Fall 2008 Exam 4a key

I posted a key for last year's Exam 4a, but I don't have access to my mnstate.edu web page right now, so I wasn't able to put a link up yet or draw "nice" Lewis or VSEPR structures. Here's a direct link to the .pdf file

http://www.mnstate.edu/bodwin/courses/genchem1x/c150ce4ak.pdf

Electronegativity and Lewis Structures

Today we looked at electronegativity as a measure of a bond's polarity and a way to determine bond type. Then we started looking at Lewis structures, finishing up in the middle of NO2. We'll finish up NO2 on Friday and look at some other Lewis structures, including the Lewis structures of polyatomic ions. I know, you're on the edge of your seats...

There's a new MC posted.

Electron configs and properties...

We've been looking at how we can use electron configurations to predict physical properties including size, charge, ionization energy and electron affinity.

If anyone's interested in some more YouTube videos, the following links are a 3-part Bill Nye The Science Guy episode dealing with light and color.

http://www.youtube.com/watch?v=msf-DbYif-4&feature=related

http://www.youtube.com/watch?v=hEUtp2htuEs&feature=related

http://www.youtube.com/watch?v=Sbjuv-meMh8&feature=related

Bill Nye usually does a very good job of explaining scientific concepts in a (hopefully) entertaining way, and this show is no exception. If you're struggling with questions from lab this week, watching this might help.

There's a new Mastering Chemistry posted.

Quantum chemistry and orbitals...

Looks like I missed a couple days...We've been talking about quantum chemistry including the nature of light and radiation. This is so we can look at electrons a little more closely and understand their behavior as particles with wave-like character. A direct result of this wave-like character is that electrons exist in regions of space that are called orbitals. Understanding orbitals helps us interpret the chemistry that we have seen and allows us to predict various properties of atoms, ions and molecules.

There is a new Mastering Chemistry assignment posted, due Nov. 9.

Exam Monday

We reviewed in class on Friday. A couple things people are having trouble with:

1. Changing {delta}H when you change a reaction. Let's look at an example, consider the reaction A-->B with {delta}H = 25kJ/mol. Positive {delta}H means this reaction is endothermic, so we have to add 25kJ of energy to convert each mol of A into a mol of B. If we reverse the order of the reaction, B-->A, the conversion of a mol of B into a mol of A will release 25kJ of heat, making it exothermic with {delta}H = -25kJ/mol. What if we triple the reaction, 3A-->3B? For each mol of A that is converted to a mol of B, the reaction still requires 25kJ of heat, but now the reaction as written is converting 3 mols of A into 3 mols of B, so {delta}H = 3(25kJ/mol) = 75kJ/mol.

2. Heat capacity and {delta}H for a phase change. When a substance freezes, melts, boils or condenses, it releases or absorbs heat without a change in temperature. This means that if I have a pot full of water on the stove and I start to boil it, no matter how hot I make the burner the boiling water will be 100 degC until all of it has boiled away. If you're looking at a problem that goes through a phase change and changes temperature, you can break the problem down into a couple smaller problems. For changing temperature without changing phase, it's a heat capacity problem; for changes in phase, it's {delta}H of phase change problem.

Remember, if you're looking at old exams we have not talked about quantum numbers or electron configuration or wavelengths yet, that's coming up in Chapter 7.

If you're in the mood for a study break, today at 4pm the MSUM Dragon volleyball team will be Digging for a Cure to raise money for cancer treatment and research. Wear purple and cheer on your Dragons as they crush Upper Iowa. Purple T-shirts will be sold at the game for $10.

Let me know if you have questions, I will answer to the blog.

Fuels and enthalpy...

Today we looked at the enthalpy involved in the combustion of a variety of fuels. "Combustion" or "burning" means reacting a fuel with oxygen to produce carbon dioxide and water for hydrocarbon fuels. We will look over the numbers a little more on Wednesday.

For EVERYONE in lab this week, you may have noticed that there is no experimental procedure included in your lab manual. The procedure is posted on my mnstate.edu website under the Chem 150L link. This is for everyone in my labs AND Dr. Marasinghe's labs.

There's a new MC assignment posted, due Sunday. See you all in class on Wednesday.

Back to work...

Welcome back from the long weekend. I handed back exams today, if you weren't in class I will bring the exams to class on Friday.

We finished looking at the enthalpy of reaction problem that we started last week, on Friday we will look a little more closely at quantifying heat exchange.

As I mentioned in class today, don't forget that advising starts this week. If you have any questions about advising, choosing a major, or classes for next semester, let me know, I'm happy to try to answer them.

I've updated the SI times (at left...). I will also be putting up a new Masting Chemistry assignment either tonight or tomorrow morning, the MC system is down for scheduled maintenance right now.

Seminar Speaker

Tomorrow (Friday, October 9th) at 1:30pm the Department of Chemistry will host Dr. Darrell Eyman from the University of Iowa for a seminar. Dr. Eyman's talk is entitled "Grafted Active Site Catalysts: Synthesis, Characterization and Applications". The seminar will be held in SL118.

Extra exam question...

I did not give your exams back today because I wanted to give everyone a chance to earn a few more points. In class, I handed out an extra exam question worth up to 20 additional points. If you were not in class today, stop by my office and pick one up. I will also have them in lab tomorrow. This extra question is due by noon on Friday.

Today in class we started talking about thermodynamics and heat transfer, including specific heat capacity and how to calculate energy transfer from temperature change and vice versa. We stopped in the middle of a problem, we'll continue it on Friday.

Next Wednesday from 8am until noon there will be a representative from the University of Minnesota College of Pharmacy. If you are interested in a career in pharmacy or just want some more information, stop by and check out the info. He will be in HA405. UMinn's College of Pharmacy is a very highly ranked program (3rd in the nation the last time I checked...) and there are a number of MSUM alumni who are currently attending the college.

Last questions?

How can you look at a molecular formula and know if it's a strong acid or not? like in the question 4 from fall 2007, which of the following the the strongest acid? KOH CHLO4 HC2H3O2 H20 or NH3

We didn't specifically talk about identifying strong acids or bases this semester, so this type of question will not be on the exam. Strong acids and strong bases are those which "completely" ionize in solution. For bases it's a little easier because the strong bases are those that are soluble, so they follow the solubility rules for hydroxides (alkali metal hydroxides and soluble alkali earth metal hydroxides are strong bases). For acids, it's probably a matter of memorizing the strong acids (perchloric, sulfuric hydrochloric, hydrobromic, hydroiodic, nitric) and all the rest of the acids are weak. But again, we didn't specifically address this so it won't be on the exam.

For winter 2006 number 12, I took (1.62M)(50.00mL) = C2(500.0mL) and figured out C2, but I don't get the same answer as the test. Is there another step I'm missing?

That question asks for the concentration of potassium ions, not the concentration of potassium carbonate. The concentration of potassium carbonate is 0.162M, but for every mol of potassium carbonate that is dissolved there are 2 mols of potassium ions dissolved, so the concentration of potassium ions in the solution is twice the concentration of potassium carbonate, 0.324M.

Email question...

For exam 2a from fall 2008 # 13, where or how do you get n2 = (14.5mols)(0.875) = 12.7mols. I get where the 14.5 mols is coming from, but where is the 0.875 come from? and how do I get to that point?

The problem states that 12.5% of the gas is lost to a leak as the balloon rises, so 87.5% of the gas remains in the balloon.

Email questions...

There are a bunch, I'll answer them one at a time...

I have some questions on the old exams. One is balancing equations. I know we've talked about that alot in class, but I'm still kind of confused. I understand how to balance the numbers out when we're given the equation, but I get confused when starting from scratch. For example, number 7 for fall 2006 exam: Magnesium hydroxide solution + Lead (IV) nitrate solution --> Lead (IV) hydroxide + Magnesium nitrate. I know we have to know the polyatomic ions, but does the charge have anything to do with the numbers behind of the element? Why is hydroxide OH2 in the answer, when it's a OH- polyatomic ion? Does that have to do with balancing?

