35)
Ionization equilibrium of HC5H5+ is
HC5H5N+(aq) + H2O(l) <--------> C5H5N(aq) + H3O+(aq)
Ka= [C5H5N][H3O+] /[HC5H5N+]
Ka= Kw/Kb = 1.00 ×10-14/1.7 ×10-9 = 5.88 ×10-6
at equilibrium
[HC5H5N+] = 0.80 - x
[ C5H5N] = x
[H3O+] = x
so,
x2/ ( 0.80 - x) = 5.88 ×10-6
solving for x
x = 0.002166
[H3O+] = 0.002166M
pH = -log[H3O+]
pH = - log(0.002166)
pH = 2.66
36)
Ionization equilibrium of C7H5O2- is
C7H5O2-(aq) + H2O(l) <------> HC7H5O2(aq) + OH-(aq)
Kb = [HC7H5O2][OH-]/[C7H5O2-]
Kb = Kw/Ka = 1.00 ×10-14/6.4×10-5 = 1.56×10-10
at equilibrium
[C7H5O2-] = 0.050 - x
[HC7H5O2-] = x
[OH-] = x
so,
x2/(0.050 - x) = 1.56 ×10-10
solving for x
x = 2.793×10-6
[OH-] = 2.793 ×10-6M
pOH= -log[OH-]
pOH = -log(2.793 ×10-6)
pOH = 5.56
pH = 14 - 5.56
pH = 8.44
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