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?

Asked by Pragya Singh | 1 year ago |  142

##### Solution :-

C6H5OH       C6H5O + H+

Initial  0.05M     0             0

After dissociation    0.05-x ,     x,          x

$$K_{}\alpha = \dfrac{x^{2}}{0.05-x}$$

Because the value of the ionization constant is very small, the value of x is very small. Accordingly, w e may ignore x in the denominator.

$$x = \sqrt{1\times 10^{-10}\times 0.05}$$

x =
x = 2.2×10-6M

In presence of 0.01  sodium phenolate(C6H5Na), suppose y is the amount of phenol dissociated, then at equilibrium

[C6H5OH] = 0.05 -y

[C6H5OH] ≈ 0.05

[C6H5O] = 0.01 + y  ≈ 0.01M,

[H+]=yM

$$K_{\alpha }=\dfrac{(0.01)(y)}{0.05}$$

$$=1.0\times 10^{-10}$$

y = 5 × 10-10

$$\alpha = \dfrac{y}{c}α= \dfrac{5\times 10^{-10}}{5\times 10^{-2}}$$

α = 10-8

Answered by Abhisek | 1 year ago

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