Question

Calculate the pH for each of the cases in the titration of 25.0 mL of 0.200 M pyridine, C5H5N(aq) with 0.200 M HBr(aq).

a)  before addition of any HBr

b) after addition of 12.5 mL of HBr

c)  after addition of 15.0 mL of HBr

d) after addition of 25.0 mL of HBr

e) after addition of 29.0 mL of HBr

Answer 1

Ans:

Pyridine is a weak base.

Kb of pyridine = 1.7 x 10^-9(data collected)

pKb of pyridine = -log Kb = 8.77

Reaction: C₅H₅N + HBr --> (C₅H₅NH)⁺ Br^- (pyridinium bromide)

	Taken	Taken (converted to millimole) Millimole = Molarity x volume (mL)	What happens?	Remark
a	C5H5N = 25.0 mL, 0.20 M	C5H5N = 5 mmol		Weak base
b	C5H5N = 25.0 mL, 0.20 M HBr = 12.5 mL, 0.20 M	C5H5N = 5 mmol HBr = 2.5 mmol	2.5 mmol of HBr reacts with 2.5 mmol of pyridine to give 2.5 mmol of pyridinium bromide. So, after the reaction we will have Pyridine = 2.5 mmol Pyridinium bromide = 2.5 mmol	BASIC BUFFER
c	C5H5N = 25.0 mL, 0.20 M HBr = 15.0 mL, 0.20 M	C5H5N = 5 mmol HBr = 3 mmol	3.0 mmol of HBr reacts with 3.0 mmol of pyridine to give 3.0 mmol of pyridinium bromide. So, after the reaction we will have Pyridine = 2.0 mmol Pyridinium bromide = 3.0 mmol	BASIC BUFFER
d	C5H5N = 25.0 mL, 0.20 M HBr = 25.0 mL, 0.20 M	C5H5N = 5 mmol HBr = 5 mmol	5.0 mmol of HBr reacts with 5.0 mmol of pyridine to give 5.0 mmol of pyridinium bromide. So, after the reaction we will have Pyridine = 0.0 mmol Pyridinium bromide = 5.0 mmol	SALT
e	C5H5N = 25.0 mL, 0.20 M HBr = 29.0 mL, 0.20 M	C5H5N = 5 mmol HBr = 5.8 mmol	5.0 mmol of HBr reacts with 5.0 mmol of pyridine to give 5.0 mmol of pyridinium bromide. So, after the reaction we will have Pyridine = 0.0 mmol Pyridinium bromide = 5.0 mmol HBr = 0.8 mmol	SALT + ACID

	After the reaction	Volume of the solution = Volume of pyridine + volume of HBr	New Molarity = mmol/volume(mL)	Remark
a	Pyridine = 5.0 mmol	25.0 mL	0.2 M	Weak base
b	Pyridine = 2.5 mmol Pyridinium bromide = 2.5 mmol	37.5 mL	Pyridine = 0.067 M Pyridinium bromide = 0.067	BASIC BUFFER
c	Pyridine = 2.0 mmol Pyridinium bromide = 3.0 mmol	40.0 mL	Pyridine = 0.05 M Pyridinium bromide = 0.075 M	BASIC BUFFER
d	Pyridine = 0.0 mmol Pyridinium bromide = 5.0 mmol	50.0 mL	Pyridinium bromide = 0.1 M	SALT
e	Pyridine = 0.0 mmol Pyridinium bromide = 5.0 mmol HBr = 0.8 mmol	54.0 mL	Pyridinium bromide = 0.1 M HBr = 0.015 M	SALT + ACID

Kb of pyridine = 1.7 x 10^-9 (data collected)

pKb of pyridine = -log Kb = 8.77

	New Molarity = mmol/volume(mL)	Remark	pH
a	Pyridine = 0.2 M (C)	Weak base	[OH-] = (Kb. C)½ = (1.7 x 10-9 x 0.2)½ = 1.844 x 10-5 M pOH = -log[OH-] = 4.734 pH = 14-pOH = 9.266
b	Pyridine = 0.067 M Pyridinium bromide = 0.067	BASIC BUFFER	Henderson’s equation pH = 14-pOH = 5.23
c	Pyridine = 0.05 M Pyridinium bromide = 0.075 M	BASIC BUFFER	Henderson’s equation pH = 14-pOH = 5.054
d	Pyridinium bromide = 0.1 M (C)	SALT	It is a salt made up of weak base and strong acid.
e	Pyridinium bromide = 0.1 M HBr = 0.015 M (C)	SALT + ACID	Though we have salt and acid, contribution of salt to the pH is insignificant in comparison to the acid contribution. So pH will be majorly decided by acid. [H+] = C pH = -log[H+] = -log(0.015)= 1.824