The ionization constant of phenol is `1.0 xx 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?

Explanation

i. `{:(C_(6)H_(5)OH,hArr,C_(6)H_(5)O^(Θ),+,H^(o+)),(C(1-alpha),,Calpha,,Calpha):}`
`K_(a)` of phenol`=1.0xx10^(-10) (Phenol is a W_(A))`
`:. alpha=sqrt(K_(a)/C)=sqrt((1.0xx10^(-10))/0.05)=sqrt((10^(-10)xx100)/5)`
`sqrt(10^(-10)xx20)=4.47xx10^(-5)M`
ii. Phenate ion, `[C_(6)H_(5)O^(Θ)]=Calpha`
`=(0.05xx4.47xx10^(-5))`
`=2.2xx19^(-6)M`
iii. `pH=- log(2.2xx10^(-6))M=5.65`
iv. When the concentration of phenate is `0.01 M` (salt) and concentration of phenol (acid) is `0.05 M`. It forms acidic buffer.
`:. pH=pK_(a)+log [("Salt")/("Acid")]`
`=10+log(0.01/0.05)`
`=10-0.7=9.3`