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  • Calculate the degree of ionization of 0.05M acetic acid if its pK_a value is 4.74. How is the degree of dissociation affected when its solution also contains (a) 0.01M

7.53     Calculate the degree of ionization of 0.05M acetic acid if its pKa value is 4.74. How is the degree of dissociation affected when its solution also contains

(a)     0.01M 

Answers (1)

best_answer

We have,
C = 0.05 M
pK_a = 4.74 = -\log(K_a)
By taking antilog on both sides we get,
(K_a) = 1.82\times 10^{-5} = C.(\alpha)^2
from here we get the value of \alpha= \sqrt{\frac{1.82\times 10^{-5}}{5\times10^{-2}}}
                                                      = 1.908\times 10^{-2}

After adding hydrochloric acid, the concentration of H^+ ions increases and due to that the equilibrium shifts towards the backward direction. It means dissociation will decrease.

(i) when 0.01 HCl is taken
                           CH_{3}COOH\rightleftharpoons CH_{3}COO^- +H^+
 Initial conc.                  0.05                     0                              0
after dissociation        0.05 - x               0.001+x                 x

As the dissociation is very small.
So we can write 0.001+x  \approx 0.001 and  0.05 - x  \approx 0.05

Now, K_a= \frac{[CH_{3}COO^-][H^+]}{[CH_{3}COOH]}
                  \\=\frac{(0.001)(x)}{(0.05)}\\ =\frac{x}{50}

So, the value of x = \frac{(1.82\times10^{-5})(0.05)}{0.01}

Now degree of dissociation  = (amount dissociated) /(amount taken)

                                             = \frac{1.82\times 10^{-3}(0.05)}{0.05}
                                             = 1.82\times 10^{-3}

Posted by

manish

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