For cations that have ambiguous charge, the oxidation state is given by Roman numerals after the name. Something like sodium is pretty much always +1, but the transition metals and main group metals (like lead, tin, etc) can have a number of different stable charges, so these are specified. For this question, we need 2 hydroxides in the formula of magnesium hydroxide and 4 hydroxides in the formula of lead(IV) hydroxide. Hopefully it's "(OH)_2_" in the answer and not "OH2"...

Another question I have is number 11, winter exam 2006: How many grams of hydrogen are required to make 34.061g of ammonia by the following reaction? xH2(g)+ yN2(g) --> z NH3(g). I have no idea how to do this problem. Do you have to do something with mole ratio?

Yes, you need the mol ratio. First, balance the equation. Once you have correct numbers for x/y/z, then convert34.061g of ammonia into mols, use the mol ratio (x/z in this case) to convert mols of ammonia to mols of hydrogen, then use the molar mass of hydrogen to convert to grams.

How do you do concentration problems like numbers 12: 50.00mL of a 1.62 M potassium carbonate solution is diluted to 500.0mL. What is the concentration of potassium ions in the resulting solution, [K+]? and 13: what is the concentration of a perchloric acid stock solution if 21.53 mL of 1.054M Mg(OH)2(aq)is required to titrate 15.00mL of HClO4(aq) to the equivalence point in the following reaction?: a HClO4 (aq) + b Mg(OH)2 (aq) --> c H2O(aq) + Mg(ClO4)2(aq).on winter 2006 exam?

Hmm, this is a 2-fer. When you are diluting a solution of known concentration, use the formula C1V1 = C2V2 where C's are concentrations and V's are volumes. In this case, plugging in numbers gives:
(1.62M)(50.00mL) = C2(500.0mL)

The second one is a titration problem, which is just a specific type of stoichiometry problem. Write a balanced chemical equation, convert 21.53mL of 1.054M Mg(OH)2(aq) to mols, use the ratio from the balanced equation to convert mols Mg(OH)2 to mols HClO4, then use the given volume to convert mols HClO4 to concentration (mols/L).

Number 14, winter exam 2006: 50,00mL of 1.119M Co(NO3)3 (aq) is combined with 60.00mL 1.821 M Na2Co (aq). 2.946g of precipitate is recovered from this reaction. I understand parts a and b, but I don't understand c: What is the percent yield of this product?

Percent yield is the actual yield divided by the theoretical yield time 100%. Actual yield is the amount you collect or "recover" from the reaction, theoretical yield is the maximum possible amount you could produce if you use all of the limiting reagent to make product.

On exam fall 2006, number 9 is assigning oxidation numbers to each element: AgNO3. Ag is +1, N is +5 and O is -2. How are we suppose to know that? N's charge is -3, why is it's oxidation number +5? Also, how do you know the charge of transition elements?

If nitrogen were just some random nitrogen ion, we'd probably expect it to have a charge of -3, in that case it would be a "nitride". In this case, nitrogen is part of the polyatomic nitrate ion. The sum of all the oxidation numbers of the atoms in a polyatomic ion is equal to the charge of the polyatomic ion. For nitrate, we'd expect the oxygens to have oxidation numbers of -2, so:
(Ox# nitrogen) + 3(Ox# oxygen) = (charge of nitrate)
(Ox# nitrogen) + 3(-2) = (-1)
(Ox# nitrogen) = +5
For transition and main group metals (see one of the answers above), the Ox# will either be given as a Roman numeral, or it will have to be determined from a given formula. In this example, the charge of nitrate is -1, so if the given formula is "AgNO3", then the silver must have a charge of +1.

I hope this helps, I'll probably check in again a little later...

Email question...

I am having problems figuring out the questions that give volume, Temp, and Pressure. Like for example number 10 on fall 08 exam 2a, a 2.65 L steel tank contains an ideal gas at 15.83 degrees C and 1.15 atm. what is the temp if the pressure changes to 1.48 atm? I tried using the formula P1V1/T1 = P2V2/T2 and I'm having no luck getting an answer. How do I figure this problem?

That approach is close, but if this is a rigid steel tank then the volume is probably not going to change. From the comparative form of the ideal gas law, V and n cancel (because they don't change), so we're left with P1/T1 = P2/T2. Plugging in values from the problem:

(1.15atm) / (288.98K) = (1.48atm) / T2

T2 = 371.90K = 98.75 degC

Problem set key

I've also posted a key for the problem set we did in class, it's on my Fall 2009 Chem 150 page.

Keys posted

The keys for last year's Exam 2 are posted on my web page under "Previous Gen Chem I". If you have other questions, email them, I'll be responding to the blog tomorrow morning.

If you're in need of a study break, the volleyball team is at home this weekend, 7pm tonight and 4pm tomorrow afternoon.

Good luck studying.

Gases

On Monday we started talking about gases and gas laws including kinetic-molecular theory of gases, the "simple" gas laws, and finally the ideal gas law. We'll look at gases again on Wednesday and probably start reviewing for the exam next Monday.

There's a new MC, due Sunday.

I got a question in email about the Redox assignment in MC. It's a little terminology issue that can cause some confusion so let me try and handle it here. If a substance is oxidized it loses electrons. (That's the "OIL" part of "OIL RIG" or the "LEO" part of "LEO GER".) Because oxidation and reduction cannot take place independently (if you have one, then you must have the other), you could say that if substance A loses electrons and is oxidized, then it will cause something else, let's say substance B, to be reduced. So if A is oxidized, then it is acting as a reducing agent toward B. Stated simply:

If something is oxidized, then it is a reducing agent or reductant.

If something is reduced, then it is an oxidizing agent or an oxidant.

These definitions are all about the cause-and-effect relationship between oxidation and reduction.

If you have any specific questions/topics you'd like to review for the exam, let me know.

Stoichiometry practice...

Friday in class we looked at a problem set to practice stoichiometry. It's copied below. There is a new MasteringChemistry assignment posted, due Friday. Next week we will look at gas laws and continue with stoichiometry. Your next exam is October 5th.

Chem 150 – Fall 2009 – Problem Set #2

You are studying the reaction of 1.132M potassium phosphate solution with 1.275M barium nitrate solution.

1. Write a balanced chemical equation for this reaction.

a. How many grams of precipitate could you make if you completely react 125.0mL of the potassium phosphate solution?

b. How many grams of precipitate could you make if you completely react 175.0mL of the barium nitrate solution?

c. How many mL of the barium nitrate solution is required to react completely with 125.0mL of the potassium phosphate solution?

c. How many mL of the potassium phosphate solution is required to react completely with 175.0mL of the barium nitrate solution?

d. What is the theoretical yield of precipitate (in grams) if you react 125.0mL of the potassium phosphate solution with 175.0mL of the barium nitrate solution?

e. What is the limiting reagent in part d? How many moles of the excess reagent remain after the reaction is complete?

f. Write the balanced net ionic equation for this chemical process.

2. These questions deal with concentrations of the solutions used above:

a. What is the concentration of potassium ions in the 1.132M stock solution? Phosphate ions?

b. What is the concentration of barium ions in the 1.434M stock solution? Nitrate ions?

c. How many grams of potassium phosphate are present in 125.0mL of the 1.132M stock solution? How many grams of barium nitrate are present in 175.0mL of the 1.275M stock solution?

d. If you dissolved the mass of potassium phosphate in part c in enough water to make 300.0mL of solution, what would the concentration be? If you dissolved the mass of barium nitrate in part c in enough water to make 300.0mL of solution, what would the concentration be?

e. If you dilute 125.0mL of the potassium phosphate stock solution to a total volume of 300.0mL, what will be the “new” concentration of potassium phosphate? If you dilute 175.0mL of the barium nitrate stock solution to a total volume of 300.0mL, what will be the “new” concentration of barium nitrate?

Things that go boom...

Today we looked at a couple more reactions including redox. The key to redox reactions is being able to figure out oxidation numbers for substances from their chemical formulas and it takes practice.

