Question

Explain with equations and calculations, when necessary, whether an aqueous solution of each of these salts is acidic, basic, or neutral: $\left(a\right)\mathit{C}\mathit{r}{\left(N{O}_3\right)}_{\mathbf{3}}\mathit{C}\mathit{r}{\left(N{O}_2\right)}_{\mathbf{3}}\mathbf{}\mathbf{;}$$\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathit{N}\mathit{A}\mathit{H}\mathit{S}\mathit{N}\mathit{a}\mathit{H}\mathit{S}\mathbf{}\mathbf{;}$$\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathit{Z}\mathit{n}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COO}\mathbf{\right)}}_{\mathbf{2}}$

Short answer

Answer

(a) Acidic

(b) Basic

(c) Acidic

Step-by-step solution

Explanation of the Concept

Acidic salt: An acidic salt is an aqueous solution of a salt that contains the anion of the strong acid and the cation of the weak base. The anion of the weak base reacts with water to form the hydronium ion, causing the solution to become acidic.

A basic salt is an aqueous solution of a salt that contains the cation of the strong base and the anion of the weak acid. The cation of the strong acid reacts with water, resulting in the formation of the hydroxide ion. As a result, the solution is simplified.

Neutral salt: An aqueous solution of a salt containing the cation of the strong base and the anion of the weak base.

Disintegration of salt in water.

$\left(a\right)\mathit{C}\mathit{r}{\left(N{o}_3\right)}_{\mathbf{3}}\mathbf{}\mathit{C}\mathit{r}{\left(N{o}_3\right)}_{\mathbf{3}}$will disintegrate in water:

$\mathit{C}\mathit{r}{\mathbf{\left(}{\mathbf{No}}_{\mathbf{3}}\mathbf{\right)}}_{\mathbf{3}}\mathbf{\to }\mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}\mathbf{+}\mathbf{3}\mathit{N}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}$

$\mathit{C}{\mathit{r}}^{\mathbf{3}\mathbf{+}}$will form a hydrate and react with water.

$\mathit{C}\mathit{r}{\left(H_{2}O\right)}_{\mathbf{6}}^{\mathbf{3}\mathbf{+}}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathit{C}\mathit{r}{\left(H_{2}O\right)}_{\mathbf{5}}\mathbf{}\mathit{O}{\mathit{H}}^{\mathbf{2}\mathbf{+}}\mathbf{}{\mathit{H}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}}$

From the appendix$\mathit{C}\mathbf{,}\mathbf{}\mathit{t}\mathit{h}\mathit{e}\mathbf{}{\mathit{K}}_{\mathbf{a}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}\mathit{r}{\left(H_{2}O\right)}_{\mathbf{6}}^{\mathbf{3}\mathbf{+}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$

$\mathit{N}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}$is an anion of a powerful acid $\left(HN{O}_3\right)$. $\mathit{I}\mathit{t}\mathit{s}\mathbf{}{\mathit{K}}_{\mathbf{a}}$is very large, as a result NO${\mathit{K}}_{\mathbf{b}}$will be quite small.

Since, ${\mathit{K}}_{\mathbf{a}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}\mathit{r}{\left(H_{2}O\right)}_{\mathbf{6}}^{\mathbf{3}\mathbf{+}}$will be significantly greater than the ${\mathit{K}}_{\mathbf{b}}$of $\mathit{N}{\mathit{O}}_{\mathbf{3}}^{\mathbf{-}}$It will result in an acidic solution.

Disintegration of salt in water

$\left(b\right)\mathit{N}\mathit{a}\mathit{H}\mathit{S}\mathit{N}\mathit{A}\mathit{H}\mathit{S}$will disintegrate in water.

$\mathit{N}\mathit{a}\mathit{H}\mathit{S}\mathbf{\to }\mathit{N}{\mathit{a}}^{\mathbf{+}}\mathbf{+}\mathit{H}{\mathit{S}}^{\mathbf{-}}$

$\mathit{N}{\mathit{a}}^{\mathbf{+}}$is an ion derived from a strong base $\left(NaOH\right)\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}\mathit{H}{\mathit{S}}^{\mathbf{-}}$is a weak acid's ion $\left(H_{2}S\right)$.

Since, basic solution is produced by combining a salt with a cation of a strong base and an anion of a weak acid.

Disintegration of salt in water

$\left(c\right)\mathit{Z}\mathit{n}{\left(C{H}_{3}COO\right)}_{\mathbf{2}}$will in water

$\mathit{Z}\mathit{n}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COO}\mathbf{\right)}}_{\mathbf{2}}\mathbf{\to }\mathit{Z}{\mathit{n}}^{\mathbf{2}\mathbf{+}}\mathbf{+}\mathbf{2}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}{\mathit{O}}^{\mathbf{-}}$

$\mathit{Z}{\mathit{n}}^{\mathbf{2}\mathbf{+}}$will react with water to form a hydrate $\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}{\mathit{O}}^{\mathbf{-}}$will react with water as well

$$\begin{gathered} \mathit{Z}\mathit{n}{\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}}_{\mathbf{6}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathit{Z}\mathit{n}{\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}}_{\mathbf{5}}\mathit{O}{\mathit{H}}^{\mathbf{+}}\mathbf{+}{\mathit{H}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}} \\ \mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}{\mathit{O}}^{\mathbf{-}}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{}\mathbf{}\boxed{\xcancel{}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}\mathbf{+}\mathit{O}\mathit{H} \end{gathered}$$

We can get the following values C from the appendix:

$$\begin{gathered} {\mathit{K}}_{\mathbf{a}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{Z}\mathit{n}{\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}}_{\mathbf{6}}^{\mathbf{2}\mathbf{+}}\mathbf{=}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}} \\ {\mathit{K}}_{\mathbf{a}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}\mathit{O}\mathit{H}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}} \end{gathered}$$

Let us solve the problem${\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}{\mathit{O}}^{\mathbf{-}}$,

$$\begin{gathered} {\mathit{K}}_{\mathbf{w}}\mathbf{=}{\mathit{K}}_{\mathbf{a}}\mathbf{\times }{\mathit{K}}_{\mathbf{b}} \\ {\mathit{K}}_{\mathbf{b}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{1}\mathbf{.}\mathbf{8}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}} \\ {\mathit{K}}_{\mathbf{b}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{56}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}} \end{gathered}$$

Because, of the ${\mathit{K}}_{\mathbf{a}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{Z}\mathit{n}{\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}}_{\mathbf{6}}^{\mathbf{2}\mathbf{+}}\mathbf{>}{\mathit{K}}_{\mathbf{b}}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{C}{\mathit{H}}_{\mathbf{3}}\mathit{C}\mathit{O}{\mathit{O}}^{\mathbf{-}}\mathbf{}\mathbf{,}\mathbf{}\mathit{Z}\mathit{n}{\mathbf{\left(}{\mathbf{CH}}_{\mathbf{3}}\mathbf{COO}\mathbf{\right)}}_{\mathbf{2}}$will result in an acidic solution.