The NH⁺₄ ion forms acidic solutions, and the CH₃COO^- ion forms basic solutions. However, a solution of ammonium acetate is almost neutral. Do all of the ammonium salts of weak acids form neutral solutions? Explain your answer.

We know that $\mathit{N}{\mathit{H}}_{\mathbf{4}}^{\mathbf{+}}$and $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}{\mathit{O}}^{\mathbf{-}}$are both text is both a weak acid and a base. Weak acids and bases rarely form neutral solutions, especially when their concentrations vary. ${\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{K}}_{\mathbf{a}}$The solution of the salt formed by $\mathit{N}{\mathit{H}}_{\mathbf{4}}^{\mathbf{+}}$and $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}{\mathit{O}}^{\mathbf{-}}$on the other hand, is almost neutral. This is due to the fact that they have to underline nearly similar ${\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{K}}_{\mathbf{a}}$.

We can retrieve their ${\mathit{K}}_{\mathbf{a}}$text and ${\mathit{K}}_{\mathbf{b}}$text values from Appendix C text.

$$\begin{gathered} {\mathit{K}}_{\mathbf{a}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{N}{\mathit{H}}_{\mathbf{3}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{76}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}} \end{gathered}$$

By solving,

${\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{K}}_{\mathbf{a}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{N}{\mathit{H}}_{\mathbf{4}}^{\mathbf{+}}$

We get,

$$\begin{gathered} {\mathit{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{NH}}_{\mathbf{3}}\mathbf{\right]}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}}{\left[{\mathrm{NH}}_4^{+}\right]} \\ {\mathit{K}}_{\mathbf{b}}\mathbf{=}\frac{\mathbf{\left[}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COOH}\mathbf{\right]}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]} \end{gathered}$$

Where it is shown in the equations ${\mathit{K}}_{\mathbf{a}}$and ${\mathit{K}}_{\mathbf{b}}$also indicate the amount of ${\mathit{H}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}}\mathbf{-}\mathit{O}{\mathit{H}}^{\mathbf{-}}$produced. Because they are derived from a salt, the denominator of both equations will have the same value (concentration), and the reaction will produce the same amount $\mathit{I}\mathit{f}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathit{a}\mathit{n}\mathit{d}\mathbf{\left[}\mathbf{OH}\mathbf{\right]}\mathbf{,}$,$\mathit{I}\mathit{f}\mathbf{\left[}{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\mathbf{\right]}\mathbf{=}\mathbf{\left[}\mathbf{OH}\mathbf{\right]}$and the salt solution becomes neutral.

Hence, because of the varying values of ${\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{K}}_{\mathbf{c}}$, ammonium salts of weak acids will not form neutral solutions. The ammonium salt of acetic acid is nearly neutral only because the${\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{K}}_{\mathbf{c}}$values are nearly equal.

The variation in ${\mathit{K}}_{\mathbf{a}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{K}}_{\mathbf{b}}$values will change the $\mathit{p}\mathit{H}$of the solution. If the ${\mathit{K}}_{\mathbf{a}}\mathbf{>}{\mathit{K}}_{\mathbf{b}}$, then the pH will definitely be $\mathbf{<}\mathbf{7}\mathbf{.}\mathbf{0}$. If the ${\mathit{K}}_{\mathbf{a}}\mathbf{<}{\mathit{K}}_{\mathbf{b}}$, then the $\mathit{p}\mathit{H}$will definitely be $\mathbf{>}\mathbf{7}\mathbf{.}\mathbf{0}$.