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$\left({\text{K}}_{\text{en}}\right)$?

Short 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-by-step solution

Step 1(a): Find the [CH3O - ]  in pure (CH3OH)

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${\text{K}}_{\text{en}}\text{=}\text{4}\times {\text{10}}^{\text{- 16}}.$