Question

The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.33. 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

Answer 1

a)

since pKb1 >> pKb2

then

[OH-] = sqrt(Kb) = sqrt(10^-2.1) = 0.089 approx

pH = 14 + log(OH) = 14 + log(0.089 ) = 12.949

b)

mmol of base = MV = 50*0.7 = 35 mmol

mmol of aicd = MV = 0.7*25 = 17.5

this is the FIRST HALF equivalence point

so

pOH = pKb1 + log(BH+/B)

pOH = 2.10 + log(35 -17.5)/(17.5)

pOH = 2.10

pH = 14- 2.10

pH = 11.9

c)

this is the FIRST equivalence point

meaning that

pOH = 1/2*(pKb1 + pKb2)

pOH = 1/2*(2.10+7.33)

pOH = 4.72

pH = 14-4.72 = 9.28

d)

V = 75 mL

this is the SECOND half equivalnece point... then

pOH = pKb2 + (HBH++ /BH+)

pOH = 7.33

pH = 14-7.33

pH = 6.67

e)

this is equivalence point

BHH++ + H2O <-> bH+ + H3O+

Ka = [BH+][H3O+]/[BHH+]

(10^-14)/(10^-7.33) = x*x/(M)

x = sqrt((10^-14)/(10^-7.33)

x = 0.000462

pH = -log(0.000462) = 3.34