Practice? Yes indeed! There's a new Mastering Chemistry assignment, due Tuesday.

Introducing...Stoichiometry!!

Today we looked at the relationship between mols of reactants and products, stoichiometry. For any stoichiometry problems, there are a logical set of steps to follow:

1. Balance the equation.

2. Convert the quantity you know into mols.

3. Convert mols of what you know to "mols of interest" using the coefficients (ratios) in the balanced equation.

4. Convert "mols of interest" into a useful number. That useful number might be a mass or a volume or a number of other things, it all depends upon the problem.

We looked at a precipitation reaction and an acid-base reaction today, we'll look at a couple other reactions on Wednesday.

There is a new MasteringChemistry assignment posted, due Sept. 28th.

More on recognizing reaction types

Friday we talked about a couple classes of reactions and how to recognize them. We looked at precipitation reactions which required use to develop solubility rules to identify ionic compounds which we wouldn't expect to be soluble in water. Next, we looked at some molecule-forming reactions, starting with acid-base reactions, so we talked about acids and bases which react to form a salt and water (a molecule). We also looked at some gas-forming reactions, those that form a gas molecule when they react. We will continue looking at classes of reactions Monday, and there will be a new MC assignment on Monday.

Balancing equations...

Today I handed back exams, the average was around 125/150. For those of you who didn't do as well as you might have liked or missed the exam, remember that your low exam score is dropped in determining your final score for the course, so whether it's an exam that you miss or just the lowest score out of your 4 exams, it will not impact your course score directly.

We started talking about balancing reactions today. Your textbook introduces balancing reactions in Chapter 3 and the really gets into it in Chapter 4. We introduced things today and will really get into it on Friday. See you then. (Or tomorrow if you're in one of my labs...)

Speaking of labs, if you are in my lab make sure you remember to look at the online pre-lab and take the quiz on D2L.

Exam #1, Fall 2008

I posted a key for last year's exam if you'd like to check your answers.

www.mnstate.edu/bodwin, in the left panel click on "Gen Chem I" under the "Previous" heading.

Review...

Today in class we reviewed for Monday's exam. We went over the questions from the in-class problem set, the answers were: cobalt has a +3 charge, the alloy is 85% iron.

If you have other questions, let me know and I will answer them to the blog. Don't forget about your MasteringChemistry assignments.

Exam reminders:

- At least 1 empty seat between everyone

- No graphing calculators, no cell phone calculators, and turn off your phones

See you all Monday.

Naming organics

An email question...

what do the numbers in front of the name mean? (for instance 2 pentene or 4 nonene)

The numbers indicate the location of the functional group. Many (most?) organics exist as chains of carbon atoms so if we want to draw a correct picture of the molecule based only on the name, it's necessary to specify where on the chain to put functional groups. For alkenes, the molecule contains a double bond, so if we start with pentane:

CH3-CH2-CH2-CH2-CH3

and want to make one of those bonds a double bond (and remove 2 hydrogens) to form pentene, we could either make the first bond in the chain a double or the second, so we need to specify whether it is 1-pentene or 2-pentene. The same is true for other functional groups, 1-propanol has the -OH group on the first carbon, 2-propanol has the -OH group on the second carbon.

This comes up in your MasteringChemistry assignment, but it is not a super-important distinction for us right now. The chemical formula of 1-propanol is C3H7OH, and the chemical formula of 2-propanol is C3H7OH, so for purposes of calculating mass or moles they can be treated the same. When we start looking more closely at properties of molecules, we will start to see differences between them, but that's a few chapters away at this point.

Exam questions

And a reminder, we will have time to review on Friday so if you have questions you would like me to go over, let me know ahead of time or bring them with you. If we don't have questions, we will start the next chapter -- balancing equations......

Almost exam time...

Today in class we talked about naming organic compounds using roots and suffixes. The classes of organic compounds we covered were: alkanes, alkenes, alkynes, alcohols, amines, and acids. {Hmm, they all start with "a"...} There's a table in the book that lists others, but I think the "a" team is enough for us at this point.

We also looked at the difference between empirical formulas and molecular formulas. Molecular formulas are always multiples of empirical formulas (even if you're multiplying by "1" in some cases).

A few people have contacted me about the lab quiz on D2L. You will only have a D2L lab quiz if you are in my lab classes; I don't believe Dr. Marasinghe is using D2L lab quizzes. Labs will meet in the same rooms as last week to get started (BR263 for my 9am Thursday lab, SL118 for all others.

We also worked on a problem set in class, I've copied it below.

Chem 150 – Fall 2008 – Problem Set #1
1. You have found a bottle in your lab that is labeled “cobalt sulfite”, but the charge of the cobalt is not listed. After analysis, you find that the contents have the following composition: % Co = 32.91; % S = 26.86. What is the correct formula of this compound and what is the charge of the cobalt?
2. You have received a sample of an iron-nickel alloy (a mixture of metals) that contains exactly 1.00mol of metal atoms. This sample has a mass of 56.267g. What is the percent iron in this sample? {Hint: This is a weighted average problem, just like isotopic abundance, but it’s using 2 different elements instead of 2 different isotopes.}

OK, this post is getting kind of long. You have a new MC assignment posted, due Sunday.

Chemical formulas...

Today we talked about ionic and molecular compounds, including naming, writing formulas, and determining formulas from percent composition data. You will be expected to know all of the polyatomic ions listed in Table 3.5 of your textbook (page 95). Flashcards might not be a bad way to learn them...

There's a new MC assignment, due next Friday.

For those of you in my labs, be sure to check D2L. I will have info posted some time today or tomorrow, including a pre-lab "quiz" that will be due Wednesday.

Have a fun and safe weekend, I'll see you all on Wednesday.

The mole

Today we talked about predicting charges on ions and we used the mole to relate microscopic amounts to macroscopic masses.

There's a new MC, due Sept. 9th.

Periodic Table

Today we talked about the origin of the Periodic Table and element symbols.

For labs this week, you will be meeting in the following rooms. Bring your lab manuals, goggles and notebooks.

Wednesday 2:30pm (w/Marasinghe) SL118

Thursday 9:00am (w/Bodwin) BR263

Thursday 12:00pm (w/Marasinghe) SL118

Thursday 3:00pm (w/Bodwin) SL118

There is a new MC assignment posted, due the 8th.

New Mastering Chemistry

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

I wanted to add a couple notes here about MC policy/procedure:

1. MC assignments are due at 11:59pm on the due date.

2. Late assignments drop 50% per day.

3. I do not intentionally pre-list assignments. If a MC assignment is active and visible, it is something that we have already covered in class. That means when a new assignment is posted, you should look at it as soon as you have time. Don't wait until the day it's due!

4. As I mentioned in class, I intend to post shorter assignments often, rather than long assignments less frequently. That means that there will probably always be an active MC assignment.

See you all Monday.

First Friday...

Today we dove into Chapter 2 and atomic theory, through Rutherford's gold foil experiment.

Jean has times set for SI, info is listed to the left <--. I would encourage you to attend if your schedule allows.

Have a good weekend, I'll see you all on Monday.

Change, units, and approaching problems...

The bookstore has ordered more books and more MC access codes, I'm not sure when they'll arrive. If your finanacial aid and funding situation doesn't require you to use the MSUM bookstore, you can also purchase MC access directly from the publisher, and probably find a good deal on the textbook somewhere online.

Today in class we talked about physical vs. chemical changes, and then moved into some math including metric/SI units, unit conversions, and the problems solving strategy used in your textbook. As I said in class, the best thing you can do is to try and think about what a problem or calculation means and develop a reasonable and reasoned approach rather than memorizing a formula to plug in numbers.

If you didn't buy a lab manual and googles today, the Chem Club will be selling again before class on Friday. Manuals = $5; goggles = $5.

