The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.83. Calculate
ID: 870233 • Letter: T
Question
The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.83. 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).
(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____
Extra Informaiton:
(a) Since Kb1 >> Kb2, you can ignore the second ionization. However, you'll need to use either the quadratic equation or successive approximations to solve for [H ].
(b) 25.0 mL is half way to the first equivalence point, so pOH = pKb1.
(c) 50.0 mL is the first equivalence point. BH is amphoteric, so use pH = 1/2(pKa1 pKa2), where pKa = 14.00
Explanation / Answer
a.
B + H2O < -- > BH+ + HO- [BH+ ] = [HO-]
Kb = ( [BH+ ] [HO-])/(0.5M- [HO-]) = x2 /(0.5-x) = 10-2.1 = 0.008
x2 = 0.004
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.