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Characteristics of Equilibrium Constant

Concentration...

Question

Concentration of $C H_{3} C O O^{-}$ is $0.001 M$ , when $0.1$ moles of $C H_{3} C O O H$ were dissolved in $1 L$ water. $K_{a}$ of $C H_{3} C O O H$ is:

2 \times 10^{- 5}

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10^{- 5}

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10^{- 6}

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2 \times 10^{- 4}

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Solution

The correct option is B $10^{- 5}$
$C H_{3} - C O O H = C H_{3} - C O O^{-} + H^{+}$
Initial concentration $[C H_{3} - C O O H]_{0} = \frac{0.1 m o l}{1 L} = 0.1 M$
Equilibrium concentration $[C H_{3} - C O O H] = 0.1 - 0.001 \approx 0.1 M$
Equilibrium concentration $[C H_{3} - C O O^{-}] = [H^{+}] = 0.001 M$
Note: $[H^{+}]$ due to auto ionisation of water is neglected.
$K_{a} = \frac{[C H_{3} - C O O^{-}] [H^{+}]}{[C H_{3} - C O O H]}$
$K_{a} = \frac{0.001 \times 0.001}{0.1} = 10^{- 5} M$

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