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Auto Protolysis of Water

The concentra...

Question

The concentration of $[H^{+}]$ and concentration of $[O H^{-}]$ of a $0.1 M$ aqueous solution of $2$ M ionised weak acid is: [ionic product of water $= 1 \times 10^{- 14}]$ .

0.02 \times 10^{- 3} M

and

5 \times 10^{- 11} M

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1 \times 10^{- 3} M

and

3 \times 10^{- 11} M

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2 \times 10^{- 3} M

and

5 \times 10^{- 12} M

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3 \times 10^{- 2} M

and

4 \times 10^{- 13} M

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Solution

The correct option is A $2 \times 10^{- 3} M$ and $5 \times 10^{- 12} M$
Degree of ionization,
$α = \frac{[H^{+}] \times 100}{[H A]}$

$[H^{+}] = \frac{α \times [H A]}{100}$

$= \frac{2 \times 0.1}{100}$

$[H^{+}] = 2 \times 10^{- 3} M$ .

$K_{W} = [H^{+}] [O H^{-}]$

$10^{- 14} = 2 \times 10^{- 3} \times [O H^{-}]$

$[O H^{-}] = \frac{10^{- 14}}{2 \times 10^{- 3}}$

$[O H^{-}] = 5 \times 10^{- 12} M$ .

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The concentration of [H+] and concentration of [OH−] of a 0.1 M aqueous solution of 2M ionised weak acid is: [ionic product of water =1×10−14].

The correct option is A 2×10−3M and 5×10−12MDegree of ionization,α=[H+]×100[HA][H+]=α×[HA]100=2×0.1100[H+]=2×10−3M.KW=[H+][OH−]10−14=2×10−3×[OH−][OH−]=10−142×10−3[OH−]=5×10−12M.

The concentration of $[H^{+}]$ and concentration of $[O H^{-}]$ of a $0.1 M$ aqueous solution of $2$ M ionised weak acid is: [ionic product of water $= 1 \times 10^{- 14}]$ .