Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.44. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.80 M B(aq) with 0.80 M HCl(aq).

a)before any HCl is added

b)25ml HCl

c)50ml HCl

d)75ml HCl

e)100ml HCl

Answer 1

Since the base B is dibasic it will need 2 equivalents of acid to completely neutralize

Since we have pKb1 = 2.10 and pKb2 = 7.44

The pH corresponds to the first pKb1 = 2.10 Kb1 will be 10^-2.1 = 7.9 x 10^-3

in water the base B gives

B + H₂O BH⁺ + OH^-

At equilibrium 7.9 x 10^-3 = x²/0.8-x

x = 7.95 x 10^-2 This is the concentration of OH^-

[H⁺] = Kw/[OH^-]

[H⁺] = 1 x 10^-14/7.95 x 10^-2 = 1.26 x 10^-13

pH = -log[H⁺]

pH = 12.9 before any HCl is added

At any point between the initial point and the first End point, we have a buffer containing B and BH+. The pH is calculated from the HH equation for the weak acid, BH+, with Ka2 (for BH2+) = Kw/Kb1 = 10^-14/7.9 x 10^-3 = 1.26 x 10^-12

pKa2 = 11.89

pH = PKa2 + log [B]/[BH+]

If Va (volume of titrant acid) = 25mL, and Ve (amount of acid to reach the first E.P.) = 50 mL, then

pH = 11.89 + log (50-25)/25

pH = 11.89 after adding 25 mL 0.8 M HCl.

When 50 mL of 0.8 M HCl has been added we reach the first end point

At the first end point., B has been converted into BH+, the intermediate form of the diprotic acid, and BH+ is both an acid and a base.

For an amphoteric system the pH has been simplied for calculation as

pH = 1/2 [(14-pKb2)+(14-pKb1)]

pH = 1/2[(14-2.1)+(14-7.44)]

pH = 9.23 at 50mL 0.8 mL HCl.

d) At any point between after the first end point to the second end point there is a buffer containing BH+ (the base) and BH₂²⁺ (the acid).

pH = pKa1 + log[BH⁺]/[BH₂²⁺] pKa1 = 17-pKb2 = 14-7.44 = 6.56

If Va (volume of titrant acid) = 75mL, and Ve (amount of acid to reach the second E.P.) = 100 mL, then

so the concentration [BH⁺] and [BH₂²⁺] are now equal

pH = 6.56 + log 1

pH = 6.56

I am answering 4 subsections of this question