Ionisation constant of CH_3COOH is 1.7 times | vedclass

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Question

The ionization reaction for $\mathrm{CH}_{3} \mathrm{COOH}$ follows:

$\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\m

Vedclass · Accepted Answer

$6.8 	imes {10^{ - 3}}$

Vedclass · Answer

The ionization reaction for $\mathrm{CH}_{3} \mathrm{COOH}$ follows:

$\mathrm{CH}_{3} \mathrm{COOH} ightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}$

At eqm. $\mathrm{x} \quad 3.4 	imes 10^{-4} \quad 3.4 	imes 10^{-4}$

The ionization constant for the above reaction is written as:

$\mathrm{k}=\frac{\left[\mathrm{CH}_{3} \mathrm{C} \mathrm{OO}^{-}ight]\left[\mathrm{H}^{+}ight]}{\left[\mathrm{CH}_{3} \mathrm{COOH}ight]}$

Putting values in above equation, we get:

$1.7 	imes 10^{-5}=\frac{\left(3.4 	imes 10^{-4}ight)\left(3.4 	imes 10^{-4}ight)}{\mathrm{x}}$

$\mathrm{x}=6.8 	imes 10^{-3}$

Hence, option D is correct.

Ionisation constant of $CH_3COOH$ is $1.7 \times 10^{-5}$ and concentration of $H^+$ ions is $3.4 \times 10^{-4}$. Then find out initial concentration of $CH_3COOH$ Molecules

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