Wizard
Witch
SUSB-011

1. Preparation of Standard KHP Solution

KHP is a monoprotic acid of known molar mass = 204.22 g/mol = 204.22 mg/mmol.  One mole of the acid reacts with one mole of NaOH.  We make a solution of precisely known volume (250.0 mL) from an accurately weighed sample of KHP.   The concentration of that solution is known and fixed.  E.g.,

            Weight of vial + KHP                                                                        18.7956 g
            Weight of vial - sample of KHP                                                          16.3612 g

            Weight of sample of KHP                                                                   2.4344 g

            mmol KHP in sample     2434.4 mg / 204.22 mg/mmol    =               11.920 mmol      (5 sig figs)

            Volume of solution                                                                             250.0 mL          (4 sig figs)

            Concentration of KHP solution   11.920 mmol / 250.0 mL = 0.04769 mmol/ml = 0.04769 M     (4 sig figs)

Note that the entire remainder of the exercise depends on the accuracy of this concentration!

2. Standardization of NaOH

Step 1 produces a solution of an acid of known concentration (0.04769 M).  We now use that solution to determine the concentration of our NaOH solution.  In this exercise, we do this by using two burets -- one to deliver the standard acid solution (KHP) and one to deliver the NaOH solution whose concentration is to be determined.
The only buret readings that matter are the initial and final readings.  Intermediate readings made if we backtitrate are irrelelevant to the results.

                                                                                             KHP Buret                    NaOH Buret
            Final buret reading                                                      38.76 mL                         15.94 mL

            Initial buret reading                                                       4.53 mL                           2.84 mL

            Net volume                                                                34.23 mL                         13.10 mL    

            mmol KHP    34.23 mL * 0.04769 mmol / mL  =           1.632 mmol    (4 sig figs)  

           mmol NaOH ( same as mmol KHP)                     =           1.632 mmol    (4 sig figs)

            Concentration of NaOH   1.632 mmol / 13.10 mL =      0.1246 mmol / mL = 0.1246 M  (4 sig figs)

Suppose two more standardizations are conducted, yielding the values 0.1228 M and 0.1260 M respectively.

We calculate the mean (average) of the three concentrations    

                               Cavg = ( 0.1246 + 0.1228 + 0.1260 ) / 3 = 0.1245 M   (4 sig figs)

The average deviation is given by

            (  | 0.1246 - 0.1245 | + | 0.1228 - 0.1245 | + | 0.1260 - 0.1245 | ) / 3 = 0.0011 M         (2 sig figs)

The result can be written as                              Cavg0.1245 M ± 0.0011 M

The percentage deviation is 100 * 0.0011 / 0.1245 = 0.88 %            (2 sig figs)

We have used our standard KHP solution to determine the concentration of the NaOH solution that we will use to titrate the vinegar!  Both the accuracy and precision of this step will limit the accuracy with which we can do the next step.  The following aspects of the standardization can give rise to poor precision:

  • inaccuracy in the preparation of the KHP solution
  • inconsistency in recognizing the end point
  • incomplete mixing of the KHP solution


3. Determination of the Concentration of an unknown acetic acid solution

Now we use our standardized NaOH solution to titrate a solution of acetic acid (vinegar) of unknown concentration. We prepare our sample by pipeting 5.00 mL of the unknown solution into a clean 250.0 mL Erlenmeyer flask and dilute the solution with approximately 40 mL of distilled water. (As noted in the lecture notes - the precision of the pipet limits the overall precision.)
 

Run 1
Run 2
Run 3
Final buret reading
21.84
22.64
22.25
Initial buret reading
0.67
1.39
0.88
Net volume of NaOH
21.17
21.25
21.37
Concentration of NaOH (from standardization)
0.1245
0.1245
0.1245
mmol NaOH
2.635
2.646
2.661
mmol Acetic Acid (same as mmol NaOH)
2.635
2.646
2.661
Volume of Acetic Acid sample titrated (mL)
5.00
5.00
5.00
Concentration of Acetic Acid (mmol/mL = mol/L)
0.527
0.529
0.532

The average of the results is ( 0.527 + 0.529 + 0.532 )/3 = 0.529    [Note that we have only 3 significant figures in the concentration.]


The average deviation is (0.002 + 0.000 + 0.003)/3 = 0.002

The percent deviation is 100 * 0.002/0.529 = 0.4%

4. Determination of the Acid Ionization Constant, Ka, of acetic acid

The acid ionization constant is defined as the equilibrium constant for the acid ionization reaction,

HOAc = H+ + OAc-                   Ka =   [ H+ ] [ OAc- ] / [ HOAc ]

To calculate this quantity, we note that for an acid that is weak ( but stronger than pure water ), the virtually sole source of hydrogen ions is through this ionization. Therefore, the concentrations of H+ and OAc- will be equal. Also, since acetic acid is weak, only a small amount of it will ionize by definition. We suppose that the concentration of the acid is sufficient that we can ignore the decrease in the concentration of the undisoociated acid, HOAc, due to ionization compared to the nominal concentration of the acid.

Measuring the pH of the acid solution gives us a value for [ H+ ]   ( [ H+ ] = 10-pH),   and therefore, also [ OAc- ]. The concentration of the acid is the value that we found through the titrations, 0.529 M corrected for the amount whixh is dissociated. Suppose the pH is measured to be 2.45. That gives us [ H+ ] = [ OAc- ] = 10-2.45 = 3.6 X 10-3. So, the correction to the undissociated form is 0.0036

so

Substituing our values in the expression for Ka gives us:

Ka = ( 3.6 X 10-3 ) ( 3.6 X 10-3 ) /( 0.529 -.0036 ) = 2.47 X 10-5

 

CHE 133 HOME PAGE

Campus Home Page
Chemistry Department
Solar
Blackboard
OSCER
Academic Calendar
 
 
Last Update: RFS 2016-10-26