Mastering Chemistry

For those of you who need to purchase a separate access code for Mastering Chemistry, you can purchase directly from the publisher. Go to www.masteringchemistry.com and click on the "New Students" button under "Register". That should get you to a page that asks if you have an access code with the options "Yes, I have an access code" or "No, I need to purchase access online now". The "No" option will ask you about purchasing an eBook, that cost extra so if you already have a textbook you will probably not want to use that option.

Again, the course ID is: C150F09JB

Good luck and see you tomorrow.

Lab...

Also, Chem 150L will not meet this week, but lab manuals and goggles will be sold before class on Wednesday and Friday. Manuals are $5, goggles are $5, you will need both...

First day...

The semester has started!!

To sign up for Mastering Chemistry, you will need a course ID, it's listed on my web page, it is:

C150F09JB

Get signed up and look at those assignments as soon as you can.

Today in class we looked at a bunch of Chapter 1 info, we'll wrap that on Wednesday.

Welcome to Fall 2009!

Welcome to Dr. Bodwin's General Chemistry blog for Fall 2009. Check back here for class announcements and information, including answers to questions I get via email. If you really want to dig in, you can check out the blog archive (at left) to see what was posted to the blog during last fall's Chem 150 class.

I will also post announcements on the blog for things that are not necessarily chemistry-specific. If you're involved in a campus organization/team and would like to me to post your event/performance/game, let me know.

Question again...

In exam #3 question #1 how do you find the H⁺ or OH^- from pOH of 6.113?

pH + pOH = pK_w = 14 @25degC
So, if pOH = 6.113, pH = 7.887
[H⁺] = 10^-pH = 10^-7.887 = 1.30x10^-8 M
[H⁺] [OH^-] = K_w = 10^-14 @25degC so
(1.30x10^-8)[OH^-] = 10^-14
[OH^-] = 7.71x10^-7 M
As a check,
pOH = -log[OH^-] = -log(7.71x10^-7) = 6.113

Questions...

On exam #4 question #6: For each of the following reactions, predict whether the sign if delta S will be positive or negative and explain your answer. What are you looking for as an answer for the explanation part?

In the first one, I expected to see something like "A solid is forming from two solutions, so the products are more ordered/less disordered than the reactants." For the second, 2 molecules of gas are forming; for the third, the number of gas particles is changing. A number of people were giving explanations like "positive because the reaction is getting more disordered." That's not an explanation. If your answer to an "explain" question does not address the question "Why?", then it's probably not really explaining the answer.

In exam #3 question #1 could you explain again how to find H⁺ and OH^- as well as how to find K_a and K_b?

[H⁺] and [OH^-] are related by K_w. K_a and K_b are also related by K_w.
[H⁺] [OH^-] = K_w
(K_a)(K_b) = K_w
At 25degC, K_w = 10^-14

In exam #3 question #2: 1.28 mol K₂SO₃ + 1.24 mol HCl, can you explain why this does not result in an effective buffer?

This will not make an effective buffer because adding 1.24mols of HCl will almost bring you to the equivalence point in this titration, it will not give you a buffer. You could make a buffer by adding 0.64mol of HCl(aq) to 1.28mols of K₂SO₃(aq) because the resulting solution would contain 0.64mols of SO₃^2-(aq) {a weak base} and 0.64mols of HSO₃^-(aq) {its weak conjugate acid}. You could also make an effective buffer by adding 1.92mols of HCl to 1.28mols of K₂SO₃(aq) because the resulting solution would contain 0.64mols of HSO₃^-(aq) {a weak base} and 0.64mols of H₂SO₃(aq) {its weak conjugate acid}.

In exam #2 could you run through #12: Graphite (solid carbon) reacts with oxygen gas to form carbon monoxide gas. You have sealed 11.37g of graphite and 18.61g of oxygen in a 6.00L vessel and allowed the system to reach equilibrium at 73.91C. if the equilibrium constant value is 9.42x10^-6 at this temperature, what are the equilibrium concentrations of all reactants and products?

Start with a balanced chemical equation:
C(s) + O₂(g) <=> CO₂(g)

Since carbon is a solid in this reaction, it does not appear in the equilibrium constant expression, so we can put together a table {again, formatting tables in this program is rough, so I'll simply list the concentrations}:

[O₂]_initial = 18.61g / 31.998g/mol / 6.00L = 0.0969M
[CO₂]_initial = 0M

[O₂]_change = -xM
[CO₂]_change = +xM

[O₂]_equilibrium = (0.0969-x)M
[CO₂]_equilibrium = xM

Kc = [CO₂]_equilibrium / [O₂]_equilibrium = (x) / (0.0969-x) = 9.42x10^-6

This isn't a horrible equation to solve directly, but let's see if we can make an assumption. If "x" is much smaller than 0.0969M, then this expression simplifies to:

(x) / 0.0969 = 9.42x10^-6
x = 9.13x10^-7
{"x" is indeed much smaller than 0.0969, so our assumption is valid.}

[O₂]_equilibrium = 0.0969 M
[CO₂]_equilibrium = 9.13x10^-7 M

Sorry, I had a network problem...

Could you go over #2 & #3 from summer 2009 problem set #11?
Thanks

#2 is an algebra problem.
NO₂(g) + 2 H₂S(g) <=> NS₂(g) + 2 H₂O(g)
ΔG_rxn = (1)(-ΔG_f(NO₂(g))) + (2)(-ΔG_f(H₂S(g))) + (1)(ΔG_f(NS₂(g))) + (2)(ΔG_f(H₂O(g)))
From the problem and the tables in your book, you know all of the values in this equation except ΔG_f(NS₂(g)). Plug in and solve for ΔG_f(NS₂(g)).

#3 has a couple parts. First, write out a balanced net ionic equation, it will be difficult to handle this problem if you use a full molecular equation. This is just a backwards K_sp process, so the net ionic equation is:
Pb²⁺(aq) + 2 Cl^-(aq) <=> PbCl₂(s)
Calculate ΔG for this reaction at standard conditions from tabulated values. {I don't have a book with tabulated thermodynamic values handy so I can't calculate an exact value for you...} Once you know ΔG^o, plug in to calculate under non-standard conditions.

ΔG = ΔG^o + RTlnQ

Q = 1 / {[Pb²⁺][Cl^-]²}

Given the numbers in the problem, the correction term {RTlnQ} will probably only be a few kJ/mol. Remember to use kelvin temperatures and be consistent with kJ and J.

Could you go over #2 & #3 from summer 2009 problem set #11?
Thanks

#2 is an algebra problem.

NO2(g) + 2 H2S(g) <=> NS2(g) + 2 H2O(g)

ΔGrxn = (1)(-ΔGf(NO2(g))) + (2)(-ΔGf(H2S(g))) + (1)(ΔGf(NS2(g))) + (2)(ΔGf(H2O(g)))

From the problem and the tables in your book, you know all of the values in this equation except ΔGf(NS2(g)). Plug in and solve for ΔGf(NS2(g)).

#3 has a couple parts. First, write out a balanced net ionic equation, it will be difficult to handle this problem if you use a full molecular equation. This is just a backwards Ksp process, so the net ionic equation is:

Pb2+(aq) + 2 Cl-(aq) <=> PbCl2(s)

Calculate ΔG for this reaction at standard conditions.

ΔG = ΔGo + RTlnQ

ΔG = ΔGo + RTlnQ

Q = 1 / {[Pb2+][Cl-]2}

Questions...

Can you explain again why certain values are negative or positive like in the spring/winter exam reaction #2
Fe(CO)6(l) + ClO2(g) FeCl3(s) + CO2(g)
ΔHrxnº = (783.5 kJ/mol) + 3(-102.5 kJ/mol) + (-399.49 kJ/mol) + 6(-393.509 kJ/mol) = -2284.5 kJ/mol

The ΔH and ΔG values you find in the tables are formation values. If something is being formed in a reaction, the values are correct as you find them in the table. If something in being consumed/desstroyed/used up in a reaction, the magnitude of the value in the table is correct, but the sign is wrong. If something is a reactant, change the sign; if it's a product, the sign in the table is correct.

I was wondering if you could answer #3 from problem set 10?

