Determine the pH of a 0.125 M Na2CO3 solution. (Ka of H2CO3 = 5.6 x 10-11)
Sol . As Initial Concentration of Na2CO3 = [Na2CO3] = 0.125 M
So , [ Na2CO3 ] = [ CO32- ] = 0.125 M
Now ,
Reaction of CO32- with H2O :
CO32- + H2O <----> HCO3- + OH-
Initial 0.125 0 0
Change - x + x + x
Equilibrium 0.125 - x x x
Also , Kb = Kw / Ka = 10-14 / ( 5.6 × 10-11 )
= 0.1785 × 10-3
So , Kb = [HCO3-] [OH-] / [CO32-]
0.1785 × 10-3 = x2 / ( 0.125 - x )
As Kb is very small , so , 0.125 - x = 0.125
Therefore , 0.1785 × 10-3 × 0.125 = x2
x2 = 0.223125 × 10-4
x = 0.4724 × 10-2
So , [OH-] = x = 0.4724 × 10-2 M
Now , pOH = - log[OH-] = - log (0.4724 × 10-2 )
= 2.32
Therefore , pH = 14 - pOH = 14 - 2.32 = 11.68
Determine the pH of a 0.125 M Na2CO3 solution. (Ka of H2CO3 = 5.6 x 10-11)
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