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## Q3 : Week 10 – 4/8 – 4/12

#### Refresh this page every time you arrive!  I update this daily!!!

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

– Regarding Homework assignments – IF you cannot complete the assigned homework before school the next day you must email me otherwise all late assignments will now be zeros.  No late work be accepted UNLESS I GET AN EMAIL!!!

Make Sure Powerschool grades are accurate!!

_____________________________________________________                                                                  Jump toMonday Homework     4/8 – 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:

1. Review of Half Life / Arhennius Concepts – (use the Rate Law Presentation)

2. Quick Review of Integrated Law equations with selected problems and a classwork worksheet

Kinetics 1 – Rate Law Key.pdf
Kinetics 2 – rate law KEYp .pdf

Rate Law FRQ Examples – Connections .pdf

Rate Law FRQ Examples – Connections – College Board Key .pdf

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.  Hydrogen peroxide decomposition –> Elephants Toothpaste!

Period 4:

1. – 4 above.

Rate Law Presentation:

_____________________

TODAY’s NOTES:

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. This the Half – Life Formula for 1st order! They only give you this one!

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
R = K [X]2

–       =  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

Half Life of 0, 1st, and 2nd Order Reactions:

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 this next FRQ 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).

________________________________________________________________________                                Jump toTuesday Homework / TOP  4/9 – Tuesday – B Day – 2, 3b/4 Lab

Main focus –

a) To apply Equilibrium concepts to Acid Base Reactions

b) To derive the Ka x Kb = Kw formula

Period 2:

1.  Point 1 – Equilibrium acid base problems

2.  Point 3 – Acid Base continues/buffers-    Ka x Kb = Kw

3. Review Free Response Homework

Acid Base 3 – 2007 .pdf

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

4.  Lab 28 – Titration of a strong base with a polyprotic acid
Period 3b/4:

1.  Complete the classwork of Rate Law Connections:

Rate Law FRQ Examples – Connections .pdf

Rate Law FRQ Examples – Connections – College Board Key .pdf

2.  Point 1 – Equilibrium acid base problems

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

4. Review Free Response Homework

5.  Lab 28 – Titration of a strong base with a polyprotic acid

______________________________________________________________                                                                                                                 Lab 28 – H3PO4 Titration with NaOH  – weak acid (polyprotic) and strong base

Titrant: 0.10 M NaOH

Analyte : 40.0 ml of H3PO4

Objectives: Complete on the graph
1. To determine the concentration of the of the weak acid.
2. To write the 3 net ion reactions for the titration.
3. To ⇔ the Ka1 of  H3PO4 , Ka2 of H2PO4, and Ka3 of HPO4-2for the three dissociation reactions.
4. Mark all half equivalence and equivalence points and provide the % of the acid and conjugate base that
exits in each position,
5.  To verify the
a) Initial pH: Equilibrium problem
b) Final pH
c) the first 2 equivalence points, volume (x) and pH (y): Equilibrium problem
d) The conjugate acid/base pair AND its ratio that will buffer an aqueous solution at a pH of 7.

Ka1:    7.1 x 10-3                H3PO4    +     H2O    ⇔    H2PO41-   +    H3O+

Ka2:   6.3 x 10-8               H2PO41-  +   H2O     ⇔    HPO42-   +     H3O+

Ka3:   4.5 x 10-14             HPO42-   +   H2O     ⇔     PO43-      +     H3O+

Video Example:
In this example below, 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. OUR Lab uses a different concentration that is Titration.

How to complete the Lab:

4/9 – 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).

Acid Base Titration Prediction Practice with Class Notes .pdf

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

____________________________________________________________________________                     Jump to: Wednesday Homework / top 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. Used the graph of the H3PO4 Titration

a) to derive the formula Ka x Kb = kw

b) to compare Ka’s and their Kb from their conjugate bases = buffers

c) to calculate the pH at equivalence –> the conjugate base/acid reacting with water

d) calculate original pH.

e) calculate percent ionization of each Ka

2. 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:

4. Complete the ACID/BASE TEST

Period 4:

Number 1 above:

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

______________________________________________________________                                                                                                                 Lab 28 – H3PO4 Titration with NaOH  – weak acid (polyprotic) and strong base

Titrant: 0.10 M NaOH

Analyte : 40.0 ml of H3PO4

Objectives: Complete on the graph
1. To determine the concentration of the of the weak acid.
2. To write the 3 net ion reactions for the titration.
3. To ⇔ the Ka1 of  H3PO4 , Ka2 of H2PO4, and Ka3 of HPO4-2for the three dissociation reactions.
4. Mark all half equivalence and equivalence points and provide the % of the acid and conjugate base that
exits in each position,
5.  To verify the
a) Initial pH: Equilibrium problem
b) Final pH
c) the first 2 equivalence points, volume (x) and pH (y): Equilibrium problem
d) The conjugate acid/base pair AND its ratio that will buffer an aqueous solution at a pH of 7.

Ka1:    7.1 x 10-3          H3PO4    +     H2O    ⇔    H2PO41-   +    H3O+    Kb1  = 1.30 x 10-12

Ka2:   6.3 x 10-8         H2PO41-  +   H2O     ⇔    HPO42-   +     H3O+     Kb2  = 1.59 x 10-7

Ka3:   4.5 x 10-14        HPO42-   +   H2O     ⇔     PO43-      +     H3O+     Kb3 = 22.2

Video Example:
In this example below, 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. OUR Lab uses a different concentration that is Titration.

How to complete the Lab:

_____________________

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!

