(a) Since Kb1 >> Kb2, you can ignore the second ionization. However, you\\\'ll n
ID: 1065645 • Letter: #
Question
(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.
The pKb values for the dibasic base B are pKb1 2.10 and pKb2 7.72 Calculate the pH at each of the following points in the titration of 50.0 mL of a 0.80 MB(aq) with 0.80 M HCl(aq). Number (a) before addition of any HCl Number (b) after addition of 25.0 mL of HC Number (c) after addition of 50.0 mL of HCl Number (d) after addition of 75.0 mL of HC Number (e) after addition of 100.0 mL of HC IExplanation / Answer
dibasic base = pKb1 = 2.10
pKb2 = 7.72
millimoles of base = 50 x 0.8 = 40
(a) before addition of any HCl
pOH = 1/2 [pKb1- logC]
pOH = 1/2 [pKb1 - log 0.8]
= 1/2 (2.10 - log 0.8)
= 1.10
pH + pOH = 14
pH = 14 - pOH
= 12.9
pH = 12.9
(b) after addition of 25.0 mL of HCl
millimoles of HCl = 25 x 0.8 = 20
it is half equivalence point. so
pOH = pKb1
pOH = 2.10
pH = 11.9
(c) after addition of 50.0 mL of HCl
millimoles of HCl = 50 x 0.8 = 40
it is equivalence point . at equivalence point
pOH = (pKb1 + pKb2 )/ 2
= (2.10 + 7.72) / 2
= 4.91
pH = 9.09
(d) after addition of 75.0 mL of HCl
it is second half equivalence point . so
pOH = pKb2
pOH = 7.72
pH = 6.28
(e) after addition of 100.0 mL of HCl
millimoles of HCl = 100 x 0.8 = 80
it is second equivalence point.here it is only BH2+ remains.so its concentration
BH2+ millimoles = 80
BH2+ concentration = 80 / total volume
= 80 / (100 + 50)
= 0.53 M
BH2+ + H2O ------------------> BH+ + H3O+
0.53 0 0
0.53 - x x x
Ka2 = x^2 / (0.53 -x)
5.25 x 10^-7 = x^2 / (0.53 -x)
x = 5.27 x 10^-4
[H3O+] = x = 5.27 x 10^-4 M
pH = -log[H3O+] = -log (5.27 x 10^-4)
= 3.28
pH = 3.28
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