Question

The dibasic compound B (pK_b1 = 4.03, pK_b2 = 8.06) was titrated with 1.12 M HCl. The initial solution of B was 0.112 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added: V_a = 0.00, 2.00, 5.00, 9.00, 10.00, 11.00, 15.00, 19.00,20.00, and 22.00 mL. (Assume K_w = 1.01 ✕ 10⁻¹⁴.)

Answer 1

pKb1 = 4.03

pKb2 = 8.06

B molarity = 0.112 M

millimoles of B = 0.112 x 100 = 11.2

1 ) 0.00 mL HCl added

pOH = 1/2 [pKb1 - logC]

pOH = 1/2 [4.03 - log 0.112]

pOH = 2.49

pH + pOH = 14

pH = 11.51

2) 2.00 mL added

HCl millimoles = 2 x 1.12 = 2.24

B + HCl ------------------> BH+

11.2   2.24                        0    -------------------> initial

8.96     0                          2.24 ---------------> after reaction

pOH = pKb1 + log [BH+/B]

pOH = 4.03 + log (2.24/ 8.96)

pOH = 3.43

pH = 10.57

3) 5.00 mL added

HCl millimoles = 5 x 1.12 = 5.6

B + HCl ------------------> BH+

11.2 5.6                      0    -------------------> initial

5.6    0                         5.6--------------> after reaction

pOH = pKb1 + log [BH+/B]

pOH = 4.03 + log (5.6/ 5.6)

pOH = 4.03

pH = 9.97

4) 9 .00 mL added

HCl millimoles = 9 x 1.12 =10.08

HCl millimoles remains = 10.08 - 5.6 = 4.48

BH+    + H+ ---------------> BH2+

5.6        4.48                     0

1.12       0                         4.48

pOH = pKb2 + log [BH2+]/[BH+]

pOH = 8.06 + log (4.48 / 1.12)

pOH = 8.66

pH + pOH = 14

pH = 5.34