Find the pH and Concentration of H2SO3, HSO3-, and (SO3)2-
a) 0.050M H2SO3
b) 0.050M NaHSO3
c) 0.050M Na2SO3
(a). ICE table of H2SO3 is :
.................H2SO3............+...............H2O ---------------------> HSO3-.................+................H3O+
Initial (I).....0.050 M.................................................................0.0 M...................................0.0 M
Change(C)...- y ......................................................................+ y.........................................+ y
Equilibrium (E) ..(0.050-y) M....................................................y M.........................................y M
Expression of Ka is :
Ka = [HSO3-].[H3O+] / [H2SO3]
1.3 x 10-2 = y2/(0.050-y)
y2 + 0.013 y - 0.00065 = 0
On solving, we have
y = 0.0198
So,
[H3O+] = y = 0.0198 M
pH = - log [H3O+]
pH = - log 0.0198
pH = 1.70
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(b).
ICE table of HSO3- is :
.................HSO3-............+...............H2O ---------------------> SO32-.................+................H3O+
Initial (I).....0.050 M.................................................................0.0 M...................................0.0 M
Change(C)...- y ......................................................................+ y.........................................+ y
Equilibrium (E) ..(0.050-y) M....................................................y M.........................................y M
Expression of Ka2 is :
Ka2 = y2/0.050-y
y<<<0.050, so neglect y as compare to 0.050, we have
6.3 x 10-8 x 0.050 = y2
y2 = 0.315 x 10-8
y = 5.61 x 10-5 = [ H3O+ ]
pH = - log 5.61 x 10-5
pH = 4.25
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(c). Kb1 of SO32- = 1.0 x 10-14/6.3 x 10-8 = 1.587 x 10-7
ICE table of SO32- is :
.................SO32-............+...............H2O ---------------------> HSO3-.................+................OH-
Initial (I).....0.050 M.................................................................0.0 M...................................0.0 M
Change(C)...- y ......................................................................+ y.........................................+ y
Equilibrium (E) ..(0.050-y) M....................................................y M.........................................y M
Expression of Kb1 is :
Kb1 = y2/0.050-y
y<<<0.050, so neglect y as compare to 0.050, we have
1.587 x 10-7 x 0.050 = y2
y2 = 7.93 x 10-9
y = 8.90 x 10-5 = [OH-]
pOH = - log[OH-]
pOH = -log 8.90 x 10-5
pOH = 4.05
pH = 14-pOH
pH = 14-4.05
pH = 9.95
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