A 0.371-M aqueous solution of
C5H5N
(pyridine) has a pH of 9.37.
Calculate the pH of a buffer solution that is
0.371 M in
C5H5N and
0.159 M in
C5H5NH+.
pH =
First, identify the pKb from the intitial concentration:
B+ H"O <-< HB+ + OH-
Kb = [BH+][OH-]/[B]
pOH = 14-pH = 14-9.37 = 4.63
[OH-] = 10^-pOH = 10^-4.63 =2.34410^-5
[BH+] = [OH-] =2.34410^-5
[B] = 0.371-2.34410^-5 = 0.3568707 M
Kb = (2.34410^-5)(2.34410^-5) /(0.3568707) = 0.00055940682 = 5.59*10^-4
now
pKb = -log(Kb) = -log(5.59*10^-4) = 3.2525
p
for the buffer
pOH = pKB + log(C5H5NH+/C5H5N)
pOH = 3.2525+ log(0.159/0.371) = 2.8845
pH = 14-2.8845 = 11.1155
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