As Soon As Something Said That Is Wrong, Credit Is Lost. Be Succinct, Complete, and Clear

Test Taking:

The exam grader doesn’t know you. S/he isn’t going to guess what you are saying. And if s/he can’t read your writing, you won’t receive credit.
The exam grader spends a little more than a minute to grade an entire question, short or long, 2 parts or 10, the average is 1.14 minutes, so do not ramble, get to the point, bullet point, no need to repeat the question, or work too hard making complete sentences and focus more on content
As soon as something said that is wrong, credit is lost. Be succinct, complete, and clear.
You are not penalized for guessing on the m/c.
Read the question carefully – sometimes your brain will inadvertently add information that is not in the question. And answer only the question.

Nernst Equation

Cell voltage, E, at non-standard conditions (25oC): E = E0 – ln QE = E0 – ln Q

Similar Equations

Arrhenius / / reaction rate at two different temperatures
Clasius-Claperyon / / vapor pressures of pure liquids or solids to temperature
van’t Hoff / / vapor pressure at two different temperatures
/ equilibrium constant to the change in temperature given the standard enthalpy change (ΔHo).

Kinetics

Rate expression: rate = k [A]m[B]n
Equations on AP appendix are for 1st order.

Determining rate order from graphs. Straight lines:

0-order: [A] vs. t;1st-order: ln[A] vs. t(other: 2nd-order)

Graphs:

COLLIGATIVE PROPERTIES
OLD-TEACHER SUGGESTIONS

AP Chemistry: Rules for Significant Figures

ByPeter J. Mikulecky,Michelle Rose Gilman, andKate Brutlag

Part of theAP Chemistry For Dummies Cheat Sheet

After you get an answer on the AP Chemistry exam, make sure that you provide the correct number of significant figures in your answer. Here is a summary of the rules for assigning significant figures:

Any nonzero digit is significant.Thus, 6.42 seconds (s) contains three significant figures.
Zeros sandwiched between nonzero digits are significant.Thus, 3.07 s contains three significant figures.
Zeros on the left side of the first nonzero digit arenotsignificant.Thus, 0.0642 s and 0.00307 s each contain three significant figures.
When a number is greater than 1, all digits to the right of the decimal point are understood to be significant.Thus, 1.76 s has three significant figures, while 1.760 s has four significant figures. We understand that the6is uncertain in the first measurement, but is certain in the second measurement.
When a number has no decimal point, any zeros after the last nonzero digitmay or may notbe significant.Thus, in a measurement reported as 1370 s, we cannot be certain if the "0" is a certain value, or if it is merely a placeholder. Be a good chemist. Report your measurements in scientific notation to avoid such annoying ambiguities.
Numbers resulting fromcounting(for example, one kangaroo, two kangaroos, three kangaroos . . .) or fromdefined quantities(for example, 60 seconds per 1 minute) are considered completely certain.These values are understood to have an unlimited number of significant figures, consistent with their complete certainty.
When taking the log of a number, as when calculating pH or pOH from [H+] or [OH–], only the decimal portion of the answer applies toward the significant figure count (not the preceding integer).For example, if [H+] = 0.0100Mand pH = –log[H+], then pH = 2.000. Why? 0.0100 contains three significant figures. Therefore, the decimal portion of the log answer (the mantissa) contains three significant figures. The preceding integer (the characteristic "2" in this case) does not count toward the significant figure total.

The final thing to burn into your brain about significant figures is that your final answer should always be rounded to the same number of significant figures as the least-precise number you were given in the problem. However, do not round any of the numbers you are given until the very end after you have plugged them into your equations in their full, precise glory.

KINETICS

A very strange question, but easily answered with a mindless rule I teach my students. The unit of K is M(1-r)/s (or atm(1-r)/s) . For example, a second order reaction would have a rate constant of M-1/s . ( 1 -2 = -1). Using this rule, if the unit of K is M2, then 1 - r = 2, and r = -1 .

You could also prove this to be true using factor labeling.

I should also point out that rate orders not only can be negative; they can also be fractions. There was a question on an AP test fairly recently in which one of the reactants had a rate order of 0.5 . (IIRC !!)

