Saturday, May 14, 2011
Extra Credit Opportunity
Sunday, May 1, 2011
All Solutions to Review Problems Posted
Saturday, April 30, 2011
Solutions to Day 3 and Day 4 of the AP Chemistry Test Review
- Please go the the Events Calendar. Under the date of Saturday, April 30th, 2011 are the solutions to the problems with $ signs for the day 3 and day 4 review for the AP Chemistry test review. I hope to have the Podcast solutions for remaining three problems that were optional to do from the day 4 and day 5 reviews posted under the date Sunday, May 1st, 2011 by tomorrow morning. Make sure that you are proficient at all of the types of problems reviewed last week, but especially equilibrium problems.
- As you review for the test, make sure that you review all of the problems from last week and all of the problems from the Friday morning review sessions. At the review session held at school this morning, I noticed a lot of people using review books for the AP Chemistry exam. That is great, but make sure that you use the old free response problems as study material also. The review books usually do not match the complexity of the old free response questions.
- On Monday morning, be at school by 7:30 a.m. Be at the auxiliary gym by 7:45 a.m. If you need to borrow a calculator or you think that you need batteries for your calculator, come by room #327 prior to the exam.
- The format of the test is first the multiple choice section then the free response section. You will have 90 minutes to answer 75 multiple choice questions. You will only be given a periodic table for the multiple choice exam. You will need to remember basic equations and do simple calculations without the aid of a calculator. You will have a list of equations and constants for the free response section of the test. You will have two parts to the free response section. Part (A) is the part of the free response test that calculators are allowed. You will have 55 minutes for part (A). Part (B) does not allow calculators and is when you answer conceptual questions. You will have 40 minutes to finish part (B). The multiple choice part of the test is worth 50% of your overall grade and the free response is the other 50%.
- Remember, you are not only studying for the AP Chemistry exam, but also for the final for the AP Chemistry class, which will count for 20% of your semester grade. The test will given over a total of three days. On Tuesday, May 3rd, you will take the multiple choice section of the final. On the AP multiple choice test, you get roughly 50 seconds per question. The multiple choice test for the final will be designed with 41 multiple choice questions to be completed in a 50 minute time frame. Wednesday, May 4th is a late start day, so no part of the test will be given. We will choose the class T-shirt that day. On Thursday, May 5th, the non-calculator portion of the free response part of the exam will be given. Time limitations of the real AP Chemistry test will be followed, so you will have 40 minutes that day to complete the exam. On Friday, May 6th, the calculator portion of the free response part of the exam will be given. You normally have 55 minutes for this part of the exam, but will only be given 50 minutes. I will remove parts of the original questions to make it fair terms of time. On Monday, May 9th, I am going to reserve that day for any make-up exams that need to be completed. If you have completed all exams, you will be released from class. Remember, I will curve the final and then add the extra credit that you have earned by coming to the early morning sessions and the practice exams. After studying for the real AP Chemistry exam, you should not need to day any more studying for the class final. I will make sure to choose questions that address material that we have covered extensively in class and during the review process.
- For the class final, you will have an equilibrium problem for the first part of the calculator portion of the free response section. Since we did not have a test over thermodynamics (entropy, enthalpy, Gibb's free energy) and electrochemistry, I will guarantee you questions on the free response part of the final exam covering those topics. I will also guarantee you a question covering chemical reactions like the once given on the day 3 review from the last week of class.
- Feel free to call me on my cell phone if you have any questions tomorrow as you study. You should all have my number. If not, call someone in class that may have it.
Thursday, April 28, 2011
Day 4 In Class Problems Corrections and Clarifications
- For the last question done in class,1998 #1 b) i), there was an error. We (I) forgot to square the value for the hydroxide concentration, thus the value for the zinc ion concentration was wrong. The correction is in the solution below. I caught the mistake with a student when we did the confirmation of "Q" when we plugged in the concentration values, so the lesson is always confirm your answers if you have the time.
- For part (d), 1996, there was some confusion. In the solution below, I showed how you could use Henderson-Hasselbalch to solve for the concentration of the conjugate base. Since both the pH and pKa are the same, subtracting pKa from pH would leave you with zero. To solve for "x", the inverse of log has to be performed. This would make the zero on one side of the equation "1" and possible to solve for.
- Part (d), 1996 can also be solved by knowing the concept that pH is equal to pKa when you are at the half-way point of a titration. That is when the amount of conjugate base is equal to the original acid. This ratio would be one. The log of 1 is zero, thus pH is equal to pKa. The problem gave the concentration and volume of the acid, so the millimoles of the acid at the equivalence point can be determined. For the problem, 10.0 millimoles of acid would be at the equivalence point. Therefore, half of 10.0 millimoles is 5.0 millimoles, which is the amount of NaOH that would be added to get the ratio of acid to conjugate base just right to have the desired pH.
