## Q3: Week 10  – 4/11 – 4/14

4/11 – Monday – B Day – 2, 3b/4 Lab

Main focus –

a) To review the concepts of the Arrhenius equation.

b) To introduce equilibrium in Acid/Base applications.

Period 2, 3/4:

1. Review this weekends Forms:
This weekends forms keys:
Rate Law Multiple Choice Form:

Half-Life Form:

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

3. Iodine clock reaction –

a) rate law constant (k) dependent upon Temperature and Ea
A catalyst lowers the activation energy (Ea) by moving the reaction in a different pathway or MECHANISM that normally the reactants would not travel through.  A lowered activation energy will increase the rate of reaction as more energy is available for effective collisions!

I am using an iodine ion catalyst that creates the mechanism listed in last nights homework for the hydrogen peroxide decomposition.

4.  Acid Base Basics Lesson (again!!!)
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    <->   H+       +      OH
H2O   +   H2O  <->  H3O+   +      OH

B) Acid Base Definitions (Arrhenius,  Bronsted Lowry,  Lewis)
C) pH and pOH determination, Kw = [H+/H3O+] [OH] = 1 x 10-14
D) pKw =pH  + pOH
E) 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.
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

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TODAY’s 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/

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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.

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

Acid Base 6 2003A – grodski key .pdf                                                                                                       View Download

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4/12 – Tuesday – A Day – 2/3a Lab, 4

Main focus –

a) To derive the Ka x Kb = Kw formula

b) To solve for Gibbs free energy by volts through our final equation.

Period 2/3, 4 :

1. Collect Homework – Titration Graph

2. Review Free Response Homework

Acid Base 3 – 2007 .pdf

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

3. Buffered solutions again – lesson through the notes below – all review                                                                                                                                                                                                                                                                                                       4. TITRATION DEMO’s:

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.
Titrating unknown concentration of 40.0 ml of H3PO4 with 100 ml of 0.10 M NaOH
In this titration we are using a polyprotic acid that has multiple Ka’s.
Logger Pro File for the Today’s Titration:
Table 6 PolyP Acid SB Titration good example.cmbl

Phosphoric Acid Titration Curve.pdf

Na2CO3 TitrationVirtual Lab – Demo 2:

40.0 ml of an unknown concentration of sodium carbonate is titrated with 30.0 ml of 1.00 M hydrochloric acid using a drop counter. The drop counter was calibrated at 20.3 drops per milliliter.
Titrating unknown concentration of 40.0 ml of Na2CO3 with a total 50 ml of 1.0M HCl
In this titration we are using a salt (ionic compound) Na2CO3 that is basic due to the anion CO3-2 having the ability to accept 2 protons (H+) thus the salt is dibasic.

Sodium Carbonate Titration Curve.pdf

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!

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TODAY’s NOTES:

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.

2.  Has 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).

Electrolytic Cells again!

From your Reference Tables:   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

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4/12 – Tuesday Homework: –

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

Acid Base 4 – 2005.pdf

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

Lecture on Acid Base 4 worksheet:

2. Electrochemistry Review –

Complete the following Free Response questions and review with the keys.

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

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

3. Complete the Acid/Base Form: (Previously released MC questions)

3 : Acid Base MC Form:

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4/13 – Wednesday – B Day – 2, 3b/4 Lab

Main focus –

a) To Begin Practice Test 7 – Part 1 – Multiple Choice section

*Lets take a look at the new reference tables.

Period 2:

35 minutes – Start Practice Test 7 – part 1 – MC – (75 MINUTES TOTAL)

1. Practice test AP Exam 7 – – 4th quarter Exam Grade – completing a full test for AP Grade of 1 – 5 .

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.

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

Period 3/4:

1. Start and complete Practice Test 7 – Part 1 – MC ( 75 minutes total)
2. Enter you answers with the form below.
3. Gain access to the Part 1 resource.

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4/13 – Wednesday Homework: –

Spectrophotometry FRQ practice – The AP Loves the spectrophotometry labs                                                                                                                                                                                                                                                             1. Please read through my notes below on how we completed 2 labs using the spectrophotometer.

2. Please complete the older part 2 questions that will review spectrophotometry labs.  There is no need to time                  yourself here.  I would like you to think about these questions and how they relate to the 2 spectrophotometry             labs that we completed.

