0.100 M HCl is used to titrate 20.00 ml of 0.150 M triethylamine, (CH 3 CH 2 ) 3

Question

                    0.100 M HCl is used to titrate 20.00 ml of                     0.150M triethylamine, (CH3CH2)3N (Kb=5.2x10-4).Please answer                     the following questions:                 

                    1) pH when no HCl is added;                 

                    2) pH after a total of 5.00ml of HCl is added;                 

                    3) pH change after 0.5ml 0.100M H2SO4 is added to the solution present in part 2;                 

                    4) pH at the equivalence point (stoichiometry point);                 

                    5) pH after a total of 40.00ml HCl is added.                 

Explanation / Answer

1)Poh= -1/2[pKb=log c]

=1/2(5.2x10?4 + log[0.003]0

=2.523398745

pH= 14-pOH = 11.48

2) 1 mole 0f pyridine= 1 mole 0f HCl

0.003mole = 0.003moles

but concentration of HCLleft in the solution = 0.005-0.003= 0.002 M

pH= -log[H+] = -LOG[0.002]

= 2.698970004

3) CONCENTRATION OF PYRIDINE LEFT IN THE SOLUTION= 0.003-0.00005

=0.00295 M

pOH= 2.5

pH= 11.46

4) total pyridine is neutralised by Hcl

pH= 7

5) concentration of HCl = 0.004 M

Net concentration of solution = 0.004-0.003= 0.001 M HCl left in the solution

pH= 2.397940009

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