Question

Consider the titration of 65 mL of 1.75 M diethylamine, (CH3CH2)2NH (Kb = 1.3×10-3) with 0.25 M HCl. How many mL of HCl will need to be added to the diethylamine solution to reach a pH of 11.11?

The answer is NOT 455 mL or 45.5 mL

Answer 1

We will use the relation,

pH = pKa + log([base]/[salt])

Find Ka from Kb value

We will use the relation,

pH = pKa + log([base]/[salt])

Find Ka from Kb value

Ka = 1 x 10^-14/Kb

     = 1 x 10^-14/1.3 x 10^-3

     = 7.69 x 10^-12

pKa = -log(Ka)

       = -log(7.69 x 10^-12)

       = 11.11

Feed the value in Hendersen-Gasselbalck equation to find ratio of [Base]/[Salt]

11.11 = 11.11 + log([Base]/[Salt])

[Base]/[Salt] = 1

Thus the system is at equivalence point

M of [Et2NH] = M of [HCl]

find moles of Et2NH = M x L

                                 = 1.75 M x 0.065 L

                                 = 0.114 moles

So, at equivalence point [Salt] = [H+] = 0.114 moles

L of HCl required = moles/M

                            = 0.114/0.25

                            = 0.456 L

                            = 456 ml

This is the most obvious answer to your problem. Check your answer. It has to be this asnwer.