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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 (b) 0.1M in HCl ?

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 

(b)     0.1M in HCl ?

Answers (1)

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Let the x amount of acetic acid is dissociated in this case

                                CH_{3}COOH\rightleftharpoons CH_{3}COO^- +H^+

 Initial conc.                  0.05                     0                              0
after dissociation        0.05 - x               0.1+x                 x

As the dissociation is very small.
So we can write 0.1+x  \approx 0.1 and  0.05 - x  \approx 0.05

 K_a= \frac{[CH_{3}COO^-][H^+]}{[CH_{3}COOH]}
                  \\=\frac{(0.1)(x)}{(0.05)}\\ =2x

So, the value of x = \frac{(1.82\times10^{-4})(0.05)}{0.1}

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

                                             = \frac{1.82\times 10^{-4}(0.05)}{0.05}
                                             = 1.82\times 10^{-4}

Posted by

manish

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