0.4g of NaOH is present in one litre of the solution shows that $H^{+}$ concentration of the solution is:

10^{- 2}

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10^{- 4}

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10^{- 10}

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10^{- 12}

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Solution

The correct option is C $10^{- 12}$
1 L of solution contains 0.4 g of NaOH.
The molecular weight of NaOH is 40 g.
$[O H^{-}] = \frac{W e i g h t}{m o l e c u l a r w e i g h t} \times \frac{1}{v o l u m e} = \frac{0.4}{40} \times \frac{1}{1} M = 0.01 M$
But, $[H^{+}] [O H^{-}] = 1 \times 10^{- 14}$ .
So, $[H^{+}] = \frac{1 \times 10^{- 14}}{[O H^{-}]} = \frac{1 \times 10^{- 14}}{0.01} = 1 \times 10^{- 12} M$ .

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0.4g of NaOH is present in one litre of the solution shows that H+ concentration of the solution is:

The correct option is C 10−121 L of solution contains 0.4 g of NaOH.The molecular weight of NaOH is 40 g.[OH−]=Weightmolecularweight×1volume=0.440×11M=0.01MBut, [H+][OH−]=1×10−14.So, [H+]=1×10−14[OH−]=1×10−140.01=1×10−12M.

0.4g of NaOH is present in one litre of the solution shows that $H^{+}$ concentration of the solution is: