Question

You may want to reference (Pages 795 - 812) Section 17.4 while completing this problem. Consider the following curve (Figure 1) for the titration of a weak base with a strong acid and answer each of the following questions. Figure 1 of 1 PH 0 10 20 30 40 50 Volume of acid added (ml):

Answer 1

part-A

from the titration curve, it is seen that a sudden change of pH occurs in the range from 9.5 to 2.5 and pH at equivalence point is near 6.0

part-B

from the titration curve, the corresponding volume of acid at pH-6.0 is nearly 25.0 mL of HCl. therefore, 25.00mL HCl required to achieve equivalence point.

part-C

initially, when acid did not add to the weak base, the pH of this base will be determined using initial concentration and kb of the weak base.

when 0.0mL HCl added to the weak base (BOH)

BOH ↔ B+ + OH-

Kb= [B+] [OH-]/[BOH]

pH= 14-pOH , where pOH= -log[OH-]

part-D

when titration started weak base is neutralized by strong acid HCl

BOH + HCl = B+ + H2O

at the half equivalence point, [BOH]=[B+]

applying Henderson-hasselbalch equation

pOH = pKb + log [B+]/[BOH]

pOH= pKb + log(1)

pOH=pKb

pH= 14-pOH= 14-pKb

so, at half equivalence point, when 12.5 mL HCl added, pH=14-pKb

part-E

at the equivalence point, after addition of 25.0 mL HCl, pH calculated by working an equilibrium problem based on concentration and Ka of conjugate acid.