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## Q3 : Week 9 – 4/3 – 4/6

If you have not already please join the REMIND for this class.

Make Sure Powerschool grades are accurate!!

_____________________________________________________                                                                  Jump toMonday Homework     4/3 – Monday – A Day – 2/3a Lab, 4

Main focus –

a) To Review the Integrated Rate law Equations with examples

b) To review the half-life concepts in Rate Law and all graphs

c) To review the concepts of the Arrhenius equation and values that affect                                   the Rate law constant.

b) To introduce equilibrium in Acid/Base applications.

Period 2/3a,4:

1. Quick Review of Integrated Law equations with selected problems

Kinetics 1 – Rate Law Key.pdf

Kinetics 2 – rate law KEYp .pdf

KSP FRQ TEST 2001 – Grodski & College Board Keys.pdf

2005B 2009 Kinetics quiz 2013 with college board and grodski keys.pdf                                       View Download

2022 AP Chem FRQ:

2. Review of Half Life Concepts – (use the Rate Law Presentation)

3.  Review of weekend’s Forms: ( I will not have time to do this is class)

4.  Review of the weekends Free Response Homework –
Kinetics (rate law),Ksp Free Response, 2004B, 2009 FRQ .pdf

5. Iodine clock reaction –

– Looked at the mechanism in the Rate Law presentation to explain the “clock” reaction.

– Viewed the Iodine clock Test video and observed the first 2 tests to prove the iodate ion has an order of 1.

– Used that to create a demo with synchronous clocking with different concentrations of iodate.

Iodine Clock Reaction and Lab:

Iodine Clock Demo.  Knowing what the order of the Iodate ion from the lab above is how did set this up?

Rate Law Presentation:

_____________________

TODAY’s NOTES:  from Friday

Reference Table:
 This is the first order integrated formula.         This is the second order integrated formula. There is no zero order integrated formulas on the AP. This is the Arrhenius equation that ties activation energy ( Ea) and temperature to the rate law constant.

It is usually very helpful to utilize these equations when you arrange them into a linear equation.

1st Order:                   X   –>  Y

R = K [X]1

ln[A]t – ln[A]0 =  -kt

ln[A]t  =   -kt   +    ln[A]0
y     =    mx  +      b
(negative slope)

2nd Order:                X  +   X  –>  X2

–       =  kt
[A]t    [A]0

=  kt   +    1
[A]t                [A]0

y     = mx  +   B
(positive slope)

_____________________

TODAY’s NOTES:

Understanding the factors that influence the rate law constant:

The Arrhenius equation:

K = Ae-Ea/RT

Rate LAW Constant = Frequency factor * e (Activation Energy/ Gas Law Constant * Kelvins)

– The Value of the rate law constant increases as Temperature increases (smaller negative exponent).

– The Value of the rate law constant increases as Activation Energy decreases (smaller negative exponent).

THE Value of K is thus is INDEPENDENT OF THE CONCENTRATION OF THE reactants. This must be the case because we can solve for it using ANY data from the same rate law equation if temperature stays constant!

NOW if remove the ex component of the equation we get the following:

ln K =  lnA -Ea/RT

or

ln K =  -Ea/R+  lnA

_____________________

ACID/BASE NOTES:

You must know how to complete all 5 Points on a titration curve. Every single Acid Base problem is based on these 5 points.

Guess what? I have written some notes on these points : http://mrgrodskichemistry.com/ap-chem-acid-base-titration-notes/

But Before you go to that site please try to remember the skills we learned in November.

This is a list of the Acid & Base Basics*
*. WE DID MOST OF THIS UNIT IN NOVEMBER so all we have left is the equilibrium (ICE tables) part of the unit.