These are for my own personal knowledge, not necessarily content an AP student would need to know (especially question 2).

1. On a molecular level, WHY does increasing the temperature a solvent usually increase its solubility? Increasing kinetic energy can't be enough of a reason since some things decrease in solubility or barely increase.

2.When you dissolve a non-volatile solute in a solvent such as water, you need to decrease the temperature to get it to freeze. I understand that the solute gets in the way of the solvent molecules and disrupts crystal formation, but WHY does lowering the temperature suddenly allow freezing to occur? Is it that eventually the intermolecular forces overcome the fact that the solute is in the way? The Zumdahl text (6th ed.) gives an explanation about vapor pressure of ice vs. a solution that I don't really understand.

1) Whether the solubility of a solute increases or decreases as T increases depends entirely on whether the dissolving process is endothermic or exothermic. I believe this is best understood in terms of equilibrium. energy + solid <===> solution

In a typical equilibrium with a solid dissolving in a liquid, the process is usually (though not always) endothermic. Specifically, it takes more energy to separate the solute particles (solute-solute interactions) than we get back when the solute is attracted to the solvent (solute-solvent interactions). This means that, as the temperature of the equilibrium system is increased, the system will shift to absorb energy, meaning it will shift toward the products and more solute will dissolve. The magnitude of the ∆H for the dissolving process will determine HOW MUCH the solubility will increase, but the SIGN of the ∆H will determine whether solubility increases or decreases. I’m not sure where you would find the data for this, but this implies that the ∆H of solution for NaCl (whose solubility does not change very much with temperature) would be much smaller that the ∆H of solution for KNO3(whose solubility changes dramatically as T increases).

gas <===> solution + energy

For gases dissolving in a liquid, the opposite is ALWAYS true. Because the attractive forces between the gas molecules can be ignored (because the gas molecules are so far apart that the molecules do not feel the attractions), the only thing that happens when the gas dissolves is the attractions between the gas molecules and the solvent, making the process exothermic. This means that, as T increases, the reaction will shift toward the reactants, meaning that less gas is dissolved.

I suspect that there is an entropy aspect to this as well, but I have not thought through the implications thoroughly, so I cannot give you any details. In general, however, as T increases, the entropy of a system increases, though how that impacts how much will dissolve is not immediately clear to me.

Here’s my take on point 2.

2) First of all, it is clear that a solution of a nonvolatile solute in a liquid will decrease the vapor pressure of the solution (compared to the pure solvent). Ignore the explanation of why this is true for the moment, because it is experimentally demonstrable fact. If this is the case, then clearly the boiling point of the solution MUST be higher than the boiling point of the pure solvent. To raise the vapor pressure to the value of 1 atm the temperature must be raised.

But what about the freezing point? If we look at water (with which we are very familiar) and look at its phase diagram, the melting point line is nearly vertical. In fact, most phase diagrams are not drawn to scale because then the melting point line would look vertical. The melting point line proceeds upward from the triple point (the temperature at which the vapor pressure of the liquid equals the vapor pressure of the solid).

Now what about the solution? The vapor pressure of the solution is less than the vapor pressure of the solvent (water). This means that the “triple point” of the solution (the temperature at which the vapor pressure of the solution is equal to the vapor pressure of the solid) is going to occur at a lower temperature, and since the melting point line for the solution rises from this point, the melting (freezing) point of the solution will occur at a lower temperature than that of the pure solvent.

2) Here's an alternative explanation using entropy (not original, though I can’t seem to find the documents I learned this approach from)

There is another explanation for the change in boiling point and freezing point based on Entropy. First, we understand that, for a nonvolatile solute, the vapor phase of the solution is simply the pure solvent, so the entropy of the vapor at the boiling point is exactly the same for the solvent and the solution. The same is true for the solid. If the freezing is done slowly enough, ONLY the solvent freezes, and the solute stays in solution. So the entropy of the solid solution at the freezing point is the same as that of the pure solvent.

