In this exercise, we determine the concentration of acetic acid in a vinegar sol

Question

In this exercise, we determine the concentration of acetic acid in a vinegar solution of unknown concentration by acid/base titration. We standardize the base by titratiing it with a primary standard acid, potassium hydrogen phthalate (hereafter abbreviated as KHP) Since KHP is a solid we can make a solution with a precise acid concentration by dissolving an accurately weighed sample in a precise volume of water (using a volumetric flask). The empirical formula of KHP is KCgH504 Use the molar masses on the back cover of the manual to determine its molar mass A KHP solution was prepared as follows Initial weight of KHP vial (g) Final weight of KHP vial (g) Volume of KHP solution (mL) 19.0715 15.7719 250.0 What is the concentration (M) of the KHP solution? Question 1 [Note: The answers to almost all the subsequent questions depend on the concentration of KHP. Enter Your Answer: 0.06451 A solution of sodium hydroxide was standardized by titrating known volumes of the above KHP solution. The resulting data are summarized in the table below: Initial KHP buret reading Final KHP buret reading Initial NaOH buret reading Final NaOH buret reading RUN 1 4.98 40.73 RUN 2 0.5 33.18 3.04 22.69 24.62

Explanation / Answer

5. Average deviation of concentration = 1.0752 - 1.0744 = 0.0008 M

6. % deviation of conc. = [(observed deviation)/Avg. conc.]x100 = (0.0008/1.0748)x100 = 0.074%

7. Run 1:
Titre value in run 1 = (18.14 -1.21) =16.93 mL
No. of moles of NaOH used = 1.0748 x 16.93 / 1000 = 0.01819 moles = No. of moles of acetic acid present
Conc. of acetic acid = 0.01819 x (1000/50) = 0.3638 M

8. Run 2:
Titre value in run 2 = (20.56 - 3.73) =16.83 mL
No. of moles of NaOH used = 1.0748 x 16.83 / 1000 = 0.01809 moles = No. of moles of acetic acid present
Conc. of acetic acid = 0.01809 x (1000/50) = 0.3618 M

9. Run 3:
Titre value in run 3 = (21 - 4.03) =16.97 mL
No. of moles of NaOH used = 1.0748 x 16.97 / 1000 = 0.01824 moles = No. of moles of acetic acid present
Conc. of acetic acid = 0.01824 x (1000/50) = 0.3648 M

10. Average conc. = (0.3638 + 0.3618 + 0.3648)/3 = 0.3634 M

11. Deviation from run 1 to 2 = 0.3638 - 0.3618 = 0.002
Deviation from run 2 to 3 = 0.3648 - 0.3618 = 0.003
Deviation from run 3 to 1 = 0.3648 - 0.3638 = 0.001

Avg deviation = 0.002

12. % deviation = (0.002/0.3634) x 100 = 0.55%

13. Yes.

14. Error = actual conc. - determined conc. = 0.3650 - 0.3634 = 0.0016
% error = (0.0016/0.3650) x 100 = 0.438 %

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