Question

The dibasic compound B (pK_b1 = 4.00, pK_b2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL.

Find the pH after 7.6 mL of acid are added ______

Find the pH after 20 mL of acid are added _______

Answer 1

millimoles of base = 0.100 x 100 = 10

1) millimoles of acid added = 7.6 x 1.00 = 7.6

10 - 7.6 = 2.4 millimoles base left

7.6 millimoles salt formed

we use pKb 1 for this part

[base] = 2.4 / 107.6 = 0.022 M

[salt] = 7.6 / 107.6 = 0.071 M

pOH = pKb1 + log [salt] / [base]

pOH = 4.00 + log [0.071] / [0.022]

pOH = 4.51

pH = 14 - 4.51

pH = 9.49

2) millimoles of acid added = 20 x 1.0 = 20

B + 2HCl ---------> BH2⁺² + 2Cl^-

all base becomes salt

[salt] = 10 / 120 = 0.083

pOH = 1/2 [pKw + pKb + logC]

pOH = 1/2 [14 + 8.00 + log 0.083]

pOH = 10.46

pH = 14 - 10.46

pH = 3.54