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 =
per liter the weight of acid is = 19.14 Vol = 1 L. So
M.Wt = = 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 =
per liter the weight of acid is = 19.14 Vol = 1 L. So
M.Wt = = 19.14 / 0.10985 X 1 = 174.237
DETERMINATION OF MOLECULAR WEIGHT OF AN UNKNOWN DIPROTIC ACID Use Data Set to answer the followin...
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A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The titration required 30.0 mL of base to reach the equivalence point, at which point the pH was 8.68. a) What is the molecular weight of the acid? b) What is the pKa of the acid?
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