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

(I). Let C moles of phenol be dissolved in water to form the solution and let `alpha` be the degree of ionisation of phenol. The concentration of different species at equilibrium point is :

`alpha = sqrt((K_(a))/(C)) =sqrt((1.0xx10^(-10))/(0.05)) =sqrt((1.0 xx10^(-10))/(5.0 xx 10^(-2))) = sqrt(0.2xx10^(-8))`
`=0.447 xx 10^(-4) =4.47 xx 10^(-5)`
`[PhO^(-)] Calpha = (0.05 M) xx 4.47 xx 10^(-5) =2.23 xx 10^(-6) M`
`pH =- log [H_(3)O^(+)] =- log (2.23 xx 10^(-6))`
`=(6- log 2.23 ) =6-0.3483 = 5.65`
(II). When phenol (PhOH) is mixed wiht 0.01M sodium phenate solution the ionisation may be represented as :
`PhOH hArr PhO^(-) + H^(+) ,PhONa to PhO^(-) + Na^(+)`
Conc. of `PhO^(-)` ions due to ionisation of sodium phenate (completely ionised) `=0.01 M`
Let the concentration of `PhO^(-)` ions from `PhOH =x M`
`:.` Total concentration of `PhO^(-)` ions i.e.,`[PhO_^(-)] =0.01 +x ~~ 0.01 M`
(x being very small can be neglected)
Concentration of PhOH left unionised `=0.05 -x ~~ 0.05M`
Ionisation constant `(K_(a)) " for " PhOH = [[PhO^(-) ][H^(+)]]/[[PhOH]]`
`1xx 10^(-10) =((0.01M) xx [H^(+)]]/((0.05 M))`
`" or "" "[H^(+)] =(1.10^(-10) xx0.05)/(0.01) =5.0 xx 10^(-10)`
Degree of ionisation of phenal i.e., `alpha =(5.0 xx 10^(-10))/(5.0 xx 10^(-2)) =10^(-8)`