Four grams of $N a O H$ solid dissolved in just enough water to make $1$ litre of solution. What is the $[H^{+}]$ fo the solution?

10^{- 2}

moles/litre

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10^{- 1}

moles/litre

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10^{- 12}

moles/litre

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10^{- 13}

moles/litre

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Solution

The correct option is A $10^{- 13}$ moles/litre
Option D is correct
$M a s s o f N a O H = 4 g V o l u m e o f s o l u t i o n = 1 L M o l a r i t y = \frac{M o l e n u m b e r}{V o l u m e} N o . o f m o l e = \frac{G i v e n M a s s}{M o l a r m a s s} M o l a r m a s s o f N a O H = 23 + 16 + 1 = 40 g N o . o f m o l e = \frac{4}{40} = 0.1 M o l a r i t y = \frac{0.1}{1} = 0.1 M [O H^{-}] = O .1 M [H^{+ 1}] = \frac{K_{w}}{[O H^{-}]} = \frac{10^{- 14}}{0.1} = 10^{- 13} M$

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