The pKb values for the dibasic base B are pKb1 2.10 and pKb2 - 7.37. Calculate t

Question

The pKb values for the dibasic base B are pKb1 2.10 and pKb2 - 7.37. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.50 M B(aq) with 0.50 M HCl(aq) Number For part (a), you only need to consider the first ionization because KhiKh2 (a) before addition of any HCl 12.39 Blaq)+ H,oMBH (aq)+ OH (aq) Number 0.50 M + X (b) after addition of 25.0 mL of HCl 11.9 0.50 M-x Number (c) after addition of 50.0 mL of HCI 9.27 b1 b1 0.50 M-x Number Use pKb1 to find the value of Kb1, (d) after addition of 75.0 mL of HCl 6.63 2.10 K, 10-pAb-10 b1 Numberthen solve for x, which is equal to [OH DOH-_ log[OH-], and pH-14.00-DOH (e) after addition of 100.0 mL of HCI 3.56

Explanation / Answer

a)

Kb1 = 10^-pKb1

Kb1 = 7.94 x 10^-3

B    + H2O -----------------------> BH+    +   OH-

0.50                                            0              0

0.50-x                                       x                x

Kb = x^2 / 0.50-x = 7.94 x 10^-3

x^2 + 7.94 x 10^-3 x - 3.97 x 10^-3 = 0

x = 0.0592

[OH-] = 0.0592 M

pOH = -log [OH-]

pOH = 1.23

pH + pOH = 14

pH = 12.77

e) after additon of 100 mL HCl

B millimoles = 50 x 0.50 = 25

BH2+ salt is here only remains

BH2+2 concentration = millimoles / total volume

                                   = 25 / (50 +100)

                                  = 0.167 M

BH2+2 ----------------------------> BH+ + H+

0.167 0 0

0.167 -x    x x

Ka2 = x^2 / 0.267 -x

pKa2 = 14 - 7.37 = 6.63

Ka2 = 10^-pKa2

Ka2 = 2.34 x 10^-7

2.34 x 10^-7   = x^2 / 0.167 –x

x^2 + 2.34 x 10^-7 x - 3.92 x 10^-8 = 0

x = 1.98 x 10^-4

x = [H+] = 1.98 x 10^-4 M

pH = -log [H+]

pH = 3.70

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.