Question

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.

Short answer

Answer

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

Step-by-step solution

Explanation of the Concept

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 ${\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{K}}_{\mathbf{a}}$.

Retrieving the value of Ka and Kb

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 $\mathbf{>}\mathbf{7}\mathbf{.}\mathbf{0}$.