The ionization constant of phenol is 1.0 × 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?

Given:

The ionization constant of phenol (Ka) =1.0 × 10–10


Concentration of phenolate ion (C6H5O -) in phenol = 0.05 M


Concentration of phenolate ion in sodium phenolate = 0.01 M


Ionization of phenol takes place,



∴ Using the formula,


Ka =


Ka =


As Ka = 1.0 × 10–10 (given)


∴ 1.0 × 10–10 =


⇒1.0 × 10–10 =


As the value of ionization constant is very less, x will be very small. Thus, we can ignore x in the denominator.


∴ x2 = 5 × 10-12


⇒x = √ 5 × 10-12


⇒x = 2.23 × 10-6 M


Hence [H+] = [C6H5O-] = 2.23 × 10-6 M


Thus, the concentration of the phenolate ion is 2.23 × 10-6 M


Let α be the degree of ionization of phenol in the presence of 0.01 M C6H5ONa



[C6H5OH] = 0.05 – 0.05α ≈ 0.05M


[C6H5O-] = 0.01 + 0.05α ≈0.01 M


[H+] = 0.05α


Ka =


⇒Ka =


As Ka = 1.0 × 10–10 (given)


∴ 1.0 × 10–10


⇒1.0 × 10–10 = 0.01α


⇒α = 1 × 10-8


Thus, the degree of ionization is 1 × 10-8


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