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^(+)),(C(1-alpha),Calpha,Calpha):}`
`K_(a)=1.0xx10^(-10)`
`K_(a)=([C_(6)H_(5)O^(-)][H^(+)])/([C_(6)H_(5)OH])`
or `1.0xx10^(-10)= (Calpha.Calpha)/(C(1-alpha))`
`:. Calpha^(2)=1.0xx10^(-10)` , `(.: 1-alpha~~1)`
`:. alpha^(2)=(1.0xx10^(-10))/(0.05)` , `(.: C=0.05M)`
`:. alpha= 4.47xx10^(-5)`
`:. [C_(6)H_(5)O^(-)]=C.alpha=0.05xx4.47xx10^(-5)`
`=2.2xx10^(-6)M`
If `[C_(6)H_(5)O^(-)]=0.01M, [H^(+)]=0.05xxalpha`,
`[C_(6)H_(5)OH]= 0.05(1-alpha)= 0.05`
(`.: alpha` is small due to common ion effect)
`:. 1.0xx10^(-10)= (0.05xxalphaxx0.01)/(0.05)`
`:. alpha= 1.0xx10^(-8)`