Ionisation constant of CH₃COOH is 1.7 × 10^{-5} and concentration of H^+ ions is 3.4 × 10^{-4} .

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6-2.Equilibrium-II (Ionic Equilibrium)

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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

A

$3.4 \times {10^{ - 4}}$

B

$3.4 \times {10^{ - 3}}$

C

$6.8 \times {10^{ - 4}}$

D

$6.8 \times {10^{ - 3}}$

Solution

The ionization reaction for $\mathrm{CH}_{3} \mathrm{COOH}$ follows:

$\mathrm{CH}_{3} \mathrm{COOH} \rightarrow \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+}$

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.

Standard 11

Chemistry

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(IIT-2015)