Question

What is the pH at the second eq. point (40 mL acid added) of the titration of a weak base titrated with a strong acid:
pKa1= 9.06 pKa2= 4.18
[base]=0.20 M 20 mL
[acid]= 0.10 M

Answer 1

For titration of weak base and strong acid, pH at equivalence point is given by,' pH = 7 - 1/2 pkb - 1/2 loge. C + salt concentration Now, here, pka 2 = 4.18. .: ÞKb = 14 - 4'18 = 9.82. Total volume of solution after titration = (40 mL + 20 mL) = 60 mL. Volume concentration of base = 20 mL. Initial concentration of base = 0.20m. .. Final concentration of base should be : 20 40'20 M: 0.033 M. 60 But due to neutralization it will become salt. Thus, C 0.033M. ÞH = 7 - *982 -Ź log (0.033 ) - ÞH = 7 - 4.91 - (-0.74 ) => PH: 2.83