Homework Answers
Sol :-
ICE table is :
............................H2A (aq) ................+................H2O (l) <------------------> HA- (aq)...............+..............H3O+ (aq)
Initial (I)................0.125 M..........................................................................0.50 M.....................................0.0 M
Change (C).............-y....................................................................................+y..........................................+y
Equilibrium (E)......(0.125-y) M...................................................................(0.50+y) M..................................y M
Expression of Ka1 is :
Ka1 = [HA-].[H3O+] / [H2A]
1.2 x 10-2 = (0.50+y).y / (0.125-y)
1.2 x 10-2 (0.125-y) = (0.50+y).y
y2 + 0.512 y - 0.0015 = 0
On solving
y = 0.00291
So,
Equilibrium concentration of H2A = (0.125-y) M = 0.125 - 0.00291 = 0.122 M
Equilibrium concentration of HA- = (0.50+y) M = 0.50 + 0.00291 = 0.503 M
Equilibrium concentration of H3O+ = y M = 0.00291 M
------------------------------------------------------------------------------
ICE table for HA- is :
............................HA- (aq) ................+................H2O (l) <------------------> A2- (aq)...............+..............H3O+ (aq)
Initial (I)................0.503 M..........................................................................0.0 M.....................................0.00291 M
Change (C).............-y....................................................................................+y..........................................+y
Equilibrium (E)......(0.503-y) M...................................................................y M...................................(0.00291+y) M
Expression of Ka2 is :
Ka2 = [A2-].[H3O+] / [HA-]
4.7 x 10-7 = (0.00291+y).y / (0.503-y)
4.7 x 10-7 (0.503-y) = (0.00291+y).y
y2 + 0.00291047 y - 2.36 x 10-7 = 0
On solving
y = 8.0 x 10-5 (very small)
So,
|
Equilibrium concentration of H2A = (0.125-y) M = 0.125 - 0.00291 = 0.122 M Equilibrium concentration of HA-= (0.503-y) M = 0.503 - 8.0 x 10-5 = 0.503 M Equilibrium concentration of A2- = y M = 8.0 x 10-5 M Equilibrium concentration of H3O+= (0.00291+y) M = 0.00291 M |
Because,
pH = - log [H3O+]
pH = - log 0.00291 M
pH = 2.54
| Hence, pH of the solution = 2.54 M |
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