Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.76. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.65 M B(aq) with 0.65 M HCl(aq). (

a) before addition of any HCl (.

b) after addition of 25.0 mL of HCl (

c) after addition of 50.0 mL of HCl

(d) after addition of 75.0 mL of HCl (

e) after addition of 100.0 mL of HCl

Answer 1

ans)

pKb1 = 2.10 , pKb2 = 7.76

base B millimoles = 50 x 0.65 = 32.5

HCl concentration = 0.65 M

a)

before addition of any HCl

pOH = 1/2 (pKb1 - logC)

pOH = 1/2 (2.10 - log 0.65)

pOH = 1.143

pH + pOH = 14

pH = 12.86

(b)

25 ml HCl added

it is half equivalence point

at this point pOH = pKb1

pOH = 2.10

pH = 11.90

(c)

after addition of 50.0 ml HCl

it is equivalence point .

pOH = (pKb1 + pKb2 ) / 2

pOH = (2.10 + 7.76) / 2

pOH = 4.93

pH = 9.07

(d)

after addition of 75 mL HCl

it is second half equivalence point

at this point pOH = pKb2

pOH = 7.76

pH = 6.24

(e)

after addition of 100.0 mL HCl

Base millimoles = 0.65 x 50 = 32.5

salt only remained its molarity = 32.5 / (50 + 100) = 0.216 M

it is the salt of weak base strong acid

pH = 7 - 1/2 [pKb2 +logC]

pH = 7 - 1/2 [7.76 + log 0.216]

pH = 3.45