Question

22.0-mL sample of 0.126 M diprotic acid (H2A)solution is titrated with 0.1028 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10.

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

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

Answer 1

The reaction between H2A and KOH is:

H2A + KOH$\rightarrow$ KHA+H2O This is the first equivalence reaction

Molarity= moles/volume in litre

So, moles of H2A= 0.022L* 0.126M= 0.00277moles

Since 1 mole of H2A reacts with 1 mole of KOH, moles of KOH= 0.00277moles

Therefore volume of KOH at first equivalence point= 0.00277moles/ 0.1028M = 0.02694L= 26.9mL

For second equivalence: KHA+ KOH$\rightarrow$K2A+H2O

The volume required to reach the second equivalence point will be same as that of first equivalence point because the number of moles of H2A determine the moles of HA.

Number of moles of KHA= 0.00277 moles

Number of moles of KOH= 0.00277moles

Therefore volume of KOH required = 0.00277moles/0.1028M= 26.9mL

Thus 26.9mL is again required to reach second equivalence point also. So in total 26.9+26.9= 53.8mL of KOH is required