Question

Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.120 M pyridine, C5H5N(aq) with 0.120 M HBr(aq):

Kb= 1.7x10^-7, so Ka must equal 5.88x10^-24

(a) before addition of any HBr, pH=?

(b) after the addition of 12.5mL of HBr, pH=?

(c) after the addition of 24.0mL of HBr, pH=?

(d) after addition of 25.0mL of HBr, pH=?

(e) after addition of 34.0mL of HBr, pH=?

Answer 1

a) C5H5N + H2O <--------> C5H5NH+ + OH-

Kb = [C5H5NH+][OH-]/[C5H5N] = 1.7×10^-7

at equillibrium

x^2 /( 0.120 - x ) = 1.7×10^-7

we can assume

( 0.120 - x ) = 0.120 for this Kb value

x^2/0.120 = 1.7 × 10^-7

x = 1.43×10^-4

[OH-] = 1.43×10^-4

pOH = 3.84

pH = 14 - 3.84

= 10.16

b) C5H5N + HBr --------> C5H5NH+ + Br-

this is 1: 1 reaction

25ml is equivalence poibt

12.5ml is half equivalence point

at half equivalence point

pH = pKb

pKb = - logKb = -log(1.7×10^-7)= 6.77

Therefore,

pH 6.77

c) No of mole of initial C5H5N= (0.120mol/1000ml)×25ml = 0.003

No of mole of HBr added =( 0.120mol/1000ml)×24ml = 0.00288

No of mole of C5H5N remains = 0.003 - 0.00288= 0.00012

No of mole of C5H5NH+ formed = 0.00288

Total volume = 49ml

[ C5H5N] = (0.00012mol/49ml)×1000ml = 0.00245M

[C5H5NH+ ] = (0.00288mol/49ml)×1000ml = 0.0588M

Henderson - Hasselbalch equation is

pH = pKb + log([BH+]/[B])

= 6.77 + log(0.0588M/0.00245M)

= 6.77 + 1.38

= 8.15

pH = 14 - 8.15

= 5.85

d) At equivalence point

[ C5H4NH+ ] = 0.0600M

C5H5NH+ + H2O ------> C5H5N + H3O+

Ka = [C5H5N][H3O+]/[C5H5NH+]= 5.88×10^-8

x^2/(0.0600 -x) = 5.88×10^-8

we can assume 0.0600- x = 0.0600

x^2 = 3.53×10^-9

x = 5.94×10^-5

[ H3O+ ] = 5.94 ×10^-5

pH = 4.22

e) 9ml excess HBr added

No of mole of HBr =( 0.120mol/1000ml)×9ml = 0.00108

Total volume = 59ml

[ HBr ] = (0.00108mol/59ml)×1000ml = 0.0183M

HBr is strong acid

[ HBr ] = [ H+]

Thetefore,

[ H+ ] = 0.0183M

pH = -log(0.0183)

= 1.74

Note: Please , verify your Kb value of pyridine. My calculation is based on your Kb value.