a note in case you're having a bad math day, raising something to the 0.5 power is the same as taking the square root...}

I'll answer any other questions first thing in the morning.

Questions...

A couple email questions....

Someone asked about #5 and #9 from the problem set that was handed out on the day I missed class. All the questions on that problem set are from previous problem sets, so you should be able to find answer keys posted on my web page under Previous Gen Chem II.

For #5, we will ultimately need to plug into the comparative form of the Arrhenius equation, so we need to wrestle the information given into rate law constants and temperatures. The rate law expression is given, so we can plug in the room temperature rate and the given concentrations to find "k_room". We can also plug in to get "k_unknown". With the given activation energy and the appropriate value of R (8.314 ^J/_mol.K), you can plug in all of the values in the comparative Arrhenius equation and solve for T_unknown. Make sure you pay attention to units, E_a is given in kJ and R uses J, you have to convert one of them before you solve for T_unknown or you'll have a random factor of 1000 floating around in your problem.

For #9, we're looking at integrated rate law problems. In all cases, we're using the 1st order IRL because the problem tells us that the reaction is 1st order w.r.t. Cl₂(g). For all parts, plug in what you know and solve for the variable.

Other questions, let me know and I'll post answers to the blog.....

Manipulating equilibrium...

On Friday, we talked about how changes in the way a chemical equation is written affect the form and value of the equilibrium constant. A few notable cases:
1. If the chemical equation is reversed, the value of K is inverted. If A<->B has equilibrium constant K_f, then B<->A has equilibrium constant 1/K_f.
2. If a chemical equation is multiplied by some constant, the value of the equilibrium constant is raised to that power. If A<->B has equilibrium constant K, then 4A<->4B has equilibrium constant K⁴.
3. If two (or more) chemical equations are added together, the value of the equilibrium constant is the product of the equilibrium constants for all the individual reactions. If A<->B has equilibrium constant K₁, and B<->C has equilibrium constant K₂, then A<->C has equilibrium constant K_net = (K₁)(K₂).

There's a new Mastering Chemistry assignment posted, due Tuesday. Also, we have an exam on Wednesday, so if you have questions let me know or bring them to class on Monday, we will have time to review.

Don't forget that tonight is Go Pink with the Dragons. The women's basketball game starts at 6pm and they will be raising money for the Roger Maris Cancer Center.

See you all on Monday.

Equilibrium in the semi-dark

Today we looked at a couple additional equilibrium problems and behavior including:

The similarities, differences and conversion between concentration-based equilibrium constants (K_c) and pressure-based equilibrium constants (K_p). They're related by the ideal gas law!

How do we determine if a reaction is at equilibrium? The reaction quotient has the same mathematical form as the equilibrium constant and can tell us where a reaction is relative to equilibrium.

What happens when we push around a system that's at equilibrium? If experiences stress, and it squirms around until it relieves that stress. This is LeChatelier's Principle and it can be used to explain chemical equilibria, human behavior, economic trends, population dynamics, and a whole lot more.

There's a new Mastering Chemistry assignment posted, due Monday. There's also a MC due Friday, so keep up with them.

This weekend the MSUM women's basketball team is going pink to put a full court press on breast cancer. T-shirts are available at the bookstore, and the 21st-ranked Dragons would love to see a mountain of pink in the stands at their game against Concordia (St. Paul) on Saturday. This is one of the last home games of the regular season, so come and cheer the Dragons into the post season.
http://msumdragons.com/sports/2009/1/20/WBB_0120091718.aspx?id=82

See you all on Friday, the lights should be working again in SL104 by then....

Equilibrium calculations

Today we went through a few more equilibrium calculations and looked at a couple ways equilibria can be simplified. First, the "concentrations" of pure solids and liquids do not appear in equilibrium constants because their concentrations are constant (essentially) throughout the reaction.

Next we looked at assumptions that can simplify the math involved in equilibrium calculations. In many problems, the numerical value of "x" {the change in concentration when a reaction goes to equilibrium} is so small that it can be ignored in places where it's added to or subtracted from another term. For example, if the term "1.328 - 3x" appears in your equilibrium calculation, you should examine just how big you expect "x" to be. If the value of "x" turns out to be 0.00000000168, then it really doesn't affect the term "1.328 - 3x", and that term can be simplified to "1.328". Knowing when to make these approximations takes a little practice, so always check your assumptions after you finish working through a problem.

FInally, we looked at the difference between using a "full molecular" chemical equation and a "net ionic" chemical equation to determine the equilibrium constant. In many ways, the net ionic equation will always give you a more correct picture of the actual chemistry that takes place, but as long as the equilibrium constant you calculate is consistent with the chemical equation you use, I don't mind which you choose. But I strongly encourage you to think about reactions as net ionic reactions whenever possible...

There's a new Mastering Chemistry assignment, due Friday.

Equilibrium

Today we transitioned fully into equilibrium and looked at what the equilibrium constant can tell us about a system at equilibrium. We worked though a problem that let us calculate the value of the equilibrium constant using initial concentrations, and then calculated equilibrium concentrations when we knew the value of the equilibrium constant.

Remember, equilibrium is a dynamic process, so it doesn't matter if we start with all reactants, all products, or some mixture of both, the system will always settle into conditions that give the same equilibrium constant. (As long as temperature and pressure remain constant...)

I looked over some Mastering Chemistry problems, and I didn't see any that I liked right now. A bunch of them will be much better after class on Monday, so I'll wait until then to post a new assignment.

Have a great weekend. If you're looking for something to do, the MSUM Symphony Orchestra is performing a program called "Symphonic Romance" Saturday night at 7:30pm in the Hansen Theatre. A number of your fellow Chem 210-ers will be part of the performance, so come out and show your support.

See you all Monday.

Kinetics to Equilibrium to cancelled class

On Monday we talked about reaction mechanisms and looked at how the overall rate of a reaction is determined by the slowest step in the mechanism. (Not too surprisingly, this step is called the "Rate Determining Step" or the "Rate Limiting Step", and is often noted/labeled "RDS".) Our discussion of reaction mechanisms led us to the idea that if a reaction can proceed in the forward direction, it might also be able to move in the reverse direction, establishing a condition called equilibrium.

For those of you who were in my lab yesterday this may come as no shock, but I have chosen to stay home today and try to finish recovering from this nasty little cold bug that has invaded my sinuses. I will have a problem set printed out for class today that should provide some good practice in Kinetics, I encourage you all to work on it together, but class is (technically) cancelled today.

Don't forget the current Mastering Chemistry assignment, due tomorrow. See you all on Friday.

Activation energy....

Today we worked through a kinetics/rate law/activation energy problem. We will dive into reaction mechanisms on Monday...

There's a new Mastering Chemistry assignment, due Thursday.

Basketball is at home this weekend if you're looking for something to do. Tonight our teams take on Southwest MN State, tomorrow it's MSU Mankato. Women tip at 6, men at 8.