Chapter U6, Problem C23.3RE

Interpretation Introduction

(a)

Interpretation:

The equilibrium concentration of ${\text{N}}_{\text{2}}{\text{H}}_{\text{4}}$ needs to be determined.

Concept introduction:

For an equilibrium reaction as follows:

$A+B\rightleftharpoons AB$

The expression for equilibrium constant is as follows:

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

Here, $\left[A\right],\left[B\right]\text{ and }\left[AB\right]$ is equilibrium concentration of A, B and AB.

Answer to Problem C23.3RE

0.11 M.

Explanation of Solution

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

The concentration of ${\text{N}}_{\text{2}}{\text{H}}_5^{+}\left(aq\right)$ is equal to ${\text{OH}}^{-}\left(aq\right)$ that is $3.1\times {10}^{-4}$.

The expression for the equilibrium constant is represented as follows:

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

The concentration of ${\text{H}}_{\text{2}}\text{O}\left(l\right)$ does not participate in equilibrium equation as it is present in its pure state.

Putting the values,

$8.7\times {10}^{-7}\text{=}\frac{\left(3.1\times {10}^{-4}\right)\left(3.1\times {10}^{-4}\right)}{\left[{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\right]}$

Or,

$\left[{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\right]\text{=}\frac{\left(3.1\times {10}^{-4}\right)\left(3.1\times {10}^{-4}\right)}{\left(8.7\times {10}^{-7}\right)}=1.1\times {10}^{-4-4+7}=0.11\text{ M}$

Thus, the concentration of $\left[{\text{N}}_{\text{2}}{\text{H}}_{\text{4}}\right]$ is 0.11 M.

Interpretation Introduction

(b)

Interpretation:

Whether the equilibrium mixture is acidic, basic or neutral needs to be determined.

Concept introduction:

A solution is said to be acidic if pH of solution is less than 7, it is basic if pH is more than 7 and it is neutral if pH is equal to 7.

Thus, by determining the pH of the solution, its nature can be determined.

The pH of the solution is related to its pOH as follows:

$\text{pH}=14-\text{pOH}$

Also, pOH can be calculated from hydroxide ion concentration as follows:

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Answer to Problem C23.3RE

Solution is basic in nature.

Explanation of Solution

The pOH of the solution can be calculated from hydroxide ion concentration as follows:

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$

Putting the value,

$\text{pOH}=-\log \left(3.1\times {10}^{-4}\right)=3.51$

The pH of the solution can be calculated as follows:

$pH=14-pOH$

Putting the values,

$pH=14-3.51=10.5$

Thus, pH of the solution is 10.5. Since, it is more than 7 thus, solution is basic in nature.