Q2: Week 3  – 11/29 – 12/3

                                                                                                                                                                                                Jump to:  Tuesday,   Wednesday,  Thursday, Friday                                                                                                                      ^{______________________________________________________________}

11/29 – Monday – B Day – 2, 3b/4 Lab 

Main focus –                                                                                                                                                         

                                                  

     a) To review the use of the Henderson – Hasselbach to answer point 3 questions.

     b) To identify buffer solutions as conjugate acid /base pairs

     c) To use the concept of pKa for indicators and buffer solutions.

Period 2:  Collect Lab 10, 11 if not handed in yet.

1. Review Lab 12 and complete in class.

2.  Verify all 5 points on the graph.  

3.  Buffer lesson with bicarbonate ion using notes – Maybe

                                                                             

Period 3/4: Collect Lab 10 and 11 if not handed in already

1. Same as above including the buffer lesson

2. Start the homework prediction 2.

 

Today’s Notes:

       

Strong Acid Strong Base  ~Lab 10	pH = 7 @           equivalence	only spectators that DO NOT ionize water
Weak Acid  Strong Base ~Lab 11,	pH  >7 @ equivalence	Conjugate Base produced reacts with water                                                     to produce OH-
Strong Acid  Weak Base ~Lab 12	pH  <7 @ equivalence	Conjugate Acid produced reacts with water                                        to produce H3O+

            

The  Blood buffers the blood pH so that you survive!!!  There are so many organic molecules that have acidic or basic properties that would cause death in our aqueous plasma (fluid part) of our blood. The main buffer in our blood that prevents large pH changes if the following conjugate acid base pair:    Conjugate Acid                      Conjugate Base             H2CO3            <—>          H+  +  HCO3-1    Neutralizes Bases                  Neutralizes Acids

  

        The pKa of  H₂CO₃ = 6.1   The Ka of this week acid = 10 ^–3.8   = 1.58 x 10^-4

 

Blood is buffered at pH of about 7.4.  What is the ratio of conjugate base to conjugate acid in this buffered solution?

 

                                                                   pH = pKa + log [A^–]/[HA]

 

                                                                   7.4 =  6.1  + log [HCO₃^-1]/[H₂CO₃]

 

                                                                     1.3 = log [HCO₃^-1]/[H₂CO₃]

                                                                            

                                                                   10^X to both sides of the equation

                                                

                                                                          10^1.3 = [HCO₃^-1]/[H₂CO₃]

                                                

                                                                                    =    19.9     :      1        ratio

 

                               *Normal human blood has about a 20 : 1 ratio!

 

                                     What would this look like on a titration curve?

 

 

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11/30 – Tuesday – A Day – 2/3a Lab, 4

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12/1 – Wednesday – B Day – 2, 3b/4 Lab 

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12/2 – Thursday – A Day – 2/3a Lab, 4

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12/3 – Friday – B Day – 2, 3b/4 Lab 

Main focus –                                                                                                                                                         

                                                  

    a) Lab 13 – (Formal Lab Write-up) – Lab note -taking – Lab design of a Back titration

                                         

Period 3:  

1. Review of test 1

2. Note-taking for Lab 13                                                                           

Period 4: 

1. Review of test 1

2. Note-taking for Lab 13 

3. Standardization of Strong Base with KHP   

In Tums we are dealing with the active ingredient, calcium carbonate, CaCO₃.  Watch out this antacid is an insoluble salt that is found in limestone (sediment rock) and sea shells AND CHALK!!!!   It is an antacid because of its basic anion, the carbonate ion, CO₃^-2 .  It is a base because it can accept 2 H⁺ ‘s (Bronsted-Lowry Base) or it can donate 2 pair of electrons (Lewis Base) to 2 H⁺ ‘s .

 

The reaction with carbonates involve probably the most important equilibrium reaction that we have in organic chemistry. Organic Chemistry chemical reactions that support and sustain life.  It specifically deals with the Carbon based molecules and the link to the carbon cycle, specifically CO₂ (g) as the product of combustion and the reactant in photosynthesis!