What is the molecular basis of enthalpy, entropy, and free energy? Let's start with entropy. Gaseous water is more disordered than liquid water, so we expect the entropy to be higher. For both enthalpy and free energy, gaseous water is more energetic than liquid water, so whether we're talking about heat (enthalpy) or overall energy (free energy), we expect the gas to be higher energy than liquid. Note, in this case, "higher" energy means "less negative" since all of these numbers are negative.

Other questions, let me know...

PS#11, July 15th...

Chem 210 – Summer 2009 – Problem Set #11

1. You have combined 50.0mL of 0.927M barium nitrate solution and 50.0mL of 0.899M sodium sulfate solution to form barium sulfate. You have captured all of the energy liberated by this spontaneous process to decompose water to hydrogen gas and oxygen gas. How many grams of hydrogen gas could you produce?

2. You are attempting to find the Gibb’s free energy of formation for NS₂(g). When you react nitrogen dioxide with H₂S(g) to produce NS₂(g) and gaseous water, the free energy change for the reaction is 28.7^kJ/_mol. What is ΔG°_f for NS₂(g)?

3. You have reacted lead(II) nitrate solution and sodium chloride solution under standard conditions to produce lead(II) chloride. What is ΔG° for this reaction? What is ΔG for this reaction if you use 0.85M lead(II) nitrate solution, 1.7M sodium chloride solution, and perform the reaction at 15.00°C? At 91.82°C?

PS#10 from Tuesday, July 14th

Chem 210 – Summer 2009 – Problem Set #10

1. Ethanol {C₂H₅OH(l)} reacts with oxygen gas to create carbon dioxide and water.

a. Calculate ΔH°, ΔS°, and ΔG° for this reaction. Is this reaction spontaneous?

b. How much energy is transferred between the system and the surroundings when 10.83g of ethanol is burned in excess oxygen? Is energy transferred from the system to the surroundings, or from the surroundings to the system?

2. Ideally, the hydrogen gas used in fuel cells would come from water.

a. Calculate ΔH°, ΔS°, and ΔG° for the decomposition of water to hydrogen gas and oxygen gas. Is this reaction spontaneous?

b. How much energy is transferred between the system and the surroundings when 6.28g of hydrogen gas is produced by this reaction? Is energy transferred from the system to the surroundings, or from the surroundings to the system?

3. In the previous problems, did you use liquid water or gaseous water? Why? How will changing from liquid-to-gas or gas-to-liquid change your answers? On a molecular level, explain the differences in ΔH°_f, S°, and ΔG°_f for liquid and gaseous water using the numbers found in thermodynamics tables. Why is gaseous water higher or lower than liquid water for each thermodynamic quantity?

4. If burning ethanol is a step in a reaction mechanism, the reaction may have to be multiplied by some integer.

a. Double all the coefficients in the balanced equation from #1 and calculate ΔG° for the “new” reaction. How is this value of ΔG° related to the value calculated in #1?

b. How much energy is transferred between the system and the surroundings when 10.83g of ethanol is burned in excess oxygen? Compare your answer to the value you found in #1b.

5. If 50.00g of ethanol is burned and all of the water produced by the reaction is decomposed to form hydrogen gas, what is the change in Gibb’s free energy for the whole process?

Question...

exam #3 from winter 2006

Question #2 - could you explain how to determine if it is an effective buffer for each combination listed

Thank you

2. Does the combination listed result in an effective buffer solution? (4pts each)

Yes No 0.38mol HCl(aq) + 0.38mol NaOH(aq)

Yes No 0.90mol Na₃PO₄(aq) + 1.35mol HNO₃(aq)

Yes No 1.28mol Na₂CO₃(aq) + 0.64mol HCl(aq)

Yes No 2.14mol CH₃COOH(aq) + 1.96mol CH₃COOK(aq)

Yes No 0.06mol HCN(aq) + 0.98mol LiCN(aq)

An effective buffer contains approximately equal amounts of a weak acid and its conjugate base with the concentration of each at least 100x the K_a of the conjugate acid. So let's look at each combination:

0.38mol HCl(aq) + 0.38mol NaOH(aq)

This is a strong acid and a strong base, the result of this combination will give a solution of sodium chloride in water. Not a buffer.

0.90mol Na₃PO₄(aq) + 1.35mol HNO₃(aq)

Nitric acid is a strong acid, so when 1.35mol of strong acid are added to the phosphate (a weak base), the result of this combination will be 0.45mols of HPO₄^2-(aq) and H₂PO₄^-(aq). This is a combination of a weak conjugate acid and its weak conjugate base. This will be a good buffer.

1.28mol Na₂CO₃(aq) + 0.64mol HCl(aq)

Similar to the previous example, the result of mixing these solutions together will be a combination of 0.64mols HCO₃^-(aq) and CO₃^2-(aq). This will be a good buffer.

2.14mol CH₃COOH(aq) + 1.96mol CH₃COOK(aq)

Acetate ion is the conjugate base of acetic acid (weak acid), and this is an approximately equimolar mixture of acetic acid and acetate ion. This will be a good buffer.

0.06mol HCN(aq) + 0.98mol LiCN(aq)

HCN is a weak acid and cyanide ion is its conjugate base, but the concentrations are too far apart to make an effective buffer. The concentration of the conjugate acid and conjugate base should be within a factor of 10 to make as effective buffer. This is not an effective buffer

Email question...

I'm looking at Exam 3 from Winter 2006, and for question 5, shouldn't K2SO3 be the one being added to HClO4 since it is stonger?

Perchloric acid {HClO4(aq)} is probably one of the strongest acids we'll consider, so it's pretty safe to say that any titration involving perchloric acid will have the perchloric acid in the buret.

Again, when you are setting up a titration it is always preferable to have a something strong and monoprotic in the buret.

Other questions, let me know, I'll answer them to the blog as soon as I see them...

Problem Set #9, question #5...

5. You would like to perform a titration using cyanic acid { HCNO(aq), K_a = 2x10^-4 } and potassium hydroxide. Which solution should be in the buret? You do not have a pH meter, but you have found a number of indicators. Which of the following would be most appropriate for this titration and why? Indicator: Endpoint range - 2,4-dinitrophenol: 2.8-4.0 ; phenol red: 6.7-8.1 ; thymolphthalein: 9.6-10.3

Although this question has some numbers in it, you don't actually need to do any calculations to answer it. Based upon the Ka value, HCNO(aq) is a weak acid. When performing a titration, you always want to put something strong and monoprotic in the buret. HCNO(aq) is monoprotic, but it is weak. Potassium hydroxide is monoprotic and strong, so it would be the better choice to put in the buret. Now, what about the indicator? When titrating a weak monoprotic acid with a strong monoprotic base, the equivalence point should be somewhere on the basic side, so we should choose an indicator that changes color at basic pH. The range for phenol red includes some slightly basic pH's so it should work, but it's pretty close to neutral, so thymolphthalein is probably the best bet for this titration.

We can also look at this by doing a calculation. When we get to the equivalence point, we will (in essence) have a solution of potassium cyanate in water. If we assume that the concentration of cyanate is 1M, we can calcuate an expected pH of the solution using a K_b-type of equation. {I just pulled 1M out of my hat, you can do this calculation with any concentration to check...}

CNO^-(aq) + H2O(l) <=> HCNO(aq) + OH^-(aq)

[CNO^-]_initial = 1 M

[HCNO]_initial = 0 M

[OH^-]_initial = 10^-7 M

[CNO^-]_change = -x M

[HCNO]_change = +x M

[OH^-]_change = +x M

[CNO^-]_eq = (1-x) M

[HCNO]_eq = x M

[OH^-]_eq = (10^-7+x)M

Let's assume that x is significantly smaller than 1 and significantly larger than 10^-7, then our K_b expression is:

K_b = {[HCNO]_eq [OH^-]_eq} / [CNO^-]_eq = { (x)(x) } / 1 = x² = 5x10^-11

NOTE: K_b for CNO^- is calculated from the given K_a of HCNO

x = 7.07x10^-6

Hmm, this is a pretty small number, our assumption that x is significantly larger than 10^-7 might be a little questionable. 10^-7 is only about 1.4% of 7.07x10^-6, so we should be OK...

x = 7.07x10^-6 = [OH^-]eq

pOH = -log [OH^-] = -log(7.07x10^-6) = 5.15

At 25 degC,

pH = 14 - pOH = 8.849

So it looks like the pH at the equivalence point should be basic enough that thymolphthalein will be a better indicator. For practice, redo this calculation assuming that the concentration of potassium cyanate is 0.5M at the equivalence point...