A)  Auto Ionization of water:                     H2O (l)        H+(aq)          +      OH–  (aq)

H2O (l)  +   H2O (l)        H3O+(aq)        +      OH(aq)

B)  Acid Base Definitions (Arrhenius,  Bronsted Lowry,  Lewis)

C) pH and pOH determination, Keq = Kw = [H+/H3O+] [OH] = 1 x 10-14

D)                                                              pKw  =       pH    +   pOH

E)   *NEW                                                      Kw = Ka x Kb   or   pKw = pKa  +  pKb

F) Strength of Acids = Ka – acid dissociation constant (equilibrium constant for acid conjugate base                               equilibria)

G) Strong Bases

5.  Strong Acid / Base Titration:

a) Strong Acid ,  Strong Base – No equilibrium here
25.0 ml of HCl is titrated with 0.10 M NaOH. 2.8 second = 1 ml of base added.

HCl (aq)   +   NaOH(aq)  —> NaCl (aq)  +   HOH (l)

Net Ion for Strong Acid / Strong    H+(aq)  +   OH (aq)   —-> HOH (l)                                                                                                   Base ONLY:

Strong Acid Strong Base Titration Graph.pdf

6.  Weak Acid / Base Titration:

b) Weak Acid, Strong Base – Equilibrium is need to complete the points we did not complete in                                                                                November. We will usually have a weak component in a titration!
25.0 ml of acetic acid is titrated with 0.10 M NaOH. 2.8 seconds = 1 ml of base added.

HC2H3O2      +       OH      —–>    _________   +   H2O
?M, 25 ml              0.10M

Weak Acid (Acetic- 25ml) Strong Base (0.12M) .pdf

4/3 – Monday’s Homework: –

1. You can use any piece of graph paper or a blank piece of paper if you estimate carefully and will be asked to sketch the titration curve with limited data.  You will be calculating 5 points in a weak acid and strong base titration. I have provided a blank below.

These points are:

1: initial ph of acid in a beaker
2. max buffering position (and volume).
3. a point between max buffering position and equivalence point.
4. Equivalence point.
5. final ph of solution (after a known amount of base has been added).

These 5 points need to be placed on the graph and you will sketch the graph with the supporting calculations for the 5 positions above.

Here is an example 1:
Blank:

Acid Base Blank graph for Titration Prediction 1617 p.pdf

Please watch the video with me to complete all points!  If you hand this completed
you will get a 100 on your first test in the 4th quarter!  Due tomorrow in class!

Lecture on how to complete this example 1 :

In these next 2 Free Response Questions Please THINK ABOUT WHAT POINTS IN THE GRAPH ARE WE WORKING ON!!!

2.  Please complete the Free Response Question below and Review the key or video posted below:

Acid Base 3 – 2007 .pdf

Acid Base 3 – 2007 Key grodski & college board keys p.pdf

Lecture that reviews Acid Base 3 -2007.pdf  – specifically the last question using the Henderson Hasselbalch Equation (buffer equation).

3.  Please complete the following FRQ and review with the key

Acid Base 6 – 2003A – worksheet.pdf

Acid Base 6 2003A – key AP Central .pdf

4/5 – Tuesday – B Day – 2, 3b/4 Lab

Main focus –

a) To review the Arrhenius Equation

b) To apply Equilibrium concepts to Acid Base Reactions

c) To derive the Ka x Kb = Kw formula

Period 2:

1.  Explain Arrhenius Equation using notes.  Fix my blunder from yesterday.
2Iodine Clock – again with test, what would be a 2nd order? temperature?

3Acid Base continues/buffers-    Ka x Kb = Kw

4. Review Free Response Homework

Acid Base 3 – 2007 .pdf

Acid Base 3 – 2007 Key grodski & college board keys p.pdf

Period 3b/4:

1.  Explain Arrhenius Equation using notes.  Fix my blunder from yesterday.
2.  review a few Rate Law questions that were assigned with value of K or integrated
3.  Iodine Clock – test for 1st order, what would 2nd order look like? temperature
4Acid Base continues/buffers–    Ka x Kb  = Kw                                                                                                5.  Review Free Response Homework

_____________________

TODAY’s NOTES A:

Understanding the factors that influence the rate law constant:

The Arrhenius equation:

K = Ae-Ea/RT

Rate LAW Constant = Frequency factor * e (Activation Energy/ Gas Law Constant * Kelvins)

– The Value of the rate law constant increases as Temperature increases (smaller negative exponent).

