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

Phenol , PhOH ionizes as :
`PhOH+H_2O hArr PhO^(-) + H_3O^-`
Since the value of ionization constant is small, the degree of ionization
`alpha=sqrt(K_a/c)=sqrt((1.0xx10^(-10))/0.05)`
`=4.47xx10^(-5)`
`[PhO^-]=cx=0.05xx4.47xx10^(-5)=2.23xx10^(-6)` M
`[H_3O^+]=cx=0.05xx4.47xx10^(-5)=2.23xx10^(-6)` M
`pH=-log [H_3O^+]=-log (2.23xx10^(-6))`
=-log 2.23 + 6 =-0.0348 + 6= 5.65
When the solution is 0.01 M in sodium phenate (PhONa),
`PhOH hArr PhO^(-) + H^+`
`PhONa hArr PhO^(-) + Na^+`
Concentration of `PhO^-` from phenol =x (say)
Since sodium phenate is completely ionized , the concentration of phenate from sodium phenate =0.01 M
Total `[PhO^-]`=0.01 +x
Conc. of phenol left unionized =0.05-x
Now x being very very small
`[PhO^-]`=0.01 + x =0.01
[PhOH]=0.05 - x = 0.05
`K_a=([PhO^-][H^+])/([PhOH])`
`1.0xx10^(-10)=(0.01xx[H^+])/0.05`
`[H^+]=(1.0xx10^(-10)xx0.05)/0.01=5.0xx10^(-10)`
`[H^+]=5.0xxx10^(-10)`
`pH=-log [H^+]=-log (5xx10^(-10))`
=-log 5+10 =-0.70+10=9.30