Equilibrium - Result Question 57
####59. What is the $H^{+}$ion concentration of a solution prepared by dissolving $4 g$ of $NaOH$ (Atomic weight of $Na=23 amu$ ) in $1000 mL$ ?
(a) $10^{-10} M$
(b) $10^{-4} M$
(c) $10^{-1} M$
(d) $10^{-13} M$
[1999]
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Solution:
- (d) No. of moles of $NaOH=\frac{4}{40}=0.1$
[Molecular weight of $NaOH=40$ ]
No. of moles of $OH^{-}=0.1$
Concentration of $OH^{-}=\frac{0.1}{1 \text{ litre }}=0.1 mol / L$
As we know that, $[H^{+}][OH^{-}]=10^{-14}$
$ \therefore[H^{+}]=10^{-13} \quad(\because[OH^{-}]=10^{-1}) $