From a problem set...

Problem Set #4, Question 2:

2. Carbon dioxide reacts with ammonia (NH₃) in the gas phase to produce formamide (HCONH₂) and oxygen. If 15.215g of CO₂(g) and 4.139g of ammonia are combined in a 3.50L vessel and allowed to react equilibrium, the formamide concentration is found to be 24.8mM. What is the equilibrium constant for this reaction? Is this reaction product-favored or reactant-favored?

Balanced equation:
2 CO₂(g) + 2 NH₃(g) <--> 2 HCONH₂(g) + O₂(g)

[CO₂]_initial = 15.215g / 44.009g/mol / 3.50L = 0.098778M
[NH₃]_initial = 4.139g / 17.031g/mol / 3.50L = 0.069436M
[HCONH₂]_initial = 0M
[O₂]_initial = 0M
{NOTE: Once again, I'm carrying too many sig figs here on purpose because this is the middle of the calculation. I'll round later.}

Set these up in a table, I don't have tables active in the blog so....

Change in:
[CO₂] = -2x [NH₃] = -2x [HCONH₂] = +2x [O₂] = +x

So at equilibrium,
[CO₂]_eq = (0.098778-2x)M
[NH₃]_eq = (0.069436-2x)M
[HCONH₂]_eq = (0+2x)M
[O₂]_eq = (0+x)M

The problem states that the equilibrium concentration of formamide is 24.8mM, so we can use that number to find the value of "x".
[HCONH₂]_eq = (0+2x)M = 24.8mM
x = 12.4mM = 0.0124M

Plugging in to find all equilibrium concentrations:
[CO₂]_eq = (0.098778-2x)M = (0.098778-0.0248)M = 0.07398M
[NH₃]_eq = (0.069436-2x)M = (0.069436-0.0248)M = 0.04464M
[HCONH₂]_eq = (0+2x)M = 0.0248M
[O₂]_eq = (0+x)M = 0.0124M

K_c = { (0.0248)²(0.0124) } / { (0.07398)²(0.04464)² } = 0.699
Slightly reactant-favored.

Another question...

From Summer 2007, Exam 2:

7. You have found the following value in a table of equilibrium constants:
2 C₂H₃F₃(g) + 3 Cl₂(g) 􀀧 2 C₂F₃Cl₃(g) + 3 H₂(g) K_c = 5.19x10¹⁸
What is the equilibrium constant for the reaction:
6 C₂F₃Cl₃(g) + 9 H₂(g) 􀀧 6 C₂H₃F₃(g) + 9 Cl₂(g)

We're manipulating an equilibrium constant here, let's start with the equilibrium constant expression for the original reaction:
K_c = { [C₂F₃Cl₃]² [H₂]³ } / { [C₂H₃F₃]² [Cl₂]³ } = 5.19x10¹⁸

To get the second reaction, we have to reverse the original and multiply it by 3. When we reverse the direction of the equilibrium, the roles of products and reactants change, so the equilibrium constant is inverted. When an equilibrium equation is multiplied by some constant, each of the concentrations is raised to that power, so the whole equilibrium constant expression is raised to that power. This means that the equilibrium constant for the "new" reaction in the problem is:
K_c' = { [C₂H₃F₃]⁶ [Cl₂]⁹ } / { [C₂F₃Cl₃]⁶ [H₂]⁹ } = (1 / K_c)³ = 7.15x10^-57

Others? Let me know...

Question...

From Summer 2007, Exam 2:
6. For the reaction:
CH4(g) + 2 O2(g) 􀀧 CO2(g) + 2 H2O(g)
The equilibrium concentrations have been found to be [CO2]eq = 0.568M, [H2O]eq = 0.685M,
[CH4]eq = 1.38x10-8 M, [O2]eq = 0.918 M. What is the equilibrium constant?

I think some of you might be making this harder than it needs to be. The problem is giving you the equilibrium concentrations, so you just need to plug these numbers into the equilibrium constant expression:

Kc = { [CO₂] [H₂O]²} / { [CH₄] [O₂]²} = {(0.568)(0.685)²} / {(1.38x10^-8)(0.918)²} = 2.29x10⁷

Other questions, let me know...

Equilibrium Week

This week we discussed equilibrium and a number of specific equilibrium constants including: K_c, K_p, K_sp and K_f. Although these all have specific conditions or reaction types associated with them, they are all equilibrium constants that follow all the same rules and trends.

Equilibrium is on Exam 2 for previous summer and spring exams. In some cases, there might be K_sp questions on Exam 3, but the bulk of equilibrium is on previous Exam 2's.

If you have questions, let me know, I will be checking email over the weekend and I will answer all questions to the blog so everyone can see the answers. Good luck in your studying, and I'll see you Monday.

Equilibrium II

Today I handed back exams and we continued to talk about equilibrium. The average on the exam was around 115/150.

In case you misplaced your problem set (or just want another one...), here's it is:

Chem 210 – Summer 2009 – Problem Set #4

1. Oxygen gas and chlorine gas react in the atmosphere to form chlorine dioxide, one of the substances responsible for ozone destruction. In a laboratory experiment, 18.237g of oxygen gas and 31.294g of chlorine gas are combined in a 8.00L vessel and allowed to reach equilibrium. At equilibrium, the chlorine concentration is found to be 48.9mM. What is the equilibrium constant for this reaction? Is this reaction product-favored?

2. Carbon dioxide reacts with ammonia (NH₃) in the gas phase to produce formamide (HCONH₂) and oxygen. If 15.215g of CO₂(g) and 4.139g of ammonia are combined in a 3.50L vessel and allowed to react equilibrium, the formamide concentration is found to be 24.8mM. What is the equilibrium constant for this reaction? Is this reaction product-favored or reactant-favored?

3. Graphite (solid carbon) reacts with oxygen gas to form carbon dioxide gas. You have sealed 23.84g of graphite and 53.97g of oxygen in a 12.00L vessel and allowed the system to react equilibrium at 73.91°C. If the equilibrium constant value is 294.7 at this temperature, what are the equilibrium concentrations of all reactants and products?

4. At 149.79°C, methane, CH₄(g), reacts with oxygen to form carbon dioxide and water with an equilibrium constant value of 6.43x10⁶. If you combine 6.85g of methane with 29.99g of oxygen in a 18.00L vessel, raise the temperature to 149.79°C, and allow the mixture to react equilibrium, what are the concentrations of all reactants and products?

Questions...