1. Please complete Lab 27 – Rate law Lab and Lab 28 – Titration of H3PO4

2.  Please complete the following acid /base FRQ and review with the key.

TRY to Identify the points on the graph that the question is really asking about!!!!
Acid Base 6 – 2003A – worksheet.pdf

Acid Base 6 2003A – key AP Central .pdf

3.  Please study for the Titration graph prediction Test 2 – I gave you the test silly!!!!

________________________________________________________________________                           Jump to: Thursday Homework / top 4/6 – Thursday – B Day – 2, 3b/4 Lab

Main focus –

a) To plot a titration graph of a weak base titrated with a strong acid.

b) To titrate a weak dibasic base with a strong acid.

Period 2:

1.  Take the Acid Base Titration Prediction test 2.

2.  Lab 29 – Weak Base (Dibasic), Strong Acid Titration

Period 3b, 4

1.  Take the Acid Base Titration Prediction test 2.

2.  Strength of acids and Bases Lesson

3.  Lab 29 – Weak Base (Dibasic), Strong Acid Titration

Acid / Base Final Presentation:

______________________________________________________________                                                                                                                 Lab 29 – Na2CO3Titration with HCl  – weak base (dibasic) and strong acid

Titrant: 1.0 M  HCl

Analyte : 40.0 ml of Na2CO3

Objectives: Complete on the graph
1. To determine the concentration of the of the weak base
2. To write the 2 net ion reactions for the titration.
3. To ⇔ the Ka1 of  H2CO3, and  Ka2 of HCO3for the 2 dissociation reactions.
4. Mark all half equivalence and equivalence points and provide the % of the acid and conjugate base that
exits in each position,
5.  To verify the
a) Initial pH: Equilibrium problem
b) Final pH
c) the  2 equivalence points, volume (x) and pH (y): Equilibrium problem
d) The conjugate acid/base pair AND its ratio that will buffer an aqueous solution at a pH of 7.35 (blood).
I will help with this tomorrow as the video does not explain this step.

Ka1:    4.3 x 10-7          H2CO3    +     H2O    ⇔    HCO31-   +    H3O+    Kb1  = 7.69 x 10-3

Ka2:   4.8 x 10-11         HCO31-  +   H2O     ⇔      CO42-   +     H3O+     Kb2  = 6.29 x 10-8

Video Example:
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.

How to complete the Lab:

1. Please complete the Lab 28 – H3PO4 Titration – weak acid / strong base
DUE DATE Tomorrow, 4/12!

Please use the requirement re-posted below.

2. Please complete the Lab 29 – Na2CO3 Titration – Strong acid/ weak base
DUE DATE Tomorrow, 4/12!

Please use the requirement posted above.

3. Complete all Acid/base worksheets (FRQ’s) given out this week –
They will be collected as a grade.

Acid Base 3 – 2007 .pdf

Acid Base 4 – 2005.pdf

Acid Base 6 – 2003A – worksheet.pdf

Acid Base Titration Prediction Practice with Class Notes .pdf

______________________________________________________________                                                                                                                 Lab 28 – H3PO4 Titration with NaOH  – weak acid (polyprotic) and strong base

Titrant: 0.10 M NaOH

Analyte : 40.0 ml of H3PO4

Objectives: Complete on the graph
1. To determine the concentration of the of the weak acid.
2. To write the 3 net ion reactions for the titration.
3. To ⇔ the Ka1 of  H3PO4 , Ka2 of H2PO4, and Ka3 of HPO4-2for the three dissociation reactions.
4. Mark all half equivalence and equivalence points and provide the % of the acid and conjugate base that
exits in each position,
5.  To verify the
a) Initial pH: Equilibrium problem
b) Final pH
c) the first 2 equivalence points, volume (x) and pH (y): Equilibrium problem
Please verify with a calculation the 2nd equivalence point only.  The first equivalence would require                    solving for the quadratic equation as the value of x would not be small enough to approximate away.

d) The conjugate acid/base pair AND its ratio that will buffer an aqueous solution at a pH of 7.
I will help with this tomorrow as the video does not explain this step.

Ka1:    7.1 x 10-3          H3PO4    +     H2O    ⇔    H2PO41-   +    H3O+    Kb1  = 1.30 x 10-12

Ka2:   6.3 x 10-8         H2PO41-  +   H2O     ⇔    HPO42-   +     H3O+     Kb2  = 1.59 x 10-7

Ka3:   4.5 x 10-14        HPO42-   +   H2O     ⇔     PO43-      +     H3O+     Kb3 = 22.2

Video Example:
In this example below, 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. OUR Lab uses a different concentration that is Titration.

How to complete the Lab:

__________________________________________________________________                                   Jump to: Friday Homework / top       4/12 – Friday – A Day – 2/3a Lab, 4

Main focus –

a) To Review the Integrated Rate law Equations with examples                                                    b) To determine the order of a reaction from experimental evidence – Lab

Period 2/3a,4:

1. Strength of acids / bases lesson.

2. 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 = POSITIVEEndothermic:  Electricity  +  Reactants –> ProductsAnode = Place of oxidationCathode = Place of reductionElectron flow from Anode to Cathode Voltaic/Galvanic Cell – Spontaneousvolts = positive∆G = NEGATIVEExothermic:    Reactants –> Products   +   ElectricityAnode = place of oxidationCathode = Place of reductionElectron flow from Anode to Cathode

Yesterday’s Test key:                                                                                                         Weak Base Strong Acid Titration Curve Test Key.pdf                                                                                          View Download

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

Voltaic cells – I completed this in class for both classes

2010 AP voltaic cell question with scoring guide.pdf

Electrolytic cells – I completed this for only Period 2.

2007 – Electrolytic cell question with scoring guide.pdf –

2013 – Electrochemistry Question with scoring guide.pdf