Tuesday, April 26, 2011
Solutions 2005 #8, 2005 (Form B) #2, and Molar Mass Determination
Monday, April 25, 2011
Solution to All of 2005 #2 and Part (d) of 2004 #2
Saturday, April 23, 2011
Electrochem Solutions 2001 #7; 1996 #7, and 1998 #8
Monday, April 18, 2011
Balancing Redox Reactions Quiz and Announcements
Friday, April 15, 2011
2004 (Form A and B) Free-Response Questions #2 and #7 Solutions
Thursday, April 14, 2011
1999 Free-Response Question #6 Solution
Sunday, April 10, 2011
Solutions to WebAssign Problems and Some Reminders.
Tuesday, April 5, 2011
Completed Solution - #3 Ksp Packet
Saturday, April 2, 2011
Titration / pH Determination Problem Solution
Friday, April 1, 2011
Buffer Quiz
Wednesday, March 16, 2011
Chapter 16 Test
- Know how to manipulate all equations: pH=-log[H^+] or pH=-log[H3O^+]; pOH=-log[OH^-]; pH+pOH=14.XXX; [H^+][OH^-]=1x10^-14; Ka*Kb=Kw - All equations are linked in some shape or form, so one can be used to find information that is needed for a value to be used in another problem.
- Memorize all six strong acids. Know that all strong bases are metal hydroxides.
- The concept of acid strength can be attributed to the position of the proton in relation to the nucleus of the atom that it is bonded to. HCl, HBr, HI are all strong acids because the protons are far enough from the nucleus and are shielded from the nucleus of the halogen so removal of the hydrogen ion is easier. The bonds between the hydrogen and the halogens are relatively weak. HI is the "strongest" of the strong halogen acids and HCl is the "weakest" of the strong halogen acids because of the reasoning just given. All are still strong acids because of complete dissociation. The remaining halogen, fluorine, makes the weak acid HF. Even though the shared electrons between H and F are closer to F because of its "highest" electronegativity, the bond strength between H and F is strong (4.0 EN - 2.1 EN) and the postition of H to F is close, making it harder for the H^+ to be lost or donated.
- The more oxygens within an oxy-acid, the "stronger" that acid will be. This is due to the increased electronegativity of the added oxygens causing the shared electrons between hydrogen and the element that it is bonded to being farther from hydrogen and closer to the other atom. This concept applies to other elements being added to the acid that have high electronegativities.
- Acids and bases are aqueous, therefore they shift (mess-up) waters equilibrium. Water will act as a base if an acid (strong or weak) is added. Water will act as an acid if a base (strong or weak) is added.
- In a weak base problem, hydroxide is always a product.
- A pH or pOH will give the equilibrium concentration of [H^+] or [OH^-] in a weak acid or weak base problem if asked to determine the Ka or Kb of a substance.
- If a conjugate base of a weak acid is acting as a base, the Kb can be found by dividing Kw by the Ka of the weak acid. Conversely, if a conjugate acid of a weak base is acting as an acid, the Ka can be found by dividing the Kw by the Kb of the weak base.
- The second ionization of a strong polyprotic acid (example: H2SO4) can have an effect on the pH if the initial concentration of the of the acid is relatively low. The second ionization will have little to no effect on the pH if the initial concentration is large ( 1 molar or above).
- A second equilibrium system can be effected by a primary equilibrium system. The primary equilibrium equilibrium system can produce ions that will effect the ability of the secondary equilibrium system to shift right.
- A secondary system can have an effect on pH if its Ka is close to the Ka of the primary equilibrium system and the the inititial concentration of the secondary equilibrium system is large. The secondary equilibrium system will have NO effect on pH if the Ka of the second system is much lower than the first system Ka and the initial concentration of the secondary system is relatively low or close the initial concentration of the primary equilibrium system.
- Since the test is multiple choice, answer choices will be given that are close to each other. Always check "x" values against the initial concentrations to ensure that it meets the 5% threshold. If not with 5% dissociation, use the quadratic formula to get the most acurate value.
- Consider the source of each ion in a salt (ionic compound) when determining the acidic or basic properties of a a salt. A strong base source with a weak acid source will cause a salt to be basic. A weak base source with a stong acid source will cause a salt to be acidic. A strong base source with a strong acid source will cause the salt to be neutral. If a weak base source is with a weak acid source, the solution will be neutral if the Ka and Kb values are equal. If the Ka of the conjugate acid exceeds the Kb of the conjugate base, the solution will be acidic and vice versa.
Come and see me early in the morning if you have any questions. Have a good night.
