The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.64. Calculate

Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.64. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.90 M B(aq) with 0.90 M HCl(aq).

The pKb values for the dibasic base B are pKb1-2.10 and pK62 = 7.64. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.90 M B(aq) with 0.90 M HCI(aq). Number (a) before addition of any HCI Number (b) after addition of 25.0 mL of HC 0 Number (c) after addition of 50.0 mL of HCi Number (d) after addition of 75.0 mL of HCl Number (e) after addition of 100.0 mL of HCI

Explanation / Answer

First let's calculate the moles of B:

moles of B = 0.059 L * 0.90 mol/L = 0.045 moles

a) before addition of HCl, the overall reaction is:

r: B + H2O --------> BH+ + OH-   Kb1 = 10-2.10 = 7.94x10-3 Kb2 = 10-7.64 = 2.29x10-8

i. 0.90 0 0

e. 0.90-x x x

Kb = [OH][BH] / [B]

7.94x10-3 = x2 / 0.90-x but Kb is low, so 0.90-x = 0.90

7.94x10-3 * 0.90 = x2

x = 0.085 M = [OH]

pOH = -log(0.085) = 1.07

pH = 14-1.07 = 12.93

b) After the addition of 25 mL, let's see first when the first equivalence point is reached:

MaVa = MbVb

0.90 * 50 / 0.90 = Va

Va = 50 mL.

So, at 25 mL we're at half equivalence point, this means that pKb = pOH = 2.1 so pH:

pH = 14-2.1 = 11.9

c) At the equivalence point, we have the following:

pH = pKa1 + pKa2 / 2 or is the same with: pOH = pKb1 + pKb2 / 2

pOH = 2.1+7.64 / 2 = 4.87

pH = 14-4.87 = 9.13

d) After adding 75 mL, we are at the half of the second equivalence point so:

pH = pKa2; pOH = pKb2 = 7.64

pH = 14-7.64 = 6.36

e) After adding 100 mL, we are at the second and final equivalence point, so we have the following:

BH2+ + H2O ---------> BH+ + H3O+

Ka2 = [BH][H3O] / BH2

Ka2 = 1x10-14 / 2.29x10-8 = 4.37x10-7

4.37x10-7 = y2 / (0.045/0.150)

4.37x10-7 * 0.3 = y2

y = [H3O+] = 3.62x10-4 M

pH = -log(3.62x10-4) = 3.44

Hope this helps

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