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  • 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?

7.44     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?

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best_answer

We have,
 The ionization constant of phenol is 1.0 \times 10^{-10}, and
the concentration of phenol is 0.05 M
degree of ionisation = ?

Ionization of phenol;
C_6H_5OH+H_2O\rightleftharpoons C_6H_5O^-+H_{3}O^+
At equilibrium, 
the concentration of various species are-
[C_6H_5OH] = 0.05-x
[C_6H_5O^-] = [H_3O^+]=x
As we see, the value of ionisation is very less. Also x will be very small. Thus we can ignore x.

\\K_a = \frac{[C_6H_5O^-][H_3O^+]}{[C_6H_5OH]}\\ 10^{-10}=\frac{x^2}{0.05}\\ x =\sqrt{10^{-10}\times 0.05}\\

x= 2.2\times 10^{-6}

Hence the concentration of phenolate ion is [C_6H_5O^-]= 2.2\times 10^{-6}

Let \alpha be the degree of dissociation of phenol in the presence of 0.01 M of phenolate ion.

                                       C_6H_5OH\rightleftharpoons C_6H_5O^-+H^+
Concentration                  (1 - \alpha) 0.05            0.05\alpha            0.05\alpha

So,
 [C_6H_5OH]= 0.05(1-\alpha) =0.05
[C_6H_5O^-]= 0.05\alpha+0.01 \approx 0.01M

[H_3O^+] = 0.05\alpha

therefore,

\\K_a =\frac{(0.01)(0.05\alpha)}{0.05}\\ 10^{-10}=0.01\alpha\\ \alpha = 10^{-8}
The degree of dissociation is 10^{-8}

Posted by

manish

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