a) To Take Acid/Base RAT 1

b) To View a pH Titration curve and determine the unknown concentration,                                equivalence point, half equivalence point, and the final concentration of the titrant.

 Period 2: 

1.  Test 3 (1st quarter) –  1st try.

a) Individual component – 50%

b) Group Component – 50%

Period 3/4: 

1.  Test 3 (1st quarter) –  1st try.

a) Individual component – 50%

b) Group Component – 50%

2. Pre-titration worksheet review and provide the Lab set-up that represents each                       problem.  The idea is to connect these stoichiometry problems to the lab procedure.

2.  I made a new worksheet and Video that could help with writing Net Ion Reactions for Acid Base Reactions. Please print and follow along with me.

Period 2/3a –  

1. Tie Net Ion acid base writing with determining acid/base/neutral salt determination with notes below.

2. Pre-titration worksheet review and provide the Lab set-up that represents each                        problem.  The idea is to connect these stoichiometry problems to the lab procedure.

2.  Test 3 (1st quarter) –  2nd try.

          a) Individual component – 50%

          b) Group Component – 50%                                                   

a) To identify the equivalence and half equivalence points on a pH titration curve.

b) To estimate the equivalence pH and half equivalence pH along with calculating the             final pH of the titration.

c) To calculate the initial concentration of the analyte.

1.  Complete RAT 2 – 

       – 5 minutes to complete the 2 two questions you were given an opportunity to complete.

       – 20 minutes to complete ticket in groups

2.  Review of Homework- Determining the concentration of the acid 

3.  Strong Acid/Strong Base Titration Lab 10 –   

        – complete the titration and print out graphs.    

Period 3b, 4 –        

1. Complete RAT 2 – 

       – 5 minutes to complete the 2 two questions you were given an opportunity to complete.

       – 20 minutes to complete ticket in groups

2.  Strong Acid/Strong Base Titration Lab 10 –  

a) To derive the Henderson – Hasselbach Equation

b) To identify the buffer region in a titration curve

c) To Calculate the Ka of a weak acid through a titration curve and approximate Point 3.

d) To introduce the concept of pKa and how it is used with Acid/base Indicators.

Every Titration Curve Has 5 points that need to identified or calculated AND every ACID/BASE problem represents every one of these points. They include:

We will focus on Point 2 and point 3 today. These points occur in weak acid/strong base and strong acid / weak base titrations as there are no buffers in strong acid / strong base titrations.

Period 2 – Solve for the concentration of the acid and then compare both titrations!!

1.  We will use the Lab 11 to identify the buffer region of the Titration curve and Derive the Henderson – Hasselbach                    equation.

2.   We will use Lab 11 to determine and calculate the Ka of the weak acid in the titration (acetic acid).

3.   Approximate point three of the titration curve.

4.   Compare and contrast both titrations in Lab 11 and 12 and collect both labs.

5.   Chemical Indicators and their pKa’s!!!!

Period 3b/4 – Solve for the concentration of the acid and then compare both titrations!!

1.   Same as above

2.   Collect both Lab 10 and Lab 11.   

3.  Lab activity – Not a Lab – Practice skills learned

      a) Determine the concentration of the weak acid with a Standard Base using a chemical indicator.

      b) Determine the Ka of the acid.

Today’s Notes:

The new skills that we learned from Titrations in Lab 10 and 11 are identifying the parts of a Titration curve.  The striking part of the Titration graph is the asymptotic line that occurs as the pH changes becomes exponential changes as the pH nears the equivalence point.  

 

We learned the endpoint in an Acid/Base titration is really an approximation of the equivalence point.  The endpoint refers to the volume of titrant added (Standard Base Yesterday) and pH that results when the chemical indicator changes color.  This color change will never be exactly at the equivalence point but it approximates the volume of titrant added if the indicator has a color change on the asymptotic line near the equivalence point.

 

Consider our last 2 titrations.  We used phenolphthalein which has a color change (turns from colorless to pink at pH changes of 8 – 9).  What if we used Thymol blue or Methy red?  

 

Table M

Do not forget that these indicators are themselves conjugate acid base pairs in equilibrium. In the case of Thymol blue

Notice in the first Titration which you should recognize as a strong acid / strong base titration, with a pH of the equivalence pH of 7, can utilize both Thymol blue and Methyl red as indicators as the endpoints will be on the asymptotic line.  They would have endpoints with approximately the same volume as the equivalence point.

