Question

16.5 mL of 0.168 M diprotic acid (H2A) was titrated with 0.118 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. At what added volume (mL) of base does the first equivalence point occur?

20.8 mL of 0.146 M diprotic acid (H2A) was titrated with 0.14 M KOH. The acid ionization constants for the acid are Ka1=5.2×10−5 and Ka2=3.4×10−10. At what added volume (mL) of base does the second equivalence point occur?

Consider the titration of a 92.8 mL sample of 0.122 M HC2H3O2 with 0.157 M NaOH. Ka(HC2H3O2) = 1.8x10-5 Determine the pH at one-half of the equivalence point.

Answer 1

1)

millimoles of H2A = 16.5 x 0.168 = 2.772

H2A   +   KOH    --------->   HA- +    H2O

At first equivalence point :

mmoles of H2A = mmoles of KOH

2.772 = 0.118 x V

V = 23.49 mL

volume of base for first equivalence point = 23.5 mL

2)

mmoles of H2A = 20.8 x 0.146 = 3.0368

At second equivalence point :

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

M1 V1 / n1 = M2 V2 / n2

0.146 x 20.8 / 1 = 0.14 x V2 / 2

V2 = 43.4

volume of base at second equivalence point = 43.4 mL

3)

Ka = 1.8 x 10^-5

pKa = 4.74

At half - equivalence point :

pH = pKa

pH = 4.74