The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.43. Calculate

Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.43. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.70 M B(aq) with 0.70 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

Hint: (a) Since Kb1 >> Kb2, you can ignore the second ionization. However, you'll need to use either the quadratic equation or successive approximations to solve for [H ]. (b) 25.0 mL is half way to the first equivalence point, so pOH = pKb1. (c) 50.0 mL is the first equivalence point. BH is amphoteric, so use pH = 1/2(pKa1 pKa2), where pKa = 14.00 pKb. (d) 75.0 mL is half way to the second equivalence point, so pOH = pKb2. (e) 100.0 mL is the second equivalence point. The final product, BH22 , is a weak acid with Ka1 = Kw/Kb2.

Explanation / Answer

pKb1 = 2.10

pKb2 = 7.43

B molarity = 0.70 M

(a) before addition of any HCl

B   +   H2O <----------------------> BH+   + OH-

0.70 0             0   ------------------> initial

0.70 -x                                      x              x ------------------------> equilibrium

Kb1 = [BH+][OH-]/[B]

7.94 x 10^-3 = x^2 / 0.70 -x

x^2 + 7.94 x 10^-3 x - 5.56 x 10^-3 = 0

x = 0.0707

[OH-] = 0.0707 M

pOH = -log (0.0707)

pOH = 1.15

pH + pOH = 14

pH = 12.85

(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 + 7.43) / 2

pOH = 4.765

pH = 9.24

(d) after additon of 75 mL HCl

it is second half equivalence point

pOH = pKb2

pOH = 7.43

pH = 6.57

e) after additon of 100 mL HCl

B millimoles = 50 x 0.70 = 35

BH2+ salt is here only remains

BH2+2 concentration = millimoles / total volume

                                   = 35 / (50 +100)

                                  = 0.233 M

BH2+2 ----------------------------> BH+ + H+

0.233 0 0

0.233-x    x x

Ka2 = x^2 / 0.233 -x

pKa2 = 14 - 7.43= 6.57

Ka2 = 10^-pKa2

Ka2 = 2.69 x 10^-7

2.69 x 10^-7     = x^2 / 0.233 –x

x^2 + 2.69 x 10^-7 x - 6.27 x 10^-8 = 0

x = 2.50 x 10^-4

x = [H+] = 2.50 x 10^-4 M

pH = -log [H+]

pH = 3.60

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