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Degree of Dissociation

What will be ...

Question

What will be the $H^{+}$ ion concentration of 0.01 M $H C N$ solution at equilibrium, if it is 20% ionised?

A

2 \times 10^{- 3} M

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B

3 \times 10^{3} M

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C

5 \times 10^{3} M

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D

1 \times 10^{3} M

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Solution

The correct option is A $2 \times 10^{- 3} M$
$H C N (a q) ⇌ H^{+} (a q .) + C N^{-} (a q .) I n i t i a l : C 00 E q u i l i b r i u m : C (1 - α) C α C α$
C = $0.01 M = 10^{- 2} M$
$α$ % = 20%
$α$ = 0.2
$[H^{+}] = C α$
$[H^{+}] = 10^{- 2} \times 0.2$
$[H^{+}] = 2 \times 10^{3} M$

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Degree of Dissociation

Standard XII Chemistry

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