A couple questions came in from Exam 1, Summer 2007:

12. You have prepared a solution by diluting 18.64mL of 1.16M potassium iodate solution to50.00mL with water. What is the concentration of iodate ions in the resulting solution?
a. 3.11 M
b. 0.216 M
c. 0.432 M
d. 0.242 M
e. 2.31 M
For dilutions, you can use the formula C1V1 = C2V2. C1 = 1.16M (the stock concentration); V1 = 18.64mL (the stock volume); C2 = the "new" concentration; V2 = 50.00mL (the diluted volume). Every potassium iodate unit contains 1 iodate ion, so the concentration of potassium iodate is the same as the concentration of iodate.
(1.16M)(18.64mL) = C2(50.00mL)
C2 = 0.432M

13. You have prepared a solution by dissolving 1.38 mols of sugar (C6H12O6) in 500.0g of water. What is the boiling point of the resulting solution?
a. 1.4ºC
b. 105.13ºC
c. 101.4ºC
d. 98.6ºC
e. 5.13ºC
Boiling point elevation uses molality, so:
1.38mols sugar / 0.5000kg solvent = 2.76m
{delta}T = (0.52)(2.76)(1)
NOTE: I haven't included units because they would be hard to type. 0.52 deg.C/m is the boiling point elevation constant for water, it will be given on the front of the exam. Sugar is a molecular solute, so the van't Hoff factor is 1. Solving, the boiling point will CHANGE by 1.4 deg.C, so the boiling point of the solution is 100.0+1.4 = 1.104 deg.C

14. A reaction is found to be second order with respect to reactant A and zero order with respect to reactant B. If [A]o = 0.942M, [B]o = 0.613M and k = 2.49x10-2 M-1min-1, what is the initial rate of the reaction?
a. 1.44x10-2 M/min
b. 4.31x10-2 M/min
c. 2.21x10-2 M/min
d. 9.36x10-3 M/min
e. 2.81x10-2 M/min
You don't need a time for this one, the problem is basically giving you all the parts of the rate law equation except the rate itself. If the reaction is second order w.r.t. A and zero order w.r.t. B, the rate law expression is:Rate = k [A]^2Plugging in the given values for k and [A], should give you the correct answer.

15. A reaction is found to be second order with respect to ammonium ion, a reactant. If [NH4+]o = 3.34M and k = 1.28 M-1sec-1, what will the concentration of ammonium ion be after 2.16minutes have passed?
a. 0.326 M
b. 3.02x10-72 M
c. 0.575 M
d. 6.02x10-3 M
e. 1.18 M

Use the second order integrated rate law. If you got the wrong answer on this one, check your time units, k is given in seconds and the time is given in minutes. Convert t to seconds before you start.
1/[NH4+]t = (1.28 M-1sec-1)(129.6sec) + 1/3.34 = 166.2 M-1
[NH4+]t = 6.02x10-3 M

Other questions, let me know...

Kinetics...

I've posted a few more old exams on my web page, the full set from Summer 2008 and the first couple exams from Winter 2009. If you have questions, email me and I'll post answers to the blog. Good luck in your preparation for the exam.

Day 1...

Today we looked at states of matter, solutions, and colligative properties.

Your lab experiments will be available online, please visit my web page to print them out. (www.mnstate.edu/bodwin, click on the "Chem 210L" link in the left panel, then click on the experiment name) The class syllabus is also posted on the Chem 210L page. Be sure to print and read the experiment before class.

I will be updating many things on my web page throughout the next day or two...

Summer 2009

Welcome to the blog for summer 2009. This is where I will post answers to questions people ask and make class announcements, so if you have a question about class, check here first.

Osmotic pressure...

16. A newly discovered protein has been isolated from seeds of a tropical plant and needs to be characterized. A total of 0.137g of this protein was dissolved in enough water to produce 2.00mL of solution. At 31.68°C the osmotic pressure produced by the solution was 0.134atm. What is the molar mass of the protein? (20pts)

Osmotic pressure is calculated using the equation:
P_os = cRTi
Where:
P_os = osmotic pressure
c = concentration, usually in units of molarity, M
R = Universal gas constant, 0.08206 ^L.atm/_mol.K in this case
T = absolute temperature in K
i = van't Hoff factor, the number of solute particles per formula unit
For this problem, the osmotic pressure and temperature are given. We'll assume that this "protein" is a single molecule, so i = 1. Plugging in:
0.134atm = M(0.08206 ^L.atm/_mol.K)(304.73K)(1)
Solving for molarity, M = 0.005359 ^mol/_L
We have 2.00mL of solution, that's 0.00200L, so our sample must contain (0.005359 ^mol/_L)(0.00200L) = 1.0717x10^-5 mols of protein. We measured 0.137g of protein, so the molar mass must be:
0.137g / 1.0717x10^-5 mols = 12780 ^g/_mol

Question...

From Exam 3:
2. Explain why each of the following does not result in an effective buffer? (15pts)
1.24mol NH₄NO₃(aq) + 0.03mol NH₃(aq)
0.38mol HI(aq) + 0.38mol NaI(aq)
1.28mol Na₂CO₃(aq) + 0.64mol NaOH(aq)

What makes an effective buffer solution? 1) an approximately equimolar combination of a weak conjugate acid and its weak conjugate base; 2) at concentrations at least 100x the K_a of the weak conjugate acid.
The first combination listed above is indeed a combination of a weak conjugate acid (NH₄⁺) and its weak conjugate base (NH₃), but their concentrations are not approximately equimolar. If the concentration of conjugate acid and conjugate base are not within a factor of ~10, the buffer will not be able to effectively control pH. This could be an effective buffer if an extra mol of NH₃ was added.
The second combination is equimolar, but HI(aq) is a strong acid, therefore it would not make an effective buffer. This could be an effective buffer if HF and NaF were used instead of HI and NaI.
The third combination is a mixture of a weak base with a strong base, and the resulting solution does not contain an equilmolar combination of a weak conjugate acid and its weak conjugate base, so it is not an effective buffer. This could be an effective buffer if 0.64mols of HCl(aq) was added instead of 0.64mol NaOH(aq), because the resulting solution would contain 0.64mols of HCO₃^-(aq) {a weak conjugate acid} and 0.64mols of CO₃^2-(aq) {its weak conjugate base}.

I'm here...

I am in my office right now, I plan to be here all morning so if you have questions feel free to stop in...

Email question...

Can you explain problem 11 on Spring 2009 Exam 2b

Methane (CH₄) reacts with oxygen to form carbon dioxide and water. under some set of conditions at some point in time, you find that 6.495g of methane react every minute in a 600.0 mL vessel.
a. what is the rate of methane consumption?
b. what is the rate of oxygen consumption?
c. what is the rate of carbon dioxide production?
d. what is the rate of water production?
e. what is the rate of the reaction?

Reaction rates are expressed as (change in concentration)/(change in time), so for this reaction under these conditions, the rate of methane consumption is:
{(6.495g CH₄)/(16.043g/mol)/(0.6000L)}/(1min) = 0.6748 M/min
For the rest of the rates, we'll need a balanced reaction:
CH₄(g) + 2 O₂(g) --> CO₂(g) + 2 H₂O(g)
Average rates are determined by the stoichiometry of the reaction. For every mol of CH₄ consumed, 2 mols of O₂ are consumed, so the rate of O₂ consumption must be twice as fast as the rate of CH₄ comsumption. 2x(0.6748 M/min)=1.350 M/min
Every mol of CH₄ that reacts produces 1 mol of CO₂, so the rate of CO₂ production is the same as the rate of CH₄ consumption, 0.6748 M/min.
Every mole of CH₄ that reacts produces 2 mols of H₂O, so the rate of H₂O production is twice the rate of CH₄ consumption, 1.350 M/min. {Which is the same as the rate of O₂ consumption. Note that things that have the same coefficient in the balanced reaction have the same rate.}
The rate of the reaction is also related to the stoichiometry of the reactants. Take the rate of consumption or production of anything in the reaction and divide it by its coefficient, to get the rate of the reaction, 0.6748 M/min.

Other questions, let me know...

Questions....

First a question that was posted before Exam #4.  I'm sorry I didn't see this one before the test.

I was wondering about adding for delta H, and delta S. Sometimes I mix the numbers up and put positives where there should be negatives. for example number 11 on summer 2007 test. for the delta H I got all the numbers right except for PH3(g). How come it is positive not negative?

Calculate the following values for the unbalanced reaction listed at 25ºC.
Reaction 1: P4O10(s) + H2(g)  -> PH3(g) + H2O(g)

The values listed in the table are (in order of the reaction):
{delta}H_f :  -2984 ; 0 ; +5.4 ; -241.818
S^o  :  +228.86 ; +130.684 ; +310.23 ; +188.825
{delta}G_f :  -2697.7 ; 0 ; +13.4 ; -228.572

If something is a reactant, change the sign listed in the table; if it's a product, don't change it.  PH3(g) is a product in this reaction, so use the values directly from the table, in this case they all happen to be positive numbers.  You can think of these numbers sort of like prices.  Let's say you look up the price of a calculator and it's $10.  If you're buying the calculator, the balance in your checking account will decrease by $10.  If you're selling the calculator, the balance in your checking account will increase by $10.  If you're forming something in a reaction, the values listed in the table represent what you have; if you're consuming something in a reaction, the values in the table represent what you have lost.

