Question

Construct the titration curve, for the titration of 50.0 mL of 0.125 M trimethylamine with 0.175 M HCl (aq).

A. Calculate the initial pH of the solution in the flask.

B. Calculate the pH after 15 mL of titrant is added.

C. Calculate the pH at the equivalence volume.

D. Calculate the pH after 45 mL of titrant is added.

E. Calculate the pH after 60 mL of titrant is added.

Please show all work

Answer 1

pkb of trimethylamine = 4.19

a ) initial pOH of solution

pOH = 1/2(pkb-log C)

C = concentration of trimethylamine = 50/1000*0.125 = 0.00625 mole

    = 1/2(4.19-log0.00625) = 3.2

pOH = 3.2

pH = 14-3.2 = 11.8

b) pOH of basic buffer = pkb + log(salt/base)

No of moles of trimethylamine = 50/1000*0.125 = 0.00625 mole

No of moles of HCl   = 15/1000*0.175 = 0.002665 mole.

     = 4.19 + log(0.002665/(0.00625- 0.002665))

    = 4.06

c) at equivalence point

no of moles of trimethylamine = no ofmoles of HCl

No of moles of trimethylamine = 50/1000*0.125 = 0.00625 mole

No of moles of HCl   = 0.00625/0.175 = 35.71 ml

total volume of solution = 35.71+50 = 85.71 ml

d) after equivalence point pH depends up on excess acid added

excess acid = (45-35.71)/95*0.175 = 0.0171 M

pH of acid = -log(H+) = -log(0.0171) = 1.77

e)

after equivalence point pH depends up on excess acid added

excess acid = (60-35.71)/110*0.175 = 0.04 M

pH of acid = -log(H+) = -log(0.04) = 1.4