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Archive – Q4 week 1 – 19-20

Week of 4/6 – 4/10 – REFRESH – Spring Break?
 
No class Thursday and Friday This week!
 
4/6 – Monday –  We will meet at 10:00 for extra help only!
                               Acid Base equilibrium Free response Test Tomorrow.
 
Link to today’s meeting:
 
Key to the Lab due today:
Weak Acid (Ammonium Chloride) Strong Base Titration Curve Video Key.pdf
View Download
 
Today’s assignment: 2nd virtual Titration Lab – Use the graph of the titration curve below:
Due Next Monday as Extra Credit – I will take the better grade out of the tomorrow’s Test!
NEXT YEAR – May want to revisit the blank titration quiz at this point. See bottom of the page and 2018 – 2019 
Weak Acid (KHP) Strong Base Titration Curve Video.pdf
View Download
1.  Lab:  Titration of weak Acid:  30.o ml KHC8H5O5 potassium hydrogen phthlalate (KHP) 
      with 0.06 M Ca(OH)2
 
    a) Verify the 5 points from the graph of the titration as per the guidelines for the lab below with the Google Doc  
         that will be shared with you.
                             
2.  Start the Acid Base Blank graph for Titration Prediction that will be practice for tomorrow test.
Today’s Titration: 
 
KHC8H5O5 (aq  +   Ca(OH)2 (aq)  —–>    H2(l)   +   ?
 
This is not a net ion reaction so you must determine the Net Ion Reaction.
 

 KHP or potassium hydrogen phthlalate was the acid we used to standardize the strong base earlier in the year.

We use KHP to standardize strong bases like NaOH or Ca(OH)2 because
1) KHP is not very reactive (stable)
2) KHP is a solid with a high molecular mass (204.22 g/mol)
3) The volume of water that the KHP is dissolved in does not need to
be precisely measured as long as we have moles of the acid

 

 Potassium hydrogen phthalate (KHP) lab requirements:
 
Weak Acid – Strong base Titration Lab activity- Due Next Monday, Extra credit.
25.o ml KHC8H4O4 potassium hydrogen phthalate (KHP) with 0.06 M Ca(OH)2
From the graph of the provided Titration Curve in your Google doc,  complete the following:
 
A) Write the Net Ion Reaction of the titration
 
B) Graphically identify the 5 points.
 
C) Write the net ion reaction for the titration.
D) Determine the concentration of the KHP
 
D)- Verify with calculations the following points in the graph (PLEASE NUMBER THEM in your doc!)
        1) initial pH
        2)  half equivalence point – Determine the Ka of the weak acid.
        3) a pH value  in between max buffering position and below the equivalence point @ 20 ml(this new!)
        4) pH at the equivalence point.
        5) final pH
E)- determine the appropriate indicator(s) if an ENDPOINT was to be reached from the list below.
     *Remember that an Equivalence Point (moles of H+ = moles of OH) is not the same as the Endpoint!
       The Endpoint is point in the titration that you estimate the equivalence point by using chemical acid base           indicators that are acid and conjugate bases themselves with Ka’s and pKa’s.  
 
Chemical acid/base indicators which are weak acid base equilibrium reaction are written the following way:
 
   HI          <—–>     H+        +           I
 
                                                           Conjugate acid                                           Conjugate Base
    
    If Litmus is the chemical acid/base indicator used the conjugate acid is RED while the Conjugate Base is Blue. These color changes are due to the changes in the chromophore that we have exhaustively covered.  The point here is that we can use this color change to stop a titration (ENDPOINT) If the color change occurs near the Equivalence point.   
 
When would litmus have a color change?  
 
It would occur when there is 50% conjugate acid and 50% conjugate base.  This would be at ITS HALF EQUIVALENCE POINT!   At the half equivalence point the pH = pKa!  So all you need to do is choose an appropriate chemical indicator that has a pKa at +1 or -1 of the Equivalence point.In the case of the litmus we would use this indicator if the titrations equivalence point was about 5.5 to 7.5.  
 
*Remember if the color change is occurring near the equivalence point it is occurring ON the asymptote and thus the volume determined by stopping the titration (Endpoint) will be very similar to the volume determined on the Equivalence Point. 
 