 

In the second titration you should recognize as as weak acid / strong base titration because the starting pH is greater and the equivalence point pH is NOT 7!  In this titration the asymptotic line is not quite as long thus Methy Red would no longer be appropriate to use because its color change would occur at MUCH different volume than the equivalence point volume resulting in a very poor approximation of the volume needed by the titrant to neutralize the acid.  It would lead to undervalued concentration of the acid in this case.  THymol blue however would however lead to very good approximation of the equivalence volume because its color change is closer to the equivalance point.

 

Notice both the Titrations have an equivalence volume of 50 ml of base added, illustrating that both acids are at the same concentration even though the acid in the titration on the right is a weaker acid.  The Strong Base DRIVES the weak acid to COMPLETION like the Strong Acid on the Left does naturally.

 

Remember the Volume at the equivalence point is what is normally needed to attain the concentration of the solution being titrated  (acid in both cases above and in our lab 10 and 11). This volume of added titrant (chemical with known concentration) is used to determine the concentration of the solution in the beaker under the buret.

 

The new skill that we added in our graphical analysis of acid base titrations allows us to determine the Ka of acid, or conjugate acid (produced from a base that is titrated).

 

Remember Ka is really a Keq which is a equilibrium constant that expresses whether the products or reactants are favored.  I have been surprised how many students have asked what this in the past few days?  We have discussed that Keq in terms of thermodynamics when we talked about spontaneity which if you remember is pathway that the universe supports.  A favorable pathway always is supported by a dispersion of energy from a concentrated source (increase in entropy).

 

                                                                    Reactants   —–>   Products

                                                                 Since Keq generally = [Products] / [Reactants] 

 

Then a Keq greater than  1 is a reaction that has more Products than Reactants at equilibrium. 

If there are more products present at equilibrium then THE FORWARD DIRECTION is more favorable pathway then the reverse reaction (which would build up reactants if more favorable).

 

                                            The forward reaction then is more spontaneous!

 

Strong Acids have an incredibly large Keq because virtually all of the acid dissociates to leave virtually zero reactants which would drive the Keq to a VERY LARGE NUMBER! Strong Acids dissociate very spontaneously!

 

                                                                        HCl  –>  H⁺   +    Cl^–

       

                                                         Keq  = [H⁺][Cl^–] /  [HCl]

 

                                                         Keq = VERY LARGE

 

Weak acids have an incredibly small Keq because virtually all of the acid remains undissociated and thus there are less products and more reactants. Weak acids dissociate very UNspontaneously!

 

                                                               HC₂H₃O_{2  —–>}  H⁺   +   C₂H₃O₂^–

 

                                                Keq   =    [H⁺][C₂H₃O₂^–] / HC₂H₃O₂

 

                                                             Keq  =   Very Small 

 

Now because these reactions are very similar  (HA   —->  H⁺  +   A^– ) we call them Ka! 

                                                Keq  = Ka when we deal with acids dissociating.

 

We can find the Ka graphically because of the relationship between Ka and what it equals:

 

                                                                      

Using this relationship we know that when we have equal amounts of conjugate Base (A^–) and conjugate acid (HA) the log [A^–] / [HA] will go to zero because the log of 1 = 0 !!

 

                                                        Thus when [A^–]  =  [HA]   the pH = pKa

 

and we can get the pKa of acid which is just the -log of the Ka.  Once we have the pKa we just 

 

                                  perform the 10^-x calculation to attain the Ka  of the acid.

 

How do we find the point where the conjugate Base (A^–) and conjugate acid (HA) are equal

                                                

                                                               We find the half equivalence point!!! (point 2!)

 

So once we establish the equivalence point which is in the middle of the asymptote we attain the volume at that point which in the example to the left is 20 ml.  Now we take the equivalence volume and halve it: 20 ml / 2 = 10 ml 10 ml represents the volume of titrant (base) added to neutralize half of the acid.  This point is where half of the weak acid has been converted to the conjugate base.  So if we had 1 mole of weak acid, .5 moles of the weak acid remains and  .5 moles of conjugate base has been created. This is the point where the acid = conjugate base !                              pH of this point = pKA

So in the above example the pH at the Half – equivalence point is equal to 3.  Pka of the acid = 3.

 

                                       The Ka of the acid is thus:    10^-3 =    1 x 10^-3

 

This is a Ka of a weak acid because it is less than 1 which means the reactants or the undissociated acid ( HA ) is a larger quantity because the reverse reaction is more spontaneous.                                                          

 Instead of pH ranges for the color changes for chemical indicators we can use pKa’s

 They are also acid/base equilibrium systems

                                              

                                                           HIn    <—>   H⁺    +     In^–

 

                                              and thus they also have Ka’s and pKa’s

 

            What do you think the pKa can tell us about the acid/base indicators?

        

               What is significant about the pKa of the acid base indicators in terms of of the pH ranges?