In your reference tables:   A = abc

A = absorbance (- log of Transmittance)
Transmittance is the ratio of light reaching the detector to the light being emitted from the source.
The more that light is absorbed, the lower the light hitting the detector and thus the lower the
Transmittance.  This number will be less than one so we perform a .-log on its value to make it positive                                                        and a larger that 1 value and when we do this it becomes Absorbance

a = absorptivity constant (proportionality constant like R for gas laws or k for rate law)
b = path length (distance that the source light travels to the detector)
c = Concentration of colored chemical (Molarity chromophore or crystal field)

So based on A = abc the Absorbance is dependent upon the concentration of the colored chemical and the distance that the light has to travel through the cuvette.
-Percent mass of copper in brass:
Brass shot (alloy of Cu & Zinc) was dissolved with nitric acid. The solution was blue due to the crystal field splitting of Cu+2 with water. The Zinc in the alloy (Brass) also dissolved but it has no color.  We performed a serial dilution of a known concentration of [ Cu+2 ] and we put theses known concentrations in cuvettes and determined the Absorbance of each known concentration (after we set the spectrophotometer to absorb a certain wavelength of the Cu+2  solution called the peak wavelength).We plotted a Beer’s Law graph (linear relationship of diluted Cu+2 ] with lowered Absorbance. We (logger pro) plotted the graph of measured absorbance with Known concentration (we used the keep button to add the concentrations for each measures absorbance of the known concentrations).

We made a line of best fit and we used the slope and y intercept data to help with the next part

We then took the dissolved Cu+2  from the brass shot and made a 100.0ml solution.  We put that in a cuvette and measured the absorbance. Using this absorbance, and the Beer’s Law graph (utilizing linear algebra, y = mx +b from the regression line of the graph) we determined the concentration of the dissolved Cu+2 ].  We used that to calculate moles and then percent by mass.

Love this lab!
Determining the order of  reaction (with respect to Crystal Violet) of reaction (CV) with OH_
We did not get to this lab this year but please read its procedure. It’s very simple but has some important lab concepts. In this lab we placed a known solution of [CV] with is a chemical indicator (or a chromophore!) with a known concentration of OH into a cuvette in a spectrophotometer that plotted the Absorbance decreases in real time for 30 seconds. We identified the order of the CV while keeping the [OH] constant.  (The concentration of the OH was so much larger than the dilute CV that we also assumed that its concentration did not change significantly.) From the absorbance data we collected, we assumed that the concentration of the CV was proportionate to the changes of the Absorbance. (A = abc and because a and b are fixed values for the experiment we used the concept that A is proportionate to c which is concentration) Remember that over time the cuvette changes from a purple color to clear. How much it turned clear was an indication of the RATE of the reaction!  Using the absorbance values collected for each time interval as the proportionate concentration we changed the absorbance values to ln [A] and 1/[A) respectively and then made a new graph of each.  The graph that was the most linear determined the rate order with respect for CV.

Ok I love this lab too

Please complete these FRQ’s below and review with the key and your the video below:
Also Feel free to look through. the Spectrophotometry presentation below:

2006 Free response question 5 – spec.pdf

2006 Free response scoring guidelines ques 5 spec.pdf

Spectrophotometry AP question 2 – 2003 .pdf

Spectrophotometry AP question 2 – 2003 AP Key.pdf

General Spectrophotometry Review and question 2006 review:

General Spectrophotometry Review and question 2003 review:

Spectrophotometry Presentation:

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4/14 – Thursday – A Day – 2/3a Lab, 4

Main focus –

a) To complete the Part 1 of the Practice test 3.

b) To Start the Part 2 of the Practice test 3.

Period 2/3:

1. Complete the Part 1 (45 minutes)

2. Enter you answers with the form below.
3. Gain access to the Part 1 resource.
4. Start the Practice Test FRQ.

Period 4:

1. Enter your answers from the Part 1 into the form below and await your email back.  See what pace you are on.

2. Link to the shared resource (Practice Test 7 – Multiple Choice Review) .

3. Start the Practice Test 7 – FRQ

Practice Test 7 MC Form:

C

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4/14 – Spring Break Homework: –

OPTIONAL:

I will be at the School – my room (230) to GIVE A PRACTICE TEST in its entirety                                        FROM 9:30 am – 11:45 am.

I will grade it and score it as a percentage grade (scaled) and an AP Grade and have it ready for pickup the next day

NOT OPTIONAL:

1. REVIEW YOUR PRACTICE TEST 7 MULTIPLE CHOICE (PART ONE) THAT WAS COMPLETED IN SCHOOL FRIDAY OR THURSDAY WITH THE PRACTICE TEST 7 MULTIPLE CHOICE VIDEO.  I have emailed you a private link to view the video. YOU MUST DO THIS BEFORE YOU COMPLETE THE NEXT PART OF YOUR HOMEWORK OR participate in Tuesdays Practice test.

2. Please Complete the Practice Test 5 Part 1 (MC) and Part 2 (FRQ) that was given out Thursday.

a) Enter you answers to the MC section to the form below. It is on auto reply.

b)  Please review your MC questions to Practice Test 5 with the Review video that I will link to you in your gmails.

c) Review your Part 2 (FRQ) with the scoring key and Practice 5 FRQ video that I will link to you.

d)  Score your entire test to see what AP Grade you earned using the scoring worksheet. Practice Test 5 MC Form: Form for the take – home practice test

C

Practice Test 6 MC Form: Spring Break Test (classroom) –

Part 1 and Part 2 resources have been linked to your private gmail accounts.