If you want more on this please view my notes on this topic
Please use this chart to determine best indicator to use with this titration.
Show all work on graph.

Acid Base Indicators

Lecture on how to complete the Lab:  
Play in Youtube and look under the video for the links (time code in blue) to take you where you need go in the video to complete. Most of you need points 3 and 5.
 
4/7 – Monday -Homework  Acid/Base equilibrium 10 point test Tuesday – 
 
GET OUTSIDE TODAY!!!!!!!!!!!!!!!!
 
1. Complete KHP Lab with video below if you did complete this weekend. 
Due Next Monday as Extra Credit – I will take the better grade out of the tomorrow’s Test!
 
 
2. Complete the practice acid /base equilibrium problem below:
 
Please spend some time in the evening with the practice tests.  Do not spend all day with these and miss the pleasant day outside!
 
A very good practice question for acid/base would be:
 
Acid Base 3 – 2007 .pdf
Acid Base 3 – 2007 Key grodski & college board keys p.pdf
View Download
 
Lecture that reviews Acid Base 3 -2007.pdf – (last question used equilibrium way to solve)
 
Lecture that reviews Acid Base 3 -2007.pdf  – specifically the last question using the Henderson Hasselbalch Equation (buffer equation).
End of Monday..

4/7 – Tuesday – Acid/Base equilibrium free response today (10 point question)
                            – Meet today at 9:45 to review homework question and then take Free Response Test.
I may have your Ammonium Chloride lab back to before class.
1: Homework review:
Acid Base 3 – 2007 .pdf
Acid Base 3 – 2007 Key grodski & college board keys p.pdf
View Download
 
2: Buffer Notes: I will review how to identify buffer questions: POINT 3
3.  10 point Acid Base Equilibrium Free Response Question:
        a) Think points on a titration curve!
        b) Channel Ksp!
4/7 – Tuesday – Homework:
 
1.  Please re- read the notes on Buffered solutions below.  I know that some of this is redundant but important. You need to identify when you are dealing with Point 3 (buffered solution!).
There are three ways that Point three is created in Acid/Base equilibrium problems.
 
2. Complete the Acid/Base Form: (Previously released MC questions)
 
3.  Please give yourself 25 minutes and complete the following worksheet.  Grade it with the AP Centrals key. (Notice that in this question we are just moving the other direction as a weak base is being titrated with a Strong Acid.)
 
Acid Base 6 – 2003 B.pdf
View Download
Acid Base 6 2003A – key AP Central .pdf
View Download
Acid/Base MC Form: (Its on Auto-Reply and you have 2 submissions)

Acid Base MC Form 1920

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  wil 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)!

 
” A conjugate 
 
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:

               KH2PO —->  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”!


The best Buffer solution has 

                                    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 .
 
                    It 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).
End of Tuesday..

4/8- Wednesday – We will meet Today at 10:30am – Last meeting of the week!
 
Link to today’s recorded Meeting:
 
We finished the course today!
1) Review the Test
Lecture on the Acid/Base Free Reponse Test:
2)  Virtual Lab – Demo 1:
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.
3) Virtual Lab – Demo 2:
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
Download
 
4) Electrochemistry wrap-up – 
Voltaic cells – Review with thermodynamics     
2010 AP voltaic cell question with scoring guide.pdf – I went over this one!
View Download
 
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 – I did not give this one out! 🙁
View Download
 
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.
 
From your Reference Tables:

Additional Electrochemistry practice:
 
2013 – Electrochemistry Question with scoring guide.pdf
View Download
 
2002 – Acid Base Part 2 question with Grodski and College B keys.pdf
View Download
 

 

 Electrolytic – (electroplating/electrolysis)
Non-spontaneous
volts = negative

 

∆G = POSITIVE


Endothermic:  Electricity  +  Reactants –> Products
Anode = Place of oxidation
Cathode = Place of reduction
Electron flow from Anode to Cathode

 Voltaic/Galvanic Cell –
Spontaneous
volts = positive

 

∆G = NEGATIVE


Exothermic:    Reactants –> Products   +   Electricity
Anode = place of oxidation
Cathode = Place of reduction
Electron flow from Anode to Cathode

4/8 – Wednesday Homework
 
1:  Complete the Part 2 section of the 2017 test that my students took a few years ago.
You will be given a link to a Google Doc to complete your work by uploading images into the doc just like our free response test.
Pretend that you have just sat for 90 minutes finishing the multiple choice section of the test that you did not have a calculator for.  You will be given a 10 minute break and then Part 2 (Free Response Section Begins). 
 
