Question

Find the pH and Concentration of H2SO3, HSO3-, and (SO3)2-

a) 0.050M H2SO3

b) 0.050M NaHSO3

c) 0.050M Na2SO3

Answer 1

(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 K_a is :

K_a = [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 ---------------------> SO₃^2-.................+................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 K_a2 is :

K_a2 = 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). K_b1 of SO₃^2- = 1.0 x 10^-14/6.3 x 10^-8 = 1.587 x 10^-7

ICE table of SO₃^2- is :

.................SO₃^2-............+...............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 K_b1 is :

K_b1 = 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