 

The arrows up in the atmosphere are gas molecules of CO₂ (g) while the other forms may be CO₃^-2 (aq), HCO₃ ^-1 (aq), and H₂CO₃ (aq), or CaCO₃ (s).

 

In our oceans the movement of carbon through the marine ecosystem can be stored as calcium carbonate, limestone, sedimentary rock or it becoming the shelter that marine organisms use for protection (phytoplankton, sea shells, coral reefs, etc). As the diagram illustrates these reactions are in equilibrium. The Carbon cycle in out oceans is very similar to the carbon transport system in our blood , EXCEPT we do not make an insoluble precipitate, CaCO3 (s).

In our bodies we must get rid of the carbon dioxide that is released as a by – product of cellular respiration.  We dissolve the CO₂ (g) in our plasma (water part of out blood) and because of fact that CO₂ (g) is Lewis acid it lowers our blood pH which will force us to breath.

 

Not only does equilibrium system help transport CO₂ (g) out of the body through the plasma it also serves an important blood buffer that keeps you blood at a pH of about 7.4.

 

Mechanism of how Carbonic acid  can be decomposed back into water!

 

 

Standardization of the Strong base in this back titration of TUMS.

In this analysis of the “active ingredient” of Tums ( CaCO₃ ) we will need to use a strong base to titrate the excess HCl that was used to drive the reaction to completion:

 

                Net Ion:          2H⁺ (aq)   +   CaCO₃ (s)   —>   CO₂ (g)    +     H₂O (l)   +   Ca⁺²

                                   Excess

The excess Strong Acid will drive the reaction to completion due to Le Chateliers Principle.  It also important to to have excess Strong Acid to react with all the CaCO₃ that will be mixed and “hidden” with the binder, coloring, flavor, and etc.  That is why it is important to crush the tablets (tabs!) with a mortar and pestle in order to expose all of the salt, CaCO₃ to the acid.

 

Now after we add the excess acid we will have to Titrate this excess acid with a Strong Base. This lab produces incredibly accurate results ONLY IF OUR SOLUTIONS that we use ARE ACCURATE.  We are adding 0.04 moles of protons from the 100.0 ml of a 0.4 M HCl solution.  Because we are adding about 0.01 moles of extra protons from the strong acid we need to titrate the excess protons with a strong base AND that solution needs to verified or STANDARDIZED.  

 

I made a 0.5 M NaOH solution. I was very careful to make this solution BUT because strong Bases are very reactive that solution is definitely LOWER in concentration as the hydroxide reacts with the CO₂  (Lewis Acid) in the air.  I used 20 grams of solid NaOH (.5 mole) dissolved in a 1 liter of solution to make a 0.50 M NaOH solution

BUT not all of the solid is NaOH as CO₂  from the air will cause the NaOH to become Na₂CO_{3. This also happens in solution.  Strong bases are carbon dioxide scrubbers that are used to remove it from space capsules.}

 

                                                 2NaOH(s)    +     CO₂ (g)     —–>  Na₂CO₃      +     H₂O

 

 Also NaOH is deliquescent which means it can absorb water from the air and start dissolving itself. This would add weight to the Base and dilute is quantity.

 

Because NaOH has these issues we need to verified or STANDARDIZE it with a weak acid that is not reactive and does not have these issues. A perfect solution to this (no pun intended!) is using KHP. Potassium Hydrogen Phalate.  

 

KHP, (KHC₈H₄O₄) is excellent to standardize a Strong Base with because it 

 

                                            1. Weak Acid – thus does not react unless forced to by a Strong Base

                                            2.  Is a solid –  we eliminate volume errors when making this solution

                                            3.  It has a large molecular mass (204 g/mol) – measurement errors with a large

                                                                                                                         molecular mass will not lead 

                                                                                                                         to large mole errors.

 

 

                       Net Ion:  HC₈H₄O₄^–     +     OH^–      —->    C₈H₄O₄^{– 2}       +       H₂O

 

You will find from this standardization that my solution of 0.50 M NaOH is less but we need to know its exact value to proceed with the lab.