– The Value of the rate law constant increases as Activation Energy decreases (smaller negative exponent).

THE Value of K is thus is INDEPENDENT OF THE CONCENTRATION OF THE reactants. This must be the case because we can solve for it using ANY data from the same rate law equation if temperature stays constant!

NOW if remove the ex component of the equation we get the following:

ln K =  lnA -Ea/RT

or

ln K =  -Ea/R+  lnA

_____________________

TODAY’s NOTES B:

Buffer discussion –   What are buffer solutions?

A buffer solution is one where a conjugate acid and a conjugate base ARE both present in a solution AND they resist pH changes in the solution.

*Remember that conjugate acid/base pairs are the SAME chemical plus or minus a proton (H+).

Having an acid (conjugate acid) and a base (conjugate base) will “buffer” OR maintain the pH of a solution BECAUSE the addition of an acid to “buffered solution” will neutralize the addition of an acid as the conjugate base will react with the added acid AND prevent the pH from changing.  If a base is added to a “buffered” solution the conjugate acid will neutralize added base AND prevent the pH from changing.

Lets look at ammonia NH3 (weak base) in water:

reaction 1:                              NH3            +       H2O    <—–>       NH4+          +      OH
conjugate base                          conjugate acid

Because NH3 is  weak base it will reach equilibrium and both conjugate base  and  conjugate acid  will be present in solution.  A conjugate acid /base pair will never neutralize each other because their ability to donate a proton (acid) and ability to accept a proton (base) are inversely related to each other in terms of Kw (1 x 10-14).   Lets revisit a Derivation that we have done:

Lets start with the NH3 in aqueous solution and view its corresponding equilibrium expression

Now lets view the the conjugate acid,  NH4in aqueous solution and its equilibrium expression

These 2 chemical reaction are competing when they are in water, remember that water is the chemical that is in largest quantity and that these 2 reaction are really the forward and reverse reaction of reaction 1 above.

So lets add them together to get a perspective of how the forward reaction and the reverse are related:

Wait?  Isn’t this the auto ionization of water that has a Kw = 1 x 10-14  ???

YES!!! And if we are adding the reactions we are also multiplying the Ka and Kb:

So what this means that as the conjugate acid strength increases its conjugate base strength decreases!! The ability of conjugate acid to act like an acid is inversely related to the ability of the conjugate base to act like a base.

“Stronger the conjugate acid (higher Ka) the weaker its conjugate base (lower Kb)!”
“Weaker the conjugate acid the stronger the conjugate base!”

Given the following Ka’s for 2 weak acids, which conjugate base is the strongest?

HNO2   —–>    H+    +    NO2–        Ka = 4.6 x 10-4

HF     —–>    H+    +       F           Ka = 3.5 x 10-4

answer: F–    – This means that F–   ionize water (make into OH) better than NO2– because it is a stronger base than NO2– because its conjugate acid (HF) is weaker than the conjugate acid (HNO2) of NO2 .

So now it should be clear why A conjugate acid /base pair will never neutralize each other because their ability to donate a proton (acid) and ability to accept a proton (base) are inversely related to each other in terms of Kw (1 x 10-14)!

Because the conjugate acid and base do not react with each other they are available to react with both acid or base that are added to a buffer solution. As long as there is enough of the conjugate acid or the conjugate base the buffer solution will not change it pH if an acid or a base is added.  That is what buffers do!!!!!

So if I have the conjugate acid base pair from reaction 1 above the conjugate base will neutralize the acid that is added:

NH3   +   Acid added to buffered solution     —–>    NH4 +   +    conjugate base of acid added.

And if I add a base to the buffered solution from reaction 1 above, the the conjugate acid will neutralize the base added :

NH4 +  Base added to buffered solution   —–>   NH3    +  conjugate acid of base added.

So a base or an acid added to this buffered solution will maintain this pH!

We calibrate our pH probes with a buffered solution at a pH of 7.  I use the following:
The conjugate acid is delivered by the salt:

potassium phosphate monobasic:

## KH2PO4  —->  K+  +  H2PO4-1

The conjugate base is delivered by the salt:

sodium phosphate Dibasic:

Na2HPO —–>   2Na+  +  HPO4-2

This product contains a mixture of these 2 salts in a plastic capsule that is to be dissolved in 100 ml of water.