However, the entropy of the solution is greater than the entropy of the pure solvent. So the change in entropy going from the solution to the vapor phase is LESS than the change for the pure solvent, and the change going from the solution to the solid phase is GREATER than for the pure solvent.

Since both boiling and freezing are equilibrium processes, at the boiling point and freezing point ∆G=0, and (rearranging ∆G = ∆H – T∆S)

T = ∆H/∆S.

Since the vaporization (and freezing) involve ONLY solvent molecules, AND the temperature is close to the temperature for the pure solvnt, ∆H can be assumed to change very little. However, we just asserted that ∆S for the vaporization of the solution is smaller that for the pure solvent, so ∆H/∆S is bigger (the boiling point is a higher temperature). In the same way, ∆S for the freezing of the solution is bigger than for the pure solvent, so ∆H/∆S is smaller, meaning that the freezing point is lower for the solution.

Just to build a bit on Thomas's excellent explanation. The melting point of a liquid occurs at the temperature where the vapor pressure of the solid and the vapor pressure of the liquid are equal. I don't think this point is emphasized sufficiently in textbooks. Picture a sealed container containing both ice and water. Water evaporates, ice sublimes. If they both reach equilibrium at the same water vapor pressure, then the system is stable. However, suppose the vapor pressure of the ice is greater than that of the water. Then when the water reaches equilibrium with its vapor, the ice will continue to sublime. The phase with the higher vapor pressure will disappear, and the phase with the lowest vapor pressure will form. As Thomas points out, one can see from the phase diagram, by extending the solild-vapor boundary (which is the vapor pressure curve of the solid) and the liquid vapor boundary, that at temperatures ABOVE the melting point, the solid has a higher vapor pressure than the liquid, and vice-versa.

The presence of a solute lowers the vapor pressure of the liquid, but does not change the vapor pressure of the solid. The vapor pressure of the solid is thus now higher than that of the liquid, so the solid melts. To reach phase equilibrium again we must lower the temperature until the two vapor pressure curves coincide.

In fact, you can actually derive the Kf of 1.86o by solving the Clausius-Clapyron equation for the temperature at which the vapor pressure of water in a 1 molal solution becomes equal to the vapor pressure of the ice.

Hope this is interesting to you! I love this stuff, and really regret its exclusion from the new course.

Paul

MIT OpenCourseWare: Highlights for High School Students

We teach our students to use "Claim-Evidence-Reasoning" (CER), or Claim-Data-Warrant, when writing lab conclusions and answering those "explain" and "justify" questions on FRQs (and on our tests). Our English and history teachers also use claim-data-warrant in their classes, which really helps improve student writing across disciplines.

For example, a student might answer a question about the pH of the equivalence point for a strong base-weak acid titration (that asks for justification) with "it's basic because the pH is higher than 7", or "it's basic because it is a strong base-weak acid titration". If they give evidence (the conjugate base reacts with water to form hydroxide ions, or they write the equation) then reasoning (the hydroxide ions make the solution basic), it is a more complete answer. We have seen improvement in lab conclusions, short answers on tests, and FRQ responses. Without CER I find that students sometimes simply don't know how to support their claims.

For this year's AP exam, my students told me that they were so trained to give the evidence and reasoning that sometimes they wrote it all out in the first part, then saw a prompt for the justification later and ended up re-writing parts. Not a bad thing in my opinion (except it means they aren't reading the whole question before beginning, despite my instructions to read it all first).

As a former research scientist I tend to emphasize labs. Here is a list of the ones we did this year:

Density of metals
Golf Ball Density Lab (I am an avid golfer - this lab also includes designing a way to remove a sunken golf ball (d=1.14 g/mL) from a 1 L grad cylinder without pouring out the water - next year's challenge what is the volume of a golf ball dimple?))
Vex-1 chem formula (Vex - Vernier lab manual expts)
Vex-3 molar mass volatile liq
Mex-9 - Spectroscopy using a red tide spectrometer (Mex derived from old Masterton and Slowinski lab manual)
Vex-13 Acids and Bases (including NaOH stdzn with KHP)
Mex-13 Model Building
Mex-14 White Solid Identification (based on mp, aq and hexane solubillity and conductivity)
Vex-34 vapor pressure
Vex-4 Freezing Point Depressi
Vex 25 - Fe3+/I- reaction spectrophotmetrically
Vex-35 Kinetics (OH- catalyzed CV rxn)

