In: Chemistry
The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.58. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.50 M B(aq) with 0.50 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
pKb1 = 2.10
pKb2 = 7.58
B molarity = 0.50 M and volume= 50 mL
(a)
before addition of any HCl
B + H2O <----------------------> BH+ + OH-
0.50 0 0 ------------------> initial
0.50 -x x x ------------------------> equilibrium
Kb1 = [BH+][OH-]/[B]
7.94 x 10^-3 = x^2 / 0.50 -x
On solving,
x = 0.059 M
[OH-] = 0.059 M
pOH = -log (0.059)
pOH = 1.23
pH + pOH = 14
pH =12.77
(b)
after additon of 25 mL HCl
it is first half equilivalence point .
at this point . pOH = pKa1
pOH = 2.10
pH + pOH = 14
pH = 11.90
c)
after additon of 50 mL HCl
it is first equivalence point
here : pOH = (pKb1 + pKb2)/2
pOH = (2.10 + 7.58) / 2
pOH = 4.84
pH = 9.16
(d)
after additon of 75 mL HCl
it is second half equivalence point
pOH = pKb2
pOH = 7.58
pH = 6.42
e)
after additon of 100 mL HCl
B millimoles = 50 x 0.50 = 25
BH2+ salt is here only remains
BH2+2 concentration = millimoles / total volume
= 25 / (50 +100)
= 0.167 M
BH2+2 ----------------------------> BH+ + H+
0.167 0 0
0.167 -x x x
Ka2 = x^2 / 0.167 -x
pKa2 = 14 - 7.58 = 6.42
Ka2 = 10^-pKa2
Ka2 = 3.80 x 10^-7
3.80x 10^-7 = x^2 / 0.167 -x
on solving,
x = 2.5 x 10^-4
x = [H+] = 2.5 x 10^-4 M
pH = -log [H+]
pH =3.60