Someone also asked about a Mastering Chemistry question from Assignment 5.  It has 2 parts...

The rate constant for a certain reaction is k = 3.30×10⁻³ s^-1. If the initial reactant concentration was 0.250M, what will the concentration be after 13.0 minutes?

Integrated rate laws are all about choosing the correct order and then doing some algebra.  In this case, the order of the rate law expression must be determined by the units of the rate law constant.  Remember that:
Rate₀ = k[reactant]^x
Since the rate is expressed as "M/time", then this must be a first-order rate law because (1/s)(M) = M/s.  That means we have to use the first-order integrated rate law expression
ln[reactant]_t = -kt + ln[reactant]₀
ln[reactant]_13.0min = -(3.30x10^-3 s^-1){(13.0min)(60s/min)} + ln(0.250)  =  -3.96
[reactant]13.0min = 0.0191M

A zero-order reaction has a constant rate of 4.30×10⁻⁴ M/sec. If after 40.0seconds the concentration has dropped to 5.00×10⁻² M, what was the initial concentration?

This one tells you it's a zero-order process, but it might trip you up by giving you the "constant rate" rather than a "rate constant".  We can plug the rate and the concentration at 40.0sec in to the regular rate law expression to get the rate law constant:

Rate₀ = k[reactant]⁰

(4.30x10^-4 M/sec) = k (5.00x10^-2 M)⁰

Since this is a zero-order process, it turns out that k is equal to the constant rate.  Again, plug values into the correct integrated rate law expression and have some algebra fun. For a zero-order process, the IRL is:

[reactant]_t = -kt + [reactant]₀

(5.00x10^-2 M) = -(4.30x10^-4 M/sec)(40.0sec) + [reactant]₀

Solving, the initial concentration of reactant was 0.0672M.

Other questions, let me know...

It's Monday morning...

No one has any questions before the exam?!?

Oh Nernst...

We've looked at redox reactions, voltaic cells, non-standard conditions and the Nernst equation, and converting between Eo, free energy and K. Tomorrow we will look at some electron counting (electroplating, etc.)

There is a new Mastering Chemistry posted, it will be your last MC assignment, and it's due Monday.

If you would like to take Exam 4 at any time other than Monday in class, you must let me know in advance. Since the semester has gotten shuffled around with the flood cancellations, I want to be flexible about Exam 4, but you must make arrangements with me in advance. If you don't talk to me about alternate date/time, I will assume that you will be taking the exam during class on Monday.

Exam 4

OK, the news you've all been waiting for...

After looking over the conflicts people sent me and thinking about the schedule,

Exam 4 will be on May 4th.

For the few of you who have significant other exams/quizzes/assignments/papers due on May 4th, I will be a little more flexible with allowing alternate times for the exam. Let me know if you are one of the people in that situation and we will make arrangements.

By the way, don't forget that there's a current Mastering Chemistry assignment, due Monday.

See you all tomorrow...

Redox!! It's shocking!!!!

We started talking about redox chemistry on Monday, reviewing oxidation numbers. Then we talked about a step-by-step approach to balancing redox equations. You don't always have to use every step for every redox equation, but if you run into a real "ugly" redox equation it's helpful.

Today we took another look at balancing redox and tried to decide how to determine if a redox reaction is spontaneous or not. We can physically separate the two half reactions of a redox process and assemble a voltaic cell. The direction of electron flow in the external circuit of a voltaic cell indicates the spontaneous process with the electrons flowing from the anode (oxidation) to the cathode (reduction).

I'm still trying to nail down a date for Exam 4. If you have a conflict (lots of exams, papers due, presentations, etc.) on April 29th, May 4th or May 6th, send me an email TODAY and let me know what your conflict is. I will look at everyone's conflicts and make a decision on the date tomorrow morning.

Moving heat (and energy)...

We've been talking about entropy, enthalpy, Gibb's Free Energy and thermodynamics in general.

There's a new Mastering Chemistry posted, due Monday.

Friday...

Today in class we got started on Thermodynamics by reviewing enthalpy and looking at the other quantity that influences reactions, entropy. Entropy describes and quantifies the disorder of a chemical system.

I didn't have a chance to look at your exams yet, but you will get them back on Monday. There will also be a new Mastering Chemistry assignment on Monday.

Have a great weekend.

Squeezing in one more question...

I have a question about the Winter 2006 Exam (3). I'm looking at #2, asking whether a combination of compounds result in effective buffer solutions.One combination is 1.28mol Na2CO3(aq) + 0.64mol HCl (aq). I think this does make an effective buffer solution because the strong acid converts 0.64mol of Na2CO3 (a weak base) into its conjugte acid. Therefore you'd have equal amounts of a weak base and its conjugate acid. Is this correct reasoning so far?However, another combination is 0.90mol Na3PO4(aq) + 1.35mol HNO3 (aq). Here there is more strong acid than there is weak base, so all of Na3PO4 would be converted into its conjugate acid...that wouldn't result in a good buffer, correct? The answer key says that this combination would make a good buffer. Why?

But would it stop at monohydrogen phosphate ions? After all of the phosphate is neutralized to monohydrogen phosphate (a conjugate acid/conjugate base pair), half of the monohydrogen phosphate is neutralized to dihydrogen phosphate. This is an equimolar mixture of a weak conjugate acid (dihydrogen phosphate) and its weak conjugate base (monohydrogen phosphate). For most polyprotic acids (or bases), we can make more than one effective buffer by using each of the protonation equilibria. For phosphate, there are three buffers possible at three different pH's: phosphate/monohydrogen phosphate at ~12, monohydrogen phosphate/dihydrogen phosphate at ~7, and dihydrogen phosphate/phosphoric acid at ~2.

Likewise, carbonate can make the buffer mentioned in the problem (carbonate/bicarbonate at ~10) as well as bicarbonate/carbonic acid buffer at ~6.

We're back!

Yesterday in class we reviewed for tomorrow's exam.

Another question:
What is the Kb of a base if 500.0mL of a solution containing 0.153 mol of the base and 0.191
mol of its conjugate acid has a pH of 6.114? Over what pH range would this conjugate acid/
conjugate base pair make an effective buffer?
How do you determine the pH range that the solution would be a good buffer?
A weak conjugate acid/conjugate base combination will be an effective buffer over a pH range that is within 1 unit of the pK_a of the weak conjugate acid, so we have to work through to find the K_a. Plugging the numbers from the problem into the Henderson-Hasselbalch equation,
6.114 = pK_a +log(0.153/0.191)
pK_a = 6.210
NOTE: You can calculate concentrations of the conjugate acid and base, but the "500.0mL" volume mathematically cancels out, so you can just use mols directly in the H-H equation for this specific problem.
At this point, we can answer the buffer range question. Since the pK_a of the acid is 6.210, this system will make good buffers from 5.210 to 7.210.
Once you know the pK_a of the conjugate acid, the pK_b of the conjugate base is:
14 - 6.210 = 7.790
And the K_b of the base is:
10^-7.790 = 1.623x10^-8

Mastering Chemistry Assignments 12 & 13:
A number of people have talked to me about the 2 Mastering Chemistry assignments that were due last week. I realize that the sandbagging efforts and evacuations caused some difficulties. At the same time, these assignments were posted to Mastering Chemistry and announced on the class blog before classes were cancelled so everyone should have had the opportunity to work on them before the due date. As a compromise, I have changed the grading settings in Mastering Chemistry so that the possible points on those assignments is decreased by 5% for every day they are late. This means that you can still recover most (but not all) of the points on these assignments even if you do them late.

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.