7.44 The ionization constant of phenol is 1.0 × 10–10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be its degree of ionization if the solution is also 0.01M in sodium phenolate?

Ionization of phenol:
                             C6H5OH+H2O   C6H5O- + H3O+Initial conc.                0.05                          0              0At equilibrium            0.05-x                     x              xKa =C6H5O-H3O+C6H5OHKa=x×x0.05-x   
As the value of the ionization constant is very less, x will be very small.  Thus, we can ignore x in the denominator.
x=1×1010×0.05=5×1012=2.2×106M=[H3O+]Since H3O+=C6H5O-,C6H5O-=2.2×10-6 M.
Now, let  be the degree of ionization of phenol in the presence of 
0.01 M C6H5ONa.C6H5ONaC6H5O-+Na+Conc,                         0.01
Conc.  0.05 - 0.05 α             0.05 α         0.05α
[C6H5OH]=0.050.05α ; 0.05M[C6H5O]=0.01+0.05α ; 0.01M[H3O+]=0.05αKa=[C6H5O][H3O+][C6H5OH]Ka=(0.01)(0.05α)0.051.0×1010=.01αα=1×108