⇒x=1.195×10−4 As H2CO3 is a weak acid so the concentration of H2CO3 will remain 0.034 as 0.034≫>x. x=[H+]=[HCO3−] =1.195×10−4 Now,
HCO3−
x−y
(aq)+H2O(l)⇌CO32−(aq)+H3Oy+(aq) As HCO3−is again a weak acid (weaker than H2CO3 ) with x>>y. K2=
[CO3−][H3O+]
[HCO3−]
=
y×(x+y)
(x−y)
Note : [H3O+]=H+from first step (x) and from second step (y)=(x+y) [As x>>y so x+y≃x and x−y≃x ] So, K2≃
y×x
x
=y ⇒K2=4.8×10−11 =y=[CO32−] So the concentration of [H+]=[HCO3−]=concentrations obtained from the first step. As the dissociation will be very low in second step so there will be no change in these concentrations. Thus the final concentrations are [H+]=[HCO3−]=1.195×10−4 &[CO32−]=4.8×10−11