Question

Attempt 1 Calculate the pH for each of the cases in the titration of 25.0 mL of 0.100 M pyridine, C5H5N(aq) with 0.100 M HBr(aq) .

A. before addition of any HBr 9.11

B. after addition of 12.5 mL of HBr 8.77

C. after addition of 14.0 mL of HBr 8.88

D. after addition of 25.0 mL of HBr 2.005

E. after addition of 32.0 mL of HBr 1

Answer 1

Here the given pKb is weak base and the dissociation constant for Pyridine is Kb = 1.7*10^-9

Now the dissociation reaction of Pyridine is-

C5H5N + H2O ----------> C5H5NH+ + OH-

Here Kb = [ C5H5NH+] * [ OH-] / [C5H5N]

A. before addition of any HBr

For a weak base, the pH is calculated by the Henderson Hasselblach equation-

pOH = pKb + log [C5H5NH+]/ [C5H5N]

Now for the above reaction, initial [C5H5N] = 0.100 M

Lets calculate [C5H5NH+] by forming ICE table-

	[C5H5NH+]	[C5H5NH+]	[ OH-]
Initial	0.100 M	0	0
Change	-x	+x	+x
Equilibrium	0.100 M -x	+x	+x

Thus Kb = [ C5H5NH+] * [ OH-] / [C5H5N]

1.7*10^-9 = [ +x] * [ +x] / [0.100 M -x]

1.7*10^-9 * [0.100 M -x] = x²

0.17‬*10^-9 - 1.7‬*10^-9 x = x²

x² + 1.7‬*10^-9 x - 0.17‬*10^-9 = 0

Solving this -

x = 0.00013 = 1.3 ‬*10^-4

Thus [ OH-] = x = 1.3 ‬*10^-5 M

Hence pOH = -log [ OH-] = -log [ 1.3 ‬*10^-5 ] = 5 - log 1.3 = 4.88

Hence pH = 14 - pOH = 14 - 5.88 = 9.12‬

B- after addition of 12.5 mL of HBr

Now when we add HBr, then the H+ of HBr will react with C5H5N and form more C5H5NH+. That means amount of C5H5N will decrease from the solution and amount of C5H5NH+ will increase.

Now initial C5H5N was present V1 = 25.0 mL , M1 = 0.1 M

   initial HBr was added V1 = 12.5 mL , M1 = 0.1 M

After addition, final volume V2 = 25.0 mL + 12.5 mL = 27.5 mL

Thus the new concntration of C5H5N = M2 = M1V1/V2 = 0.1 M * 25.0 mL / 27.5 mL =0.09 M

Similarly new concntration of HBr = M2 = M1V1/V2 = 0.1 M * 12.5.0 mL / 27.5 mL =0.045 M

Now forming ICE table-

	[ C5H5N]	[HBr]	[C5H5NH+]
I	0.09 M	0.045 M	0
C	- 0.045 M	- 0.045 M	+0.045 M
E	0.045 M	0	0.045 M

Thus putting these values for

pOH = pKb + log [C5H5NH+]/ [C5H5N]

= -log Kb + log 0.45]/ [0.45]

= -log (1.7*10^-9) + log 1

= 8.77 + 0

pOH = 8.77

Thus pH = 14 - pOH = 14 - 8.77 = 5.23‬

C- Similar calculations as B will be done for the rest.

initial C5H5N was present V1 = 25.0 mL , M1 = 0.1 M

   initial HBr was added V1 = 14 mL , M1 = 0.1 M

After addition, final volume V2 = 25.0 mL + 14 mL = 29 mL

Thus the new concntration of C5H5N = M2 = M1V1/V2 = 0.1 M * 25.0 mL / 29 mL =0.086 M

Similarly new concntration of HBr = M2 = M1V1/V2 = 0.1 M * 14.0 mL / 29 mL =0.048 M

Now forming ICE table-

	[ C5H5N]	[HBr]	[C5H5NH+]
I	0.086 M	0.048 M	0
C	- 0.048 M	- 0.048 M	+0.048 M
E	0.038‬ M	0	0.048 M

Thus putting these values for

pOH = pKb + log [C5H5NH+]/ [C5H5N]

= -log Kb + log [0.048]/ [0.038]

= -log (1.7*10^-9) + 0.101

= 8.77 + 0.101

pOH = 8.871

Thus pH = 14 - pOH = 14 - 8.871 = 5.129‬

The rest calculations will be done in similar way