1. ICE table of CO32- is
:
.........................CO32-
..............+...................H2O
-----------------------> HCO3-
.............+.................OH-
Initial (I)............0.115
M............................................................................0.0
M..................................0.0 M
Change (C)........- y
.....................................................................................+y
...................................y
Equilibrium (E)...(0.115 - y )
M......................................................................y
M.................................y M
Expression of Kb is :
Kb = [HCO3-].[OH-] /
[CO32-]
Kb = y2 / (0.115 - y)
2.1 x 10-4 = y2 (0.115- y)
0.00002415 - 0.00021 y = y2
y2 + 0.00021 y - 0.00002415 = 0
On solving, we have
y = 0.00481
Therefore, Concentration of OH- =
[OH-] = y = 0.00481 M
----------------------------------------------------------------------------------
We know ,
pOH = - log[OH-]
pOH = - log 0.00481
pOH = 2.32
-------------------------------------------------------------
Also,
pH = 14 - pOH
pH = 14 - 2.32
pH = 11.68
-------------------------------------------------------