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Question

The ionization constant of phenol is 1.0×1010. 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?

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Solution

The balanced chemical equation for the ionization of phenol is:
PhOH+H2OPhO+H3O+.

The initial concentrations of phenol, phenoxide ion and hydronium ion are 0.05, 0 and 0 respectively.
Corresponding equilibrium concentrations are 0.05x, x and x respectively.

Ka=[PhO][H3O+][PhOH]

1×1010=x×x0.05x
x is very small (as the ionization constant is very small) . Hence, x can be ignored in the denominator.

1×1010=x×x0.05

x=2.2×106M=[H3O+]=[PhO]
Let α be the degree of dissociation of phenol in presence of 0.01 M phenoxide.

The dissociation of phenoxide is as represented:
PhONaPhO+Na+
Also,
PhOH+H2OPhO+H3O+
[PhOH]=0.050.05α0.05 M (as the degree of dissociation is very small.)
[PhO]=0.01+0.05α0.01 M
[H3O+]=0.05α

Ka=[PhO][H3O+][PhOH]
1.0×1010=0.01×0.05α0.05
α=1×108

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