Monday, March 14, 2011
Acidic Nature of Highly Charged Metal Ions Quiz
The complex ion in the question, hexaaqua aluminum ion, would be octahedral in shape with six ligands (waters) surrounding the central aluminum ion. The hybridization of the aluminum ion would be sp^3d^2. The question asks that you write a dissociation reaction for the I.C.E table, so make sure to show Al(H2O)6^+3 losing a proton and show the resulting conjugate base.
Tuesday, March 1, 2011
Beginning of Chapter 16
Monday, February 28, 2011
Chapter 15 Equilibrium Test
Tomorrow is the test over Chapter 15, Equilibrium. The test will be all written and consist of six questions. Some of the questions will have multiple parts. You will be given the list of equations and a periodic table supplied for the AP Chemistry exam. Below is a list of what you should know for the test and a key to the LeChatelier's Worksheet.
Know how to perform all calculations from the lab. This includes absorbance to concentration calculations, I.C.E. table set-up, and "k"(equilibrium) calculations using the data from the I.C.E. table.
For all I.C.E. table values, remember that simple stoichiometry must be followed. An example would be the relationship for a balanced equation shown below.
2A + B <--> 3C
A change of "C" would be three times larger than a change in "B". A change in "A" would be twice as large as the change in "B". A change of two "A" would be equivalent of a change of three "C".
Know how to use the equation for converting a k>p to a k>eq and vise versa.
Know how to predict the shift of a system at equilibrium by stressing the system. Remember, LeChatlier's Principle states that a system will try to re-establish equilibrium by partially undoing the stress caused to it. You will have to visually show this like you did on the LeChatelier's worksheet and also be able to discuss what the system must do to re-establish equilibrium.
If you are given a problem where there are no products initially, the shift must occur to the product side to create products.
If you are given a problem where both reactants and products are present initially, you must first determine the reaction quotient, Q, to determine the shift of the reaction. The shift must cause the value of "Q" to eventually equal k>eq.
A right shift would cause you to add "x" to the product side and subtract "x" from the reactant side. A left shift would cause you to subtract "x" form the product side and to add "x" to the reactant side.
You must always keep all "x" values; therefore use of the quadradic equation may be a must OR always first look for easy squares. What is meant by "easy squares" the values of both products and reactants can have their square roots found before doing the hard math of a quadradic equation. Look at the quiz from this past weekend as an example of this.
Use the hand out from today as a guide / reason for disregarding subtraction "x" once the equilibrium values have been placed in an equilibrium expression. If the k>eq value and the initial concentrations differ by a magnitude of 10^5 or more, it is permissible to disregard subtracting "x" in the equilibrium equation to solve for "x". "x" must still be used to determine equilibrium concentration / partial pressure values.
If solving for equilibrium concentrations of reactants and products, be sure to place the values back in the equilibrium expression to confirm that the concentrations equal the equilibrium value given in the problem.
Friday, February 25, 2011
Equilibrium Reaction Quiz
Friday, February 11, 2011
Results of the Reaction Rate Test and the Kinetics Lab Write-Up
# of scores/percentage/letter grade
3 / 98.0% / A+
2 / 89.9% / A-
2 / 87.2% / B+
1 / 81.1% / B-
5 / 79.1% / C+
2 / 76.4% / C
2 / 73.6% / C
2 / 70.9% / C-
2 / 68.2% / D+
1 / 65.5% / D
1 / 62.8% / D-
1 / 52.o% / F
1 / 35.8% / F
1 / 22.3% / F
Kinetic Lab Report
Use the lab report procedure that was given to you at the beginning of the year. The entire lab report must be typed. If you do not have it, I will give a short summary of the report format and special points that I will be looking for in this lab.
- Purpose: Why did you do the lab. What was being studied.
- Theory: Please do a good job on this for this lab. Discuss all of the concepts being studied by this lab. Points of discussion should be rate of reaction (what is it), concentration affects on rate, temperature affects on rate. Discuss collision / molecular theory when discussing concentration and temperature affects on rates. Discuss activation energy and what was done to determine it.
- DO NOT RE-WRITE THE PROCEDURE! I know what you did.
- Data Table: Re-Create the data table from the hand-out. Make sure to include units and report time in just seconds.
- Analysis: Show all regression data (including "r" squared) to support which order the reaction is and which order the reaction is not. Once the order has been determined, make sure to report the rate law constants with the correct units for each trial. Show the calculations using the modified Arrhenius equation to determine the activation energy.
- Sources of Experimental Error: Please do a good job on this. Give at least one SEE of substance that could alter your results. Please state what the SEE(s) are and HOW it / they can affect the results.
- Conclusion: State if results matched theory. Did the reaction rate alter the way that it should by changing concentration and temperature. All results of the lab should be touched upon. Also, discuss any alterations to the lab that would improve the results.
Thursday, February 10, 2011
Chapter 21 and 14 Test
- Make sure that you KNOW the nuclear decay products of alpha, beta, and gamma particles. Know how each would be represented.