## Lets look at some titration curves of weak acids that we have titrated or will titrate:

-Notice that the points labeled the “Buffered Region” is where a Buffer solution is being made as this is the place in the graphs that both the conjugate acid AND conjugate base are both present.

## –Notice that the half equivalent points determine the pKa of the acid WHICH also tells us the maximum buffer pH or the median pH range of that particular buffer.  We determine appropriate based on the pKa’s or Ka’s of the conjugate acid!  This means if I need a buffer to maintain a pH at 9 I would use a buffer solution with approximately equal amount of  NH4+ and NH3 .

If we titrate “partially” meaning we add enough base to drive the reaction forward enough to produce some conjugate base but do not add enough to reach the equivalence point we create a buffer and we are in the “Buffer Region”!

## 1. Equal amounts of Conjugate Pairs–This would allow for a greater range buffering ability.                                                                                       Think about the half- equivalence point.  At that position the  solution can equally buffer                                                                                        an acidic or basic additions.

2Has the largest concentration of Conjugate Pairs This would allow
the buffer to RESIST pH changes the most. The greater the moles of the conjugate
pairs,  the greater amount of acid or base the buffer solution can neutralize before it
runs out of one the conjugate pairs and pH will drastically change.

3.  The pKa of the conjugate acid approximates the pH region you
want to buffer. – The half equivalence point will be the that region!!!

## So we make a buffered solution 3 ways:

1:  Making an aqueous solution of a conjugate acid or conjugate base .

The reaction will go to equilibrium quick making the the other conjugate partner.

Step 1 is the Not a Preferred method because the reaction will not make much of the other conjugate ion since these                           acid and base are weak.

2:  Partial titration – Stoichiometrically drive the reaction to a point that is below the equivalence point so that there is a                                    significant amount of both the conjugate base and conjugate acid.

*3:  Deliver the conjugate base or conjugate acid via a salt!(students often do not recognize
this way as they forget to look for the ion in the formula that is the conjugate partner).

Review Of Buffers and Henderson – Hasselbach – POINT 2 and 3!

4/4 – Tuesday’s Homework: –

1. Please complete the FRQ below and review with the key and or the video below:

I am collecting This Tomorrow!

Acid Base 4 – 2005.pdf

Acid Base 4 – 2005 Key grodski and college board keys p.pdf

Lecture on Acid Base 4 worksheet:

2.  Study for an in class Acid Base Predict a Titration Curve. All 5 points. Review last nights lecture on how to calculate each of the 5 points and previous homework assignment. You may want to study using the class notes on Acid and Base.  What would be a great practice would be to try another example

In the titration 25.0 ml of HC2H3O2 was titrated with 0.10M NaOH:

weak acid                                    strong base

Here is the Graph of this Titration:

USE my notes to Review the calculations that Review How to Verify Each Point.

You will be calculating 5 points in a weak acid and strong base titration. I have provided a blank below.

These points are:

1: initial ph of acid in a beaker
2. max buffering position (and volume).
3. a point between max buffering position and equivalence point.
4. Equivalence point.
5. final ph of solution (after a known amount of base has been added).

OH and you will have 20 minutes to complete tomorrow..

4/5 – Wednesday – A Day – 2/3a Lab, 4

Main focus –

a) To predict 5 points on a titration curve.

b) To identify when a buffer solution has been created in acid base equilibrium                          problems

c) To predict strength of acid from molecular structure.

Period 2/3a:

1. Review of last night’s homework frq that added a salt of the conjugate base to create a buffer.

Acid Base 4 – 2005 Key grodski and college board keys p.pdf

Collect!

