Chapter U6, Problem SI1RE

Interpretation Introduction

(a)

Interpretation:

Whether hydrazine is a strong acid, strong base, weak acid or weak base needs to be determined.

Concept introduction:

An equilibrium reaction is represented as follows:

$A+B\rightleftharpoons AB$

For the above reaction, the change in concentration of A and B is equal to that of AB.

For the above reaction, equilibrium constant can be represented as follows:

$K=\frac{\left[AB\right]}{\left[A\right]\left[B\right]}$

Here, K is equilibrium constant.

Explanation of Solution

The ionization equation for the hydrazine is represented as follows:

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_5^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The pH of 0.1 M hydrazine solution is 10. For solutions with pH value more than 7 are basic in nature thus, hydrazine is basic in nature.

Also, the given reaction is reversible thus, hydrazine does not completely react with water to release hydroxide ion. Thus, it is a weak base.

Interpretation Introduction

(b)

Interpretation:

All the compounds and ions present in 0.1 M solution of hydrazine solution needs to be listed.

Concept introduction:

An equilibrium reaction is represented as follows:

$A+B\rightleftharpoons AB$

For the above reaction, the change in concentration of A and B is equal to that of AB.

For the above reaction, equilibrium constant can be represented as follows:

$K=\frac{\left[AB\right]}{\left[A\right]\left[B\right]}$

Here, K is equilibrium constant.

Explanation of Solution

The ionization equation for the hydrazine is represented as follows:

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_5^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The pH of solution is 10, thus, hydroxide ion concentration can be calculated as follows:

From pH, pOH of solution will be:

$pOH=14-pH=14-10=4$

Thus, hydroxide ion concentration will be:

$\left[{\text{OH}}^{-}\right]={10}^{-pOH}={10}^{-4}\text{ M}$

The ICE table for the equation can be represented as follows:

$\begin{array}{l}{\text{ N}}_{\text{2}}{\text{H}}_{\text{4}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_5^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ I\text{ 0}\text{.1 - - -}\\ C-x\text{ - +}x\text{ +}x\\ E\text{ 0}\text{.1}-x\text{ - }xx\end{array}$

Here,

$x=\left[{\text{OH}}^{-}\right]={10}^{-4}\text{ M}$

Also,

$x=\left[{\text{OH}}^{-}\right]=\left[{\text{N}}_{\text{2}}{\text{H}}_5{}^{+}\right]={10}^{-4}\text{ M}$

The equilibrium concentration of ${\text{ N}}_{\text{2}}{\text{H}}_{\text{4}}$ will be:

$\left[{\text{ N}}_{\text{2}}{\text{H}}_{\text{4}}\right]=0.1-{10}^{-4}=0.099\text{ M}$

There is no change in the concentration of liquid water thus, the species present in 0.1 M solution are ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(aq\right)$, ${\text{N}}_{\text{2}}{\text{H}}_5^{+}\left(aq\right)$ and ${\text{OH}}^{-}\left(aq\right)$.

(c)

Interpretation Introduction

Interpretation:

The equilibrium equation for hydrazine needs to be written.

Concept introduction:

An equilibrium reaction is represented as follows:

$A+B\rightleftharpoons AB$

For the above reaction, the change in concentration of A and B is equal to that of AB.

For the above reaction, equilibrium constant can be represented as follows:

$K=\frac{\left[AB\right]}{\left[A\right]\left[B\right]}$

Here, K is equilibrium constant.

Explanation of Solution

The ionization equation for the hydrazine is represented as follows:

${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\rightleftharpoons {\text{N}}_{\text{2}}{\text{H}}_5^{+}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

The equilibrium constant equation for the above reaction can be represented as follows:

$K=\frac{\left[{\text{N}}_{\text{2}}{\text{H}}_5^{+}\left(aq\right)\right]\left[{\text{OH}}^{-}\left(aq\right)\right]}{\left[{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(aq\right)\right]}$

Here, water is not included in the equilibrium expression because it is in liquid form and the species in solid and liquid form are not represented in the equilibrium expression.

The given value of equilibrium constant K for the reaction of hydrazine is $8.7\times {10}^{-7}$

Thus,

$8.7\times {10}^{-7}=\frac{\left[{\text{N}}_{\text{2}}{\text{H}}_5^{+}\left(aq\right)\right]\left[{\text{OH}}^{-}\left(aq\right)\right]}{\left[{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\left(aq\right)\right]}$