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DETERMINATION OF MOLECULAR WEIGHT OF AN UNKNOWN DIPROTIC ACID Use Data Set to answer the following questions 1. Using the dat
Homework Molecular Weight of an Unknown Diprotic Acid 3. Using the second equivalence point, determine the molecular weight o
Homework Molecular Weight of an Unknown Diprotic Acid OPTIONS 1. For each data set, the volume of acid being analyzed is 10.0
DETERMINATION OF MOLECULAR WEIGHT OF AN UNKNOWN DIPROTIC ACID Use Data Set to answer the following questions 1. Using the data, sketch a titration curve for the titration of the acid with 0.1 M NaOH pH 0 volume of NaOH added 2. Using the first equivalence point, determine the molecular weight of the acid. Use the sequence of steps employed in the in-class exercise.
Homework Molecular Weight of an Unknown Diprotic Acid 3. Using the second equivalence point, determine the molecular weight of the acid. Use the sequence of steps employed in the in-class exercise
Homework Molecular Weight of an Unknown Diprotic Acid OPTIONS 1. For each data set, the volume of acid being analyzed is 10.00 mL and the concentration of NaOH is 0.100 M Data Set Conc. of Volume of acid (g/L) NaOH at 1 st pH at 1st eq. Volume of pH at 2nd eq NaOH at 2nd pt pt nt (mL 9.12 15.12 12.42 21.00 19.14 8.06 11.95 8.96 14.08 11.07 4.1 4.3 3.9 4.0 4.4 15.98 24.09 18.10 27.96 21.979.4 9.6 9.2 8.8 9.1
0 0
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Answer #1

14 0

2. At first equivalent point only one proton of diprotiec acid reacts so valance = 1 and normality of acid = Molarity / 1

For NaOH Normality = Molarity

NaOH + H2A -------------------> NaHA + H2O

N1V1 = N2V2 (N1 = normality of acid V1 = Volume of acid and N2 = normality of NaOH and V2 = volume of NaOH)

N1 x 10 = 11.07 x 0.1

N1 = 0.1107

here normality = molarity so M = 0.1107 M L-1

Molarity = Vol.inL

per liter the weight of acid is = 19.14 Vol = 1 L. So

M.Wt = Molarityol.inL = 19.14 / 0.1107 X 1 = 172.9

3. At second equivalent point only one proton of diprotiec acid reacts so valance = 1 and normality of acid = Molarity/2

But for Naormality = Molarity because Valance =1

2NaOH + H2A -------------------> Na2A + 2H2O

N1V1 = N2V2 (N1 = normality of acid V1 = Volume of acid and N2 = normality of NaOH and V2 = volume of NaOH)

N1 x 10 = 21.97 x 0.1

N1 = 0.2197

here normality = molarity/2 so M = 0.10985 M L-1

Molarity = Vol.inL

per liter the weight of acid is = 19.14 Vol = 1 L. So

M.Wt = Molarityol.inL = 19.14 / 0.10985 X 1 = 174.237

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