Consider the titration of 50ml of 010M pyridine C5H5NKb17109

Consider the titration of 50ml of 0.10M pyridine (C5H5N,Kb=1.7*10^-9) with 0.10M HCl. What is the pH of the solution at the following points along the titration curve:

1) Before any titrant is added

2) After 15ml of titrant is added

3) After 25ml of titrant is added

4) After 50ml of titrant is added

5) After 60ml of titrant is added

Solution

1) C5H5N + H2O <--------> C5H5NH+ + OH- ,

at equi [C5H4N] = ( 0.05x0.1-x)/0.05, [C5H5NH+] =[OH-] = x/0.05

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

1.7 x10^ -9 = x^2/(0.05)^2/(0.005-x)/0.05)

solving we get x = 6.5 x10^ -7 , [OH-] = 6.5x10^ -7/0.05 =1.3 x10^ -5,

pOH = -log(1.3 x10^ -5) = 4.88 , pH = 9.12

2) when 15 ml 0.1 HCl is added,

pOH = pkb + log[C5H5NH+]/[C5H5N] , HCl reacts withj pyridine to give C5H5NH+ ,

hence we have

[C5H5N] = (0.005- (0.015x0.1)]/(0.05+0.015) = 0.0538,

[C5H5NH+] = (0.015x0.1)/(0.05+0.015) = 0.023 , pKb = -log(1.7x10^ -9) = 8.77

pOH = 8.77 + log(0.023/0.0538) = 8.4

pH = 5.6,

3) 25 ml 0.1 HCl added, then [C5H5N] = (0.005-(0.025x0.1)]/(0.05+0.025) = 0.033,

[C5H5NH+] =(0.025x0.1)/(0.05+0.025) = 0.0333

pOH = 8.77+ log(0.033/0.033) = 8.77,

pH = 5.23

4) when 50 ml of 0.1M HCl is added we get equivalence point i.e moles of acid = moles of base,

0.05x0.01 = 0.05x0.01,

now we have ( 0.05x0.1) moles of C5H5NH+ ,

C5H5NH+ + H2O <----> C5H5N + H+ (aq)

at equi [C5H5NH+] = (0.005-1)/(0.05+0.05) , [H+]=[C5H5N] = x/0.1,

Ka = Kw/Kb = 10^ -14/1.7x10^ -9 = 5.88 x10^ -6,

5.88 x10^ -6 = x^2/(0.1)^2/(0.005-x)/0.1, solving we get

x = H+ moles = 5.422 x10^ -5, [H+] = 5.422x10^ -5/0.1 = 5.422 x10^ -4,

pH = -log(5.422x10^ -4) = 3.266,

5) when 60 ml HCl is added we get excess H+ ,

excess H+ = (0.06x0.1)-(0.05x0.1) = 0.001,

[H+] =(0.001)/(0.06+0.05) = 0.00909,

pH = -log(0.00909) = 2.04