The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.77. Calculate
ID: 929122 • Letter: T
Question
The pKb values for the dibasic base B are pKb1 = 2.10 and pKb2 = 7.77. Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.75 M B(aq) with 0.75 M HCl(aq).
(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 – pKb.
(d) 75.0 mL is half way to the second equivalence point, so pOH = pKb2.
(e) 100.0 mL is the second equivalence point. The final product, BH22 , is a weak acid with Ka1 = Kw/Kb2.
PLEASE ANSWER ALL PARTS. THANK YOU!
Explanation / Answer
a)
B + H2O <----------------> BH+ + OH-
0.75 0 0 -----------------> initial
0.75-x x x -------------------> equilibrium
Kb1 = [BH+][OH-]/[B]
Kb1 = x^2 / 0.75 -x
7.94 x 10^-3 = x^2 / 0.75 -x
x = 0.0733
x = [OH-] = 0.0733 M
pOH = -log [0.0733 ]
pOH = 1.14
pH + pOH = 14
pH = 14-pOH
pH = 12.86
c) after addition of 50.0 ml :
this is first equivalence point. so here
pOH = pKb1 + pKb2 / 2
= 2.1 + 7.77 / 2
= 4.94
pH = 9.06
e) addition of 100 ml:
here acid salt remains
concentration of acid salt = 0.75 x 50 / total volume
= 37.5 / (100+50)
= 0.25
BH2+ -------------------> BH+ + H+
0.25 0 0
0.25-x x x
Ka2 = [BH+][H+] / [BH2+]
5.89 x 10^-7 = x^2 / 0.25 - x
x = 3.83 x 10^-4
[H+] = 3.83 x 10^-4 M
pH = 3.42
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