2. Review weak base / strong base Titration FRQ that we completed for Monday homework:

3. Polyprotic acid Titration:  Multiple Ka’s:

Strong Base – weak Polyprotic Acid Titration (H3PO4)
Write up requirements: 40.0ml of H3PO4 is titrated with 100 ml of 0.10 M NaOH
a) determine the concentration of the Phosphoric acid solution.
b) determine the Ka1 of  H3PO4,  , Ka2 of H2PO4, Ka3 of HPO4-2
c) Write All predominant reaction in different parts of the curve.
1) Mark the percent that the chemical exists at each equivalent point.
d) Mark the equivalence points and max buffer positions on graph.
e) validate initial and final pH.

4. Complete the ACID/BASE TEST

Period 4:

1. Review of last night’s homework frq that added a salt of the conjugate base to create a buffer.

Acid Base 4 – 2005 Key grodski and college board keys p.pdf

Collect!

2. Review weak base / strong base Titration FRQ that we completed for Monday homework:

3. Complete the ACID/BASE TEST

H3PO4 Titration Virtual Lab – Demo 1:

40.0 ml of an unknown concentration of phosphoric acid was titration with 30.0 ml of 0.10 M sodium hydroxide using a pH meter and a drop-counter. The drop-counter was calibrated at 20.3 drops per milliliter. 3 drops of Phenolphthalein was also added.

Acid / Base Final Presentation:

4/5 – Wednesday Homework:

I am holding a review tomorrow with food. Please RSVP  (email)today or tonight. I will review until 5:00 pm tomorrow.

1.  Please complete the take-home test which is due tomorrow.  It is another titration graph prediction BUT this time it is a weak base / strong acid titration.  Show all work for each point!

2. Please complete the Form below. Yes you have multiple submissions.

3. Please complete the H3PO4 Titration for period 2/3A:

Strong Base – weak Polyprotic Acid Titration (H3PO4)
Write up requirements: 40.0ml of H3PO4 is titrated with 100 ml of 0.10 M NaOH
a) determine the concentration of the Phosphoric acid solution.
b) determine the Ka1 of  H3PO4,  , Ka2 of H2PO4, Ka3 of HPO4-2
c) Write All predominant reaction in different parts of the curve.
1) Mark the percent that the chemical exists at each equivalent point.
d) Mark the equivalence points and max buffer positions on graph.
e) validate initial and final pH.

How to complete the Phophoric Acid Titration –

3 : Acid Base MC Form:

4/6 – Thursday – B Day – 2, 3b/4 Lab

Main focus –

a) To determine the strength of acids.

b) To study the multiple Ka’s of a polyprotic acid

c) To determine the △H of  redox reaction using the Eº Cell (The Last formula!)

Period 2:

1.  Collect the Take-Home Acid Base Prediction test 2 – Give out key.

2.  Complete H3PO4 Titration:

a) calculate the percent ionization of each acid – A type of point 1 question.

3.  Strength of acid lesson –

4. Voltaic cells – Review with thermodynamics

2010 AP voltaic cell question with scoring guide.pdf – I went over this one!

I introduced the last formula that connected volts from electrochemistry to delta G!

If you want to see how it might appear in electrolytic cells:

Electrolytic cells
2007 – Electrolytic cell question with scoring guide.pdf –

Electrolytic cells
1) electrolysis of fused salts
2) electroplating – calculating with amperage
Faraday’s constant, New delta G Formula that ties voltage with Delta G.
WE HAVE JUST COMPLETED THE COURSE!

_____________________

TODAY’s NOTES:

Electrolytic Cells again!

 Electrolytic – (electroplating/electrolysis)Non-spontaneousvolts = negative ∆G = POSITIVE Endothermic:  Electricity  +  Reactants –> ProductsAnode = Place of oxidationCathode = Place of reductionElectron flow from Anode to Cathode Voltaic/Galvanic Cell – Spontaneousvolts = positive ∆G = NEGATIVE Exothermic:    Reactants –> Products   +   ElectricityAnode = place of oxidationCathode = Place of reductionElectron flow from Anode to Cathode

How to complete the Phophoric Acid Titration –

H3PO4 Titration Virtual Lab – Demo 1:

40.0 ml of an unknown concentration of phosphoric acid was titration with 30.0 ml of 0.10 M sodium hydroxide using a pH meter and a drop-counter. The drop-counter was calibrated at 20.3 drops per milliliter. 3 drops of Phenolphthalein was also added.

