CHEM 1211 Lab Manual - Revised 05/2017 (4 pts.) Look up the % acetic acid on a b
ID: 712906 • Letter: C
Question
CHEM 1211 Lab Manual - Revised 05/2017 (4 pts.) Look up the % acetic acid on a bottle of vinegar in your cabinet or at the po- he bottle? What is the percent error of your experimental determination from the actual ont See the formula for percent error in appendix B If your standardized NaOH were used to titrate 20.00 mL of sulfuric acid (H2SO4), a diprotic acid, what concentration of sulfuric acid would you determine if 24.66 mL of the NaOH solution were required by the titration? First write the balanced equation for the reaction so as to use the correct stoichiometry in the calculation. (8 pts.) EquationExplanation / Answer
Ans. #I. Since the question does not provide experimental % acetic acid in sample, a hypothetical value is taken for explanation.
Say, the experimental % acetic acid in your sample is 4.5 %.
Now,
% Error = [ (Actual value - Experimental value) / Actual value] x 100
= [(5.0 % - 4.5%) / 5.0 %] x 100
= 10.0 %
#II. Balanced reaction: 2 NaOH + H2SO4 ----------> Na2SO4 + 2 H2O
At equivalence point, the total number of moles of H+ from acid is equal to total number of OH- from base.
That is- x(M1V1), acid = y(M2V2), base - equation 1
Where, x = moles of H+ produced per mol acid = 2 for H2SO4
y = moles of OH- produced per mol base = 1 for NaOH
V and M are volume and molarity of respective solution.
# Given, Volume of acid = 20.0 mL
Molarity of acid = ?
Volume of NaOH = 24.66 mL
Molarity of standard NaOH = M2
Note: The molarity of standard NaOH is NOT provided in the question. The calculation is made using a hypothetical value M2 for standard NaOH. Use the exact value of standard NaOH determined by you.
# Putting the values in equation 1-
2 x (M1 x 20.0 mL) = 1 x (M2 x 24.66 mL)
Or, M1 = (M2 x 24.66 mL) / [2 x (M1 x 20.0 mL)] = 0.6165 M2
Therefore, molarity of H2SO4 = 0.6165 M2 ; where M2 = Molarity of std. NaOH