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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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