$\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}$
At eqm. $\mathrm{x} \quad 3.4 \times 10^{-4} \quad 3.4 \times 10^{-4}$
The ionization constant for the above reaction is written as:
$\mathrm{k}=\frac{\left[\mathrm{CH}_{3} \mathrm{C} \mathrm{OO}^{-}\right]\left[\mathrm{H}^{+}\right]}{\left[\mathrm{CH}_{3} \mathrm{COOH}\right]}$
Putting values in above equation, we get:
$1.7 \times 10^{-5}=\frac{\left(3.4 \times 10^{-4}\right)\left(3.4 \times 10^{-4}\right)}{\mathrm{x}}$
$\mathrm{x}=6.8 \times 10^{-3}$
Hence, option D is correct.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List $I$ (molecules/ions) | List $II$ (No. of lone pairs of $e ^{-}$ on central atom) |
| $A$ $IF_7$ | $I$ Three |
| $B$ $ICl^{-}_4$ | $II$ One |
| $C$ $XeF_6$ | $III$ Two |
| $D$ $XeF_2$ | $IV$ Zero |
Choose the correct answer from the options given below: