Question

$Mg$ can reduce $N{O}_3^{-}$ to $N{H}_3$ in basic solution as follows:

$N{O}_3^{-}+Mg\left(s\right)+H_{2}O⟶Mg\left(OH{)}_2\right(s)+O{H}^{-}(aq)+N{H}_3(g)$

A $25.0$ mL sample of $N{O}_3^{-}$ solution treated with $Mg$. The $N{H}_3\left(g\right)$ was passed into $50$ mL of $0.15N$ $HCl$. The excess $HCl$ required $32.10$ mL of $0.10M$ $NaOH$ for its neutralisation. What was the molarity of $N{O}_3^{-}$ ions in the original sample?

• A. $0.33$
• B. $0.09$
• C. $0.15$
• D. $0.17$

Answer 1

The correct option is D $0.17$

Meq. of $N{H}_3$ formed $=$ Meq. of $HCl$ used for $N{H}_3$ $=50\times 0.15-32.10\times 0.10$ $=4.29$
These meq. of $N{H}_3$ are derived using valence factor of $N{H}_3=1$ (an acid-base reaction).
In redox change valence factor of $N{H}_3$ is $8$.
$8e+N^{5+}⟶N^{3-}$
Meq. of $N{H}_3$ for valence factor $8=8\times 4.29=34.32$
Also, meq. of $N{O}_3^{-}=$ meq. of $N{H}_3$ $=8\times 4.29=34.32$
$N_{N{O}_3^{-}}=\frac{34.32}{25}=1.37$
Also, $M_{N{O}_3^{-}}=\frac{1.37}{8}=0.1716$