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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 

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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}

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manish

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