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In water, strong acids totally dissociate. Their conjugate bases aren't very strong. In water, weak acids only partly dissociate. Their conjugate foundations are solid.

The equilibrium in any acid-base reaction favours the reaction that sends the proton to the stronger base.

It is given that:

$\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{}\mathit{M}\mathbf{}{\left(C{H}_3\right)}_{\mathbf{2}}\mathit{N}\mathbf{}\mathit{H}$

Appendix C: ${\mathit{K}}_{\mathbf{b}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{9}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$

To begin, create the balanced equation for ${\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathit{N}\mathbf{}\mathit{H}$dissociation in water.

localid="1662033204023" ${\left({\mathrm{CH}}_3\right)}_2\mathrm{N}\mathrm{H}+{\mathrm{H}}_2鈬�{\left({\mathrm{CH}}_3\right)}_2\mathrm{N}{\mathrm{H}}_2^{+}+\mathrm{OH}$

Then, to get the equation for ${\mathit{K}}_{\mathbf{b}}$, build the ICE table.

localid="1662033218654" $$\begin{gathered} \begin{array}{cc}\mathbf{\left[}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{NH}\mathbf{\right]}\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{OC}\mathbf{}\mathbf{}\mathbf{鈬�}& \mathbf{[}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathbf{}{\mathbf{NH}}_{\mathbf{2}}^{\mathbf{+}}\mathbf{]}\mathbf{+}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}\\ \mathbf{I}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{M}& \mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\\ \mathbf{C}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{-}\mathbf{x}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}& \mathbf{+}\mathbf{x}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathbf{x}\\ \mathbf{E}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{M}\mathbf{-}\mathbf{x}& \mathbf{+}\mathbf{x}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{+}\mathbf{x}\end{array} \\ {\mathit{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}\mathbf{\left[}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}{\mathbf{NH}}_{\mathbf{2}}^{\mathbf{+}}\mathbf{\right]}}{[{\left({\mathrm{CH}}_3\right)}_2\mathrm{NH}]} \\ {\mathit{K}}_{\mathbf{a}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{-}\mathbf{x}} \end{gathered}$$

It is known that:

$$\begin{gathered} \mathit{x}\mathbf{=}\left[{\left(C{H}_3\right)}_{2}N{H}_2^{+}\right] \\ \mathbf{}\mathbf{}\mathbf{=}\left[O{H}^{-}\right] \end{gathered}$$

The ${\mathit{K}}_{\mathbf{b}}$of${\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{2}}\mathit{N}\mathbf{}{\mathit{H}}_{\mathbf{2}}$is quite tiny since it is a weak base. Assume that x has no influence on the denominator's $\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{}\mathit{M}$M. Then, to solve for x, insert the ${\mathit{K}}_{\mathbf{b}}$.

$$\begin{gathered} {\mathit{K}}_{\mathbf{b}}\mathbf{=}\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{0}\mathbf{.}\mathbf{070}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \\ {\mathit{x}}^{\mathbf{2}}\mathbf{=}\mathbf{\left(}{\mathbf{K}}_{\mathbf{b}}\mathbf{\right)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{)}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{2}\mathbf{\right)} \\ \mathit{x}\mathbf{=}\sqrt{\mathbf{\left(}{\mathbf{K}}_{\mathbf{b}}\mathbf{\right)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{)}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{3}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{=}\sqrt{\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{9}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{)}\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{070}\mathbf{)}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{4}\mathbf{\right)} \\ \mathit{x}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{43}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}\mathbf{}\mathit{M}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{5}\mathbf{\right)} \end{gathered}$$

We know that $\mathit{x}\mathbf{=}\mathit{O}{\mathit{H}}^{\mathbf{-}}$. Using the ${\mathit{K}}_{\mathbf{w}}$of water, solve for $\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}$.

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{K}}_{\mathbf{w}}\mathbf{=}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{6}\mathbf{\right)} \\ \mathbf{}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\frac{\left[K_w\right]}{\left[{\mathrm{OH}}^{-}\right]}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{7}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{6}\mathbf{.}\mathbf{43}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{8}\mathbf{\right)} \\ \mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{56}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathit{M}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{9}\mathbf{\right)} \end{gathered}$$

Now, solving to evaluate the value of $\mathit{p}\mathit{H}$as:

$$\begin{gathered} \mathbf{}\mathit{p}\mathit{H}\mathbf{=}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{10}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\mathit{l}\mathit{o}\mathit{g}\mathbf{(}\mathbf{1}\mathbf{.}\mathbf{56}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{)}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{11}\mathbf{\right)} \\ \mathit{p}\mathit{H}\mathbf{=}\mathbf{11}\mathbf{.}\mathbf{81}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{12}\mathbf{\right)} \end{gathered}$$

Therefore, the value of $\mathit{p}\mathit{H}$is: $\mathbf{11}\mathbf{.}\mathbf{81}$