Part 1: Multiple Choice – 50 % of the test score
Part 2 : Free Response – 50 % of the test score
 
Now you will need to take out lined paper that will be provided for you on test day in either a booklet form or the lined paper to respond will be directly on the test in between each question.  You will not have any space to answer the questions in between the questions.  I have been making space in between these question for you in our AP tests but you need to practice writing your answers on a separate paper.  Make sure you label appropriately!!
 
Example: Question 1:
 
                    a) (i)  Your work
 
                         (ii)  Your silly work
 
Do not make it hard for the AP reader to FIND your answer and do not make it hard for them to Read Your answers (make it legible!!)
 
   a) Please give yourself 1 hour and 45 minutes to complete this section. You will have the use of      
        full reference tables and your graphing calculator!
 
   b) Once you complete this part 2 section please review to see how many point your earned WITH          the AP Central Key then Review with my Key to understand how complete each question.  I  
        wrote my key days after the AP questions were released last year.
            I am not grading this. You are by looking at the Key’s ! Try to get 70% right to be on track for a 5!
            Question 1:     10 points
            Question 2:     10 points
            Question 3:     10 points
            Question 4:       4 points
            Question 5:       4 points
            Question 6:       4 points
            Question 7:       4 points
 
New Reference Tables:
 
2017 Free Response Questions:
AP Chemistry 2017 Free Response (part 2 questions).pdf
View Download
AP Central Key and Rubric:  
AP Chemistry 2017 Free Response – College Board key.pdf
View Download
 
Grodski Key:
AP Chemistry Free Response 2017 – Grodski Solutions p.pdf
View Download
 
 
2.  Complete a Part 1 – Multiple Choice test (Practice Test 1)- 
 
This test will be sent to you via a link through email.  Do not save this to your computer nor any screenshots.  You need to be logged into Google with your private email to view.
 
Please place you answers into the google form for this Practice Test (1) to get your results. The form will be on auto-grade.
 
3. Review every Multiple Choice question that you got incorrect or guessed correctly with my video posted in the review page under the Form that you used to grade your part 1.
 
Under the description, if you play in YOUTUBE, you will be able to click on the timecode link to go directly to the question you want to review.
                         
End of Wednesday.. END OF WEEK 1 !

4/9 – Thursday – We will Not Meet Today!
 
*I will be sending you the Practice Test 1 – via email today. I actually sent Wednesday afternoon.
 

4/1o – Friday – We will Not Meet Today!
4/10- Friday – Weekend Homework – 
 
1. Please have the 2017 Free Response Questions completed by Monday Morning.
2. Please have Practice Test 1 (Multiple Choice completed and reviewed with video 
     by Tuesday morning
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 
For 2021: Next year – Note for Mr. Grodski to himself.. I hope I am a good listener!
 
Lab test for writing a curve based on given data (opposite of ammonium chloride, KHP titrations):
ou will be given a piece of blank graph paper 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.
 
If you get all 5 points correct with calculations I will give you a 100 on this test and a 100 on your lowest test grade in the 4th quarter!
This test is really the complete opposite of how you would complete the lab as you are verifying the same points in the lab you did last Friday and Today.
Here is an example:
Blank:
Acid Base Blank graph for Titration Prediction 1617 p.pdf
View Download
 
Key: (I made  small error in this key as I stopped the titration at 75 ml instead of 70 ml of base – Thanks Paige!)
Acid Base Titration Prediction KEY.pdf
View Download
 
Example 2: 
Blank:
Acid Base Blank graph for Titration Prediction 2 p.pdf
View Download
 
Key:
Acid Base Blank graph for Titration Prediction 2 KEY p .pdf
View Download
Lecture on how to complete this example 1 :

 

Lecture on how to complete this example 2 :
Lecture on how to complete the Lab:  
Play in Youtube and look under the video for the links (time code in blue) to take you where you need go in the video to complete. Most of you need points 3 and 5.