9.0-mL sample of 0.128 M diprotic acid (H2A)solution is titrated with 0.1040 M K

Question

9.0-mL sample of 0.128 M diprotic acid (H2A)solution is titrated with 0.1040 M KOH. The acid ionization constants for the acid are Ka1=5.2×10?5 and Ka2=3.4×10?10.

Part A

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

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Part B

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

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9.0-mL sample of 0.128 M diprotic acid (H2A)solution is titrated with 0.1040 M KOH. The acid ionization constants for the acid are Ka1=5.2×10?5 and Ka2=3.4×10?10.

Part A

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

V =   mL  

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Part B

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

V =   mL  

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Explanation / Answer

mmoles of H2A = 9.0 x 0.128 = 1.152

concentration of KOH = 0.1040 M

H2A   +    2 KOH   ---------> K2A +   2 H2O

1 mol H2A -------> 2 mol KOH

1.152 mol H2A   -------> ??

moles of KOH = 1.152 x 2 / 1 = 2.304 mmol

0.1040 x V = 2.304

V = 22.15 mL

total volume required = 22.15 mL

part A)

volume of first - equivalence point = 11.1 mL

part B)

volume at second - equivalence point = 22.2 mL

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