Problem 7.125 A mixture of acetic acid (CH3CO2H; monoprotic) and oxalic acid (H2
ID: 528746 • Letter: P
Question
Problem 7.125
A mixture of acetic acid (CH3CO2H; monoprotic) and oxalic acid (H2C2O4; diprotic) requires 27.15 mL of 0.100 M NaOH to neutralize it. When an identical amount of the mixture is titrated, 15.05 mL of 1.59×102 M KMnO4 is needed for complete reaction.
Part A
What is the mass percent of each acid in the mixture? (Acetic acid does not react with MnO4. The equation for the reaction of oxalic acid with MnO4 was given
5H2C2O4(aq)+2MnO4(aq)+6H+(aq)10CO2(g)+2Mn2+(aq)+8H2O(l)
Enter your answers numerically separated by a comma.
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Problem 7.125
A mixture of acetic acid (CH3CO2H; monoprotic) and oxalic acid (H2C2O4; diprotic) requires 27.15 mL of 0.100 M NaOH to neutralize it. When an identical amount of the mixture is titrated, 15.05 mL of 1.59×102 M KMnO4 is needed for complete reaction.
Part A
What is the mass percent of each acid in the mixture? (Acetic acid does not react with MnO4. The equation for the reaction of oxalic acid with MnO4 was given
5H2C2O4(aq)+2MnO4(aq)+6H+(aq)10CO2(g)+2Mn2+(aq)+8H2O(l)
Enter your answers numerically separated by a comma.
CH3CO2H, H2C2O4= %SubmitMy AnswersGive Up
Explanation / Answer
5H2C2O4(aq)+2MnO4(aq)+6H+(aq)10CO2(g)+2Mn2+(aq)+8H2O(l)
mmol of base used --> 27.15*0.100 = 2.715 mmol of OH-
therefore
acetic acid + 2*oxalic acid = 2.715 (equation 1)
now,
mmol of KMnO4- = 15.05*(1.59*10^-2) = 0.239295 mmol of KMnO4-
ratio is 2:5 so
mmol of Oxalic acid = 5/2*0.239295 = 0.5982375 mmol of oxalic acid
acetic acid + 2*oxalic acid = 2.715 (equation 1)
acetic acid + 2*0.5982375 = 2.715 (equation 1)
acetic acid = 2.715 -2*0.5982375 = 1.518525 mmol of acetic acid
now
mass of acetic acid = mmol*MW = 1.518525 *60 = 91.1115 mg
mass of oxalic acid = mmol*MW = 0.5982375 *90.03 = 53.85932 mg
%mm = acetic acid / total mass = 91.1115/(53.85932 +91.1115)*100 = 62.8481%
then
oxalic acid --> 100-62.8481 = 37.1519 %