The ionization constant of phenol is `1.0xx10^(-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 `0.01 M` in sodium phenolate?

Explanation

`{:(,C_(6)H_(5)OH,hArr,C_(6)H_(5)O^(-),+,H^(+),,),("Initial",0.05M,,,,,,),("After disso.",0.05-x,,x,,x,,):}`
`:. K_(a) = (x xx x)/(0.05-x) = 1.0xx10^(-10)` (Given) or `(x^(2))/(0.05) = 1.0xx10^(-10)`
or `x^(2)=5xx10^(-12) or x = 2.2 xx 10^(-6) M`
In presence of 0.01 `C_(6)H_(5)ON a`, suppose y is the amount of phenol dissociated, then at equilibrium
`[C_(6)H_(5)OH]=0.05-y~=0.05, [C_(6)H_(5)O^(-)]=0.01+y~=0.01M, [H^(+)]=y M`
`:. K_(a)=((0.01)(y))/(0.05)=1.0xx10^(-10) ` (Given) or `y=5xx10^(-10)`
`:. alpha = (y)/(c) = (5xx10^(-10))/(5xx10^(-2))=10^(-8)`.