Q161CP - Chemistry: Molecular Nature Of Matter And Change

Question

Auto ionization occurs in methanol $\left({\text{CH}}_{\text{3}}\text{OH}\right)$  and in ethylenediamine                               (${\mathbf{\text{NH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}$). 

(a) The auto ionization constant of methanol  $\left({\text{K}}_{\text{met }}\right)$is $\mathbf{\text{2}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{ - 17}}}$. What is $\left[{\text{CH}}_{\text{3}}{\text{O}}^{\text{ - }}\right]$ in pure $\left({\text{CH}}_{\text{3}}\text{OH}\right)$ ?

(b) The concentration of ${\mathbf{\text{NH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}$  in pure ${\mathbf{\text{NH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{CH}}}_{\mathbf{\text{2}}}{\mathbf{\text{NH}}}_{\mathbf{\text{2}}}$ is $\mathbf{\text{2}}\mathbf{\times }{\mathbf{\text{10}}}^{\mathbf{\text{ - 8}}}\mathbf{\text{M}}$ . What is the auto ionization constant of ethylenediamine $\left({\text{K}}_{\text{en}}\right)$?

Step-by-Step Solution

Answer

(a)$\left[{\text{CH}}_3{\text{O}}^{-}\right]=4.47\times {10}^{-9}\text{M}$ 

(b) The auto ionization constant of ethylenediamine is ${\text{K}}_{\text{en}}\text{ = } \text{4}\times {\text{10}}^{\text{ - 16}}.$

Step 1(a): Find the [ CH 3 O  -  ] in pure ( CH 3 OH )

First, write the auto ionization reaction.

$2 {\text{CH}}_3\text{OH}\left(l\right)\to {\text{CH}}_3{\text{OH}}_2^{+}\left(\text{met }\right)+{\text{CH}}_3{\text{O}}^{-}\left(\text{met }\right)$

To write the ${\text{K}}_{\text{met, }}$ the liquid state chemical species is not included because it will remain constant.

$\begin{array}{rcl}& & {\text{K}}_{\text{met }}\text{ = } \frac{\text{[ product ]}}{\text{[ reactant ]}}\\ & & \text{ = } \left[{\text{CH}}_{\text{3}}{\text{OH}}_{\text{2}}^{\text{ + }}\right]\left[{\text{CH}}_{\text{3}}{\text{O}}^{\text{ - }}\right].\end{array}$

Note that $\left[{\text{CH}}_3{\text{OH}}_2^{+}\right]=\left[{\text{CH}}_3{\text{O}}^{-}\right]$  since they are produced equally by the auto ionization of ${\text{CH}}_3\text{OH}$ . So, we can designate the two as x.

$\begin{array}{rcl}& & {\text{K}}_{\text{met }}\text{ = } \left[{\text{CH}}_{\text{3}}{\text{OH}}_{\text{2}}^{\text{ + }}\right]\left[{\text{CH}}_{\text{3}}{\text{O}}^{\text{ - }}\right]\\ & & \text{ = } {\text{x}}^{\text{2}}\\ & & \text{ x = }\sqrt{{\text{K}}_{\text{met }}}\\ & & \text{ = } \sqrt{\text{2}\times \text{10 - 17}}\text{x }\\ & & \text{ = } \text{4.47}\times {\text{10}}^{\text{ - 9}}.\end{array}$

Since  $\text{x} \text{ = }\left[{\text{CH}}_{\text{3}}{\text{OH}}_{\text{2}}^{\text{ + }}\right]\text{ = }\left[{\text{CH}}_{\text{3}}{\text{O}}^{\text{ - }}\right]\text{, then }\left[{\text{CH}}_{\text{3}}{\text{O}}^{\text{ - }}\right]\text{ = } \text{4.47}\times {\text{10}}^{\text{ - 9}}\text{M.}$

Hence, $\left[{\text{CH}}_3{\text{O}}^{-}\right]=4.47\times {10}^{-9}\text{M.}$

Step 2(b): Find the auto ionization constant of ethylenediamine

First, write the auto ionization reaction.

${\text{2NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{2}}\text{(l)}\to {\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}_{\text{3}}^{\text{ + }}{\text{(en) + NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{NH}}^{\text{ - }}\text{(en )}$

To write the ${\text{K}}_{\text{en}}$, we will not include the liquid state chemical species because it will remain constant

$\begin{array}{rcl}& & {\text{K}}_{\text{en}}\text{ = }\frac{\text{[ product ]}}{\text{[ reactant ]}}\\ & =& \left[{\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3^{+}\right]\left[{\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}^{-}\right].\end{array}$

 

Note that $\left[{\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3^{+}\right]=\left[{\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}^{-}\right]$  since they are produced equally by the auto ionization of${\text{NH}}_2{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2$  . Now, solve for ${\text{K}}_{\text{en }}$.

$\begin{array}{rcl}& & {\text{K}}_{\text{en}}\text{ = } \left[{\text{NH}}_{\text{2}}{\text{CH}}_{\text{2}}{\text{CH}}_{\text{2}}\right.]\\ & & \text{ = } \text{4}\times {\text{10}}^{\text{ - 16}}.\end{array}$

Hence, the auto ionization constant of ethylenediamine is ${\text{K}}_{\text{en}}\text{ = } \text{4}\times {\text{10}}^{\text{ - 16}}.$