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What is the degree of dissociation of $\mathrm{0.1\ M\ CH_3COOH }$ in presence of $\mathrm{0.1\ M\ HCl }$ ?

$\mathrm{\left (Given,\ k_a\ of\ CH_3COOH = 10^{-5} \right )}$
Option: 1
$1 \%$

Option: 2
$0.01 \%$

Option: 3
$0.2 \%$

Option: 4
$0.4 \%$

Answers (1)

best_answer

HCl is a strong acid and hence,

$\mathrm{[H^+]_{HCl} = 0.1 M}$

Now, let us consider the dissociation of $\mathrm{CH_3COOH}$ ,

$\mathrm{CH_3COOH \rightleftharpoons CH_3COO^- + H^+, \ k_a = 10^{-5}}$

$\mathrm{0.1}$ $-$ $\mathrm{0.1}$

$\mathrm{(0.1-x)}$ $\mathrm{(x)}$ $\mathrm{(0.1 + x)}$

$\mathrm{\approx (0.1)}$ $\mathrm{\approx (0.1)}$

Now, from the expression of dissociation constant,

$\mathrm{10^{-5} = \frac{0.1 \times (x)}{0.1}}$

$\mathrm{\Rightarrow x= 10^{-5}}$

Thus, degree of dissociation is given by

$\mathrm{\alpha= \frac{10^{-5}}{0.1}\times 100\%= 0.01\%}$

Hence, the correct answer is Option (2)

Posted by

shivangi.bhatnagar

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