Ionisation constant of acetic acid is 1.8 × 10^{-5} . concentration at which it will be dissociated

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6-2.Equilibrium-II (Ionic Equilibrium)

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The ionisation constant of acetic acid is $1.8 \times 10^{-5}$. The concentration at which it will be dissociated to $2\%$, is

A

$1$

B

$0.045$

C

$0.018$

D

$0.45$

Solution

When acetic acid is dissolved in water, it partially dissociates ( $2 \,\%)$. Ionisation constant $K _{ a }=1.87 \times 10^{-5}$

$K _{ a }=\frac{ C \alpha \times C \alpha}{ C (1-\alpha)}$

We assume $\alpha \sim 0 \Rightarrow 1-\alpha \sim 1$

$\Rightarrow 1.8 \times 10^{-5}=\frac{ Ca ^2}{1}= C \times 0.02 \times 0.02$

$\Rightarrow C =\frac{1.8 \times 10^{-5}}{0.02^2}$

$=0.045 \,M$

Standard 11

Chemistry

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