Chapter 19, Problem 67QRT

(a)

Interpretation Introduction

Interpretation:

Oxidation state of nitrogen in the compounds present in given reaction has to be determined.

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\text{+}{\text{HNO}}_{\text{2}}\text{(aq)}\stackrel{}{\to }{\text{HN}}_{\text{3}}\text{(aq)}\text{+}{\text{2H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

Oxidation state of nitrogen in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$:

Hydrogen has an oxidation state of $+1$.  Sum of oxidation state of all the atoms in a compound will be zero.  Therefore, oxidation state of nitrogen can be calculated as shown below,

    $\begin{array}{l}2x+4(+1)=0\\ 2x=-4\\ x=-2\end{array}$

Oxidation state of nitrogen in ${\text{HNO}}_{\text{2}}$:

Hydrogen has an oxidation state of $+1$.  Oxygen has an oxidation state of $-2$.  Sum of oxidation state of all the atoms in a compound will be zero.  Therefore, oxidation state of nitrogen can be calculated as shown below,

    $\begin{array}{l}1+x+2(-2)=0\\ x-3=0\\ x=+3\end{array}$

Oxidation state of nitrogen in ${\text{HN}}_{\text{3}}$:

Hydrogen has an oxidation state of $+1$.  Sum of oxidation state of all the atoms in a compound will be zero.  Therefore, oxidation state of nitrogen can be calculated as shown below,

    $\begin{array}{l}1+3x=0\\ 3x+1=0\\ x=-1/3\end{array}$

(b)

Interpretation Introduction

Interpretation:

Oxidizing agent in the below given reaction has to be identified.

    ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\text{(l)}\text{+}{\text{HNO}}_{\text{2}}\text{(aq)}\stackrel{}{\to }{\text{HN}}_{\text{3}}\text{(aq)}\text{+}{\text{2H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

Oxidation state of nitrogen in ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ is $-2$, oxidation state of nitrogen in ${\text{HNO}}_{\text{2}}$ is $+3$ and that of ${\text{HN}}_{\text{3}}$ is $-1/3$.  Oxidizing agent is the one that gets reduced in the reaction.  This means the oxidizing agent gains electrons.  This in turn means that the oxidation state becomes reduced.  Therefore, the oxidizing agent in the given reaction is found to be ${\text{HNO}}_{\text{2}}$ because, the oxidation state of nitrogen is reduced from $+3$ to $-1/3$ by gaining electrons.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of $\text{0}\text{.010}\text{M}$ solution of ${\text{HN}}_{\text{3}}$ has to be calculated.

Explanation of Solution

Ionization constant for hydrazoic acid is given as $1.0\times {10}^{-5}$.

Equation for the reaction of hydrazoic acid with water at equilibrium can be given as,

    ${\text{HN}}_{\text{3}}\text{(aq)}\text{+}{\text{H}}_{\text{2}}\text{O(l)}\stackrel{}{\rightleftharpoons }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}\text{+}{\text{N}}_{\text{3}}{}^{\text{-}}\text{(aq)}$

Equilibrium expression for this can be given as,

    ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][N}}_{\text{3}}{}^{\text{-}}\text{]}}{{\text{[HN}}_{\text{3}}\text{]}}-------------------(1)$

Concentration of the products increases till the equilibrium is reached.

${\text{HN}}_{\text{3}}\text{(aq)}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\text{(aq)}{\text{N}}_{\text{3}}{}^{-}\text{(aq)}$

Initial Concentration ($\text{M}$)        $0.010$        $5.00$            $0.00$

Change as Reaction occurs ($\text{M}$)    $-x$        $+x$            $+x$

Equilibrium Concentration ($\text{M}$)    $0.010-x$    $+x$            $+x$

Substituting these values in equation (1),

    $\begin{array}{l}{\text{K}}_{\text{a}}\text{=}\frac{\text{(x)(x)}}{\text{(0}\text{.010-x)}}\\ 1.0\times {10}^{-5}=\frac{{\text{x}}^{\text{2}}}{\text{(0}\text{.010)}}\\ x^2=(1.0\times {10}^{-5})(0.010)\\ =\sqrt{1.0\times {10}^{-5}}\times \sqrt{0.010}=0.00316\times 0.1\\ =3.16\times {10}^{-4}\text{M}\end{array}$

Therefore, concentration of hydronium ion is $\text{3}\text{.16}\text{×}{\text{10}}^{\text{-4}}\text{M}$.

It is known that $\text{pH}$ can be calculated from the concentration of hydronium ion using the equation shown below,

    $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \text{=}\text{-log(3}\text{.2}\times {\text{10}}^{\text{-4}})\\ =3.494\end{array}$

Therefore, $\text{pH}$ of solution of hydrazoic acid is calculated to be $3.494$.