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?

Answer

[phenolate ion] `= 2.2 xx 10^(-6), alpha = 4.47 xx 10^(-5), alpha` in sodium phenolate `= 10^(-8)`

Explanation

Ionization of phenol :
`C_(6)H_(5)OH + H_(2)O harr C_(6)H_(5)O^(-) + H_(3)O^(+)`
`{:("Initial conc.",0.05,0,0),("At equilibrium",0.05-x,x,x):}`
`K_(a) = ([C_(6)H_(5)O^(-)][H_(3)O^(+)])/([C_(6)H_(5)OH])`
`K_(a) = (x xx x)/(0.05 - x)`
As the value of the ionization constant is very less, x will be very small Thus, we can ignore x in the denominator
`:. x = sqrt(1 xx 10^(-10) xx 0.05)`
`= sqrt(5 xx 10^(-12)`
`= 2.2 xx 10^(-6) M = [H_(3)O^(+)]`
Since `[H_(3)O^(+)] = [C_(6)H_(5)O^(-)]`,
`[C_(6)H_(5)O^(-)] = 2.2 xx 10^(-6) M`
Now, let `alpha` be the degree of ionization of phenol in the presence of `0.01 M C_(6)H_(5)ONa`.
`{:(,C_(6)H_(5)ONararrC_(6)H_(5)O^(-),+Na^(+)),("Conc.",,0.01):}`
Also,
`{:(,C_(6)H_(5)OH,+,H_(2)Oharr,C_(6)H_(5)O^(-),+,H_(3)O^(+)),("Conc.",0.05-0.05alpha,,,0.05alpha,,0.05alpha):}`
`[C_(6)H_(5)OH] = 0.05 - 0.05alpha, 0.05 M`
`[C_(6)H_(5)O^(-)] = 0.01 + 0.05alpha, 0.01 M`
`[H_(3)O^(+)] = 0.05alpha`
`K_(a) = ([C_(6)H_(5)O^(-)][H_(2)O^(+)])/([C_(6)H_(5)OH])`
`K_(a) = ((0.01)(0.05alpha))/(0.05)`
`1.0 xx 1.0^(-10) = .01alpha`
`alpha = 1 xx 10^(-8)`