***************************************

13. Vex-10 Equilibrium FeSCN2+

14. Vex-7 Acid Base Titration (including NaOH stdzn with KHP)

15. Vex-19 Buffer Prep and titration

16. Vex-23 Calcium Hydroxide Ksp

17.Vex 20 Voltaic Cells.

18. Vex 21 Electroplating a key with copper

I feel if the students have enough lab experience and tutelage, they should be able to respond to whatever CB throws at them!

Henderson-Hasselback, Weak Acid Ionization & Dilution

My understanding, too, is that the lower the concentration of a weak acid, the greater its percent dissociation.

The 5th edition of Zumdahl (Chemical Principles) explains it with an algebraic example for a solution of HA:

HA <--> H++ A-

[H+] = [A-] =x; assumexis negligible compared to [HA]0.

Ka= [H+][A-] / [HA]0=x2/ [HA]0

Now assume we dilute by a factor of 10 and calculateQ.

The new (non-equilibrium) concentrations are:

[H+] = [A-] =x/ 10

andQis:

Q= [H+][A-] / [HA]0=(x/10)2/ {[HA]0/10}

Q= 1/10 (x2/ [HA]0) = 1/10Ka

BecauseQis smaller thanKa, (Q= 1/10Ka) the dissociation adjusts toward the products to re-establish equilibrium.

If I understand your question correctly, you are confusing [H+] with percent dissociation. Generally, you would assume as you increase the concentration of a weak acid, more H+ will be created thus decreasing the pH. This is correct.

An oddity of weak acids is that their Ka value is very small, meaning the equilibrium "sits on the reactant side", there is more reactant that product. As you increase the concentration of your weak acid, more will dissociate. However, this increased concentration of conjugate base "pushes back" or readily reacts with H+, reacting at a faster rate than the dissociation of your weak acid. So the overall effect is less dissociated weak acid.

It is easy to see the values in an ICE chart. I will use a WA with a Ka of 1x10-6, Y will represent the original conc.

HWA <-> H+ + WA-

-x +x +x

if Y is 1, x = 1x10^3. Divide that by Y (1) to get a percent dissociation of .1%

If Y is .1, x = 3.2x10^-4. Divide that by Y (.1) to get a percent dissociation of .32% (lower concentration, higher dissociation)

Electron Configuration Questions

IF only 4s is accepted, then we are to ASSUME that the atoms are in the gaseous state and are being oxidized by the minimal energy necessary to remove the highest energy electrons. The question does NOT say, "What is the ground state electron configuration of the zinc ion."

Not missing questions

As readers will tell you, problems always begin on the left hand side of a page (an even number) and if needed will continue onto the right hand side. Therefore a problem will always be visible over the open two pages. In addition, lines for writing are provided only after the entire question has been presented. Based upon the needs of the box for the LS I am certain that question 2 encompassed two pages. I am also certain that there was no "weird" formatting here. The readers will individually see all 160000 exams that were written on Monday. We will see if there is a problem; let us see what readers say when the process begins on June 12.

Hello everyone:

Rest assured that the College Board folks have been made aware of the concerns that some of you have regarding this year's examination. What follows is their response to those concerns.

The formatting in the 2015 AP Chemistry free-response test booklet is appropriate and is consistent with the formatting practices that have been in place for many years. In the booklet, all free-response questions begin on a left (even-numbered) page. As a result, when a long question continues on to the right hand page, the entire question is visible when the booklet is open. The lines for answering a question are printed after the last part of the question (whether it appears on the first or the second page). It is also a longstanding practice for students to provide certain responses (e.g., Lewis diagrams, graphs) in boxes or using axes printed in the booklet. Responses to the other parts of a question should be written on the lines printed in the booklet beneath the last part(s) of the question.