4/6 – Spring Break Homework: –

Last Nights Take-Home Key:

1.  Please complete the Acid/Base Phosphoric Acid TITRATION. I have posted a video above to help.

2.  Please Review the Concepts Redox and Electrochemistry with the complete FRQ’s below:

Voltaic cells

2010 AP voltaic cell question with scoring guide.pdf –

Electrolytic cells

2007 – Electrolytic cell question with scoring guide.pdf –

2013 – Electrochemistry Question with scoring guide.pdf

Additional Acid/Base practice if you cannot get enough!

2002 – Acid Base Part 2 question with Grodski and College B keys.pdf

3. Please complete Practice Exam 2.  THIS is a previously released exam. IT WILL NOT BE GRADED but you need to complete it and review it for homework.  The EXAM has 2 parts, please try to keep track of your pace:

a) Complete Multiple Choice and FRQ with the time all.

Multiple Choice:

Part 1: Multiple Choice (normally 60 questions) – 90 minutes

90 minutes for 60 questions = 90 minutes/60 questions = 1.5 minutes per question is YOUR Pace!

Practice test 2 has 50 questions = 50 x 1.5 = 75 minutes needed to complete part 1.
They keep 10 MC questions private.

Free Response:  – 105 minutes needed to complete part 2

Part 2: FRQ

We will now use the newer AP Chemistry Reference table!
2013 AP Chemistry reference tables.pdf

b) Enter YOUR MULTIPLE CHOICE QUESTIONS into the Form below that is on auto-reply.

d) Calculate the AP Grade of 1 – 5 based on the worksheet given in front of the AP Scoring Guidelines.

e) Watch the Review Video for the MC and the FRQ. Use the time codes in the description of the video to move directly to the question you want the video (me) to review with you. THESE videos are private and you need to log into your private gmail accounts and youtube to view.

Practice Test 2MC Form:

Practice Test 2 – Multiple Choice Review

Practice Test 2 – FRQ Review

_______________________________                                                                                                                                                                                LAST YEARS Results on Same Test:

Practice Test 4 – 2016 Released Exam Class Data:  @2022 – 2023 Results

Practice 4 Test Data: In Class test from this week –  19 students reporting -(50 Mc questions)
mean = 56.8% ( 28.39/50 raw score)   SD = 6.93

This chart is the percent correct from this class for this test:
 1. 84 % 16. 87 % 31. 63% 46. 53 % 2. 90 % 17. 83 % 32. 79% 47. 42 % 3. 81 % 18. 74 % 33. 26% 48. 67 % 4. 90 % 19. 65 % 34. 61 % 49. 90 % 5. 42 % 20. 57 % 35. 37 % 50. 26 % 6. 42 % 21. 74 % 36. 53 % 7. 87 % 22. 61 % 37. 26 % 8. 84 % 23. 56 % 38. 42 % 9. 87 % 24. 56 % 39. 53 % 10. 87 % 25. 22% 40. 26 % 11. 52 % 26. 47 % 41. 58 % 12. 65 % 27. 32% 42. 42 % 13. 39 % 28. 79% 43. 63% 14. 57 % 29. 68 % 44. 79 % 15. 57 % 30. 84 % 45. 58 %

Multiple choice comparison – AS a class we have improved!!!!
 mean Sd Scale TEst 2-  (Spring Break) 49.8 % 8.06 69 TEst 4 – (Finished Thursday) 56.8 % 6.93 79 –
Class DATA For Practice Test 4 AP from our 29 AP Chemistry  STUDENTS reporting:
 AP Grade # of students Percent Globally in 2016(Mostly Juniors and Seniors!) 5 1 3.4 % 9.7 % 4 8 27.5 % 15.1 % 3 14 48.3 % 27.5 % 2 4 13.8 % 25.3 % 1 2 6.9 % 22.4 %

Average AP Score on this test :  3.06   –  Last July : 3.53

79 % of the class scored a 3 or better.
31 % of the class scored a 4 or 5.