Question

In: Chemistry

Calculate the pH for each of the following cases in the titration of 25.0 mL of...

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=?

Solutions

Expert Solution

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.

  


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