Question

# 22.0 mL of 0.128 M diprotic acid (H2A) was titrated with 0.1017 M KOH. The acid...

22.0 mL of 0.128 M diprotic acid (H2A) was titrated with 0.1017 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5and Ka2=3.4×10−10.

Part A

At what added volume of base does the first equivalence point occur?

Part B

At what added volume of base does the second equivalence point occur?

Volume of H2A = 22.0 mL = 0.022 L

Molarity of H2A = 0.128 M

Moles of H2A reacted = 2.816 x 10-3 mol

H2A + KOH <=====> KHA + H2O

Moles of H2A reacted = moles of KOH consumed

Volume of KOH = moles of KOH/molarity = 0.0277 L

Volume of KOH consumed at first equivalence point = 27.7 mL

First equivalence point will be reached after adding = 27.7 mL

KHA + KOH <====> K2A + H2O

Moles KHA formed = moles of H2A reacted

Moles of KHA = 2.816 x 10-3 mol

Moles of KOH = moles of KHA = 2.816 x 10-3 mol

Volume of KOH = 27.7 mL

Second equivalence point will reaching after adding = 27.7 x 2 = 55.4 mL

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