SUSB011 

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 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.
KHP Buret
NaOH Buret 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 C_{avg} = ( 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 C_{avg} = 0.1245 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:
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.)
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 percent deviation is 100 * 0.002/0.529 = 0.4% 4. Determination of the Acid Ionization Constant, K_{a}, of acetic acid The acid ionization constant is defined as the equilibrium constant for the acid ionization reaction, HOAc = H^{+} + OAc^{} K_{a} = [ 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 K_{a} gives us: K_{a} = ( 3.6 X 10^{3} ) ( 3.6 X 10^{3} ) /( 0.529 .0036 ) = 2.47 X 10^{5}

Last Update: RFS
20161026