- Be able to write a balanced nuclear decay emissions reaction. Make sure that you know how to balance the atomic numbers and atomic mass numbers of the parent and daughter isotopes.
- Know that all nuclear decay rates are first-order. A first-order integrated rate law applications can be used. Original isotope amounts, final isotope amounts, rate constants, and time (including half life time) can be solved for.
- Know how to determine rate order from concentration and time data. This is the integrated rate law which will require you to determine which mathematical relationship ([X] vs time OR ln[X] vs time OR 1/[X] vs time) gives the straightest line.
- The slope of the integrated rate law graph is the rate constant. All rate law constants are positive in value. Zero and first order integrated rate laws have a negative sign in front of the "k" value because of the negative slopes of each graph.
- Each order of reaction has its own units. Look at the packet given in class that summarizes rate law constant units.
- Know how to write an overall rate law.
- Remember, the sum of all of the orders of reactants is the overall rate order. Example: A reactant that is first order and another reactant that is second order will have an overall rate order of 3. An example for the units for the rate constant would be L^2/mol^2 x seconds.
- Know the differential method for determining orders of reactants and rate laws.
- **Be aware of the method for determining order of reactants that was presented in class with the 1997 Free Response question from work packet #2. Once one order is found, that order can be used in subsequent ratios of experiments. Thereby, cancelling concentrations of the reactant that has the known order is not necessary.
- Know how to determine the overall enthalpy (energy) of a reaction and the activation energy from a graph.
- Know that a catalyst lowers the activation energy needed to form an activated complex.
- A catalyst can be used in a reaction to form an intermediate, but is released in subsequent steps of a mechanism, therefore never consumed in the reation. A catalyst used in such a way is a homogeneous (same state of matter as reactant) catalyst. The intermediate is a substance that is later consumed in a later step of the mechanism.
- A heterogeneous catalyst (different state of matter than the reactants) usually controls the orientation of one of the reactants. This means that fewer collisions between molecules will result in the correct orientation for the reaction to occur. Fewer collisions needed for the reaction to occur results in a lower activation energy.
- For a reaction to occur, the particles must strike with: (1) sufficient energy and (2) the correct orientation.
- Know the theories behind why reaction rates change. A change in temperature changes the rate of collisions. Fewer collisions means a slower rate of reaction. More collisions means a faster rate of reaction. Temperature changes alter the value of the rate constant for a reaction.
- Changing the concentrations of reactants changes rate without changing the rate constant. A higher concentration means more collisions will happen between reactants, thus the rate of the reaction will go up. The reverse is true for a decrease in concentration. Fewer molecules, less chance of having a collision. The analogy of being in the hall at school during passing period (high probablility of bumping into someone) creating a high reaction rate OR being in the hall at school during a period (very low probability of bumping into someone) creating a low reaction rate is good when thinking about this concept.
- Be able to solve for any variable from any integrated rate law. (zero, 1st, 2nd)
- A change in concentration of a zero order reactant has NO EFFECT on the rate of reaction. A doubling of a concentration of a 1st order reaction doubles the rate of reaction, the halfing of a concentration of a 1st order reaction cuts the rate in half, etc. Doubling the concentration of a second order reactant will quadruple the reaction rate. Halfing the concentration of a second order reactant will cut the rate by 1/4th.
- The sum of all of the reactants and products of all of the steps of a mechanism must equal the balanced equation. Intermediates and catalysts must cancel out.
- The rate determining step of a mechanism is the slow step in a mechanism. The rate determining step must contain all or parts of the reactants that are part of the rate law.
- Reactants in a fast step can be part of the rate law if they are DIRECTLY responsible for making an intermediate that is consumed in a rate determining (slow) step of a mechanism.
- Know the concept of how the Arrhenius equation can be used to determine the activation energy of a reaction.
Remember, tomorrow morning (Friday, February 11th, 2011) is a review day for the AP Chemistry exam. Also, the rate lab performed today (Thursday, February 10th, 2011) will be a formal lab write-up that will be due Monday, February 14th, 2011. See me before the test if you have any final questions.
Saturday, January 22, 2011
Chapter 13 Test
- All nitrate ionic compounds are soluble.
- All alkali ion (group 1A of the periodic table) ionic compounds are soluble.
- All ammonium (NH4^+1) ionic compounds are soluble.
- All chloride (Cl^-1), bromide (Br^-1), and iodide (I^-1) ionic compounds are soluble except for silver ion (Ag^+1) and lead (II) ion (Pb^+2) compounds.
- All acetate (C2H3O2^-1 OR CH3COO^-1) ionic compounds are soluble except for silver ion (Ag^+1) and mercury (I) ion (Hg^+1) compounds.