Determine whether aqueous solutions of the following salts are acidic, basic, or neutral:

\((a) FeC{l_3}\)

\((b) {K_2}C{O_3}\)

\((c) N{H_4}Br\)

\((d) KCl{O_4}\)

The dissociation of a salt solution is used to assess whether it is acidic, basic, or neutral. After that, we figure out where the cation (from the base) and the anion come from (from the acid). Strong acid/base ions are resistant to hydrolysis because their conjugated acid/base is persistent—the dissociation reaction of that acid/base is not reversible. If an ion is derived from a weak acid or basic, it will hydrolyze (react with water).

\((a) FeC{l_3}\)

\(FeC{l_3}(aq) \to F{e^{3 + }}(aq) + 3C{l^ - }(aq)\)

The cation \(F{e^{3 + }}\) and the anion \(C{l^ - }\) are present.

We'll now try to figure out where these ions came from. The cation \(F{e^{3 + }}\) is derived from the base \(F e{(O H)_3}\). The anion \(C{l^ - }\) is derived from the acid \(H C l\).

Now we must determine whether the ions' acid and base are strong or weak. \(H C l\) is a powerful acid. The basic \(F e{(O H)_3}\) is a weak one. This means that while the cation \(F{e^{3 + }}\) will react with water, the anion \(C{l^ - }\)will not. The hydrolysis of the \(F{e^{3 + }}\) is as follows:

\(F{e^{3 + }}(aq) + {H_2}O(l) \to {(Fe(OH))^{2 + }}(aq) + {H^ + }(aq).\)

The ultimate result in this case is \({H^ + }\), indicating that the solution is acidic.

\((b) {K_2}C{O_3}\)

The reaction of dissociation of the salt \({K_2}C{O_3}\) is:

\({K_2}C{O_3}(aq) \to 2\;{K^ + }(aq) + CO_3^{2 - }(aq).\)

The cation \({K^ + }\) is derived from \(KOH\), and the anion \(CO_3^{2 - }\) is derived from \(CO_3^{2 - }\) is derived from \({H_2}C{O_3}\) Because \(KOH\) is a strong base, it will not react with water. The anion will react with the water since \({H_2}C{O_3}\) is a weak acid.

The hydrolysis reaction is as shown below:

\(CO_3^{2 - }(aq) + {H_2}O(l) \to HCO_3^ - (aq) + O{H^ - }(aq).\)

The final result of the hydrolysis reaction is \(O{H^ - }\), indicating that the solution is basic.

\((c) N{H_4}Br\)

\(NH_4^ + \)is derived from \(N{H_4}OH\), while \(B{r^ - }\) is derived from

\(N{H_4}Br(aq) \to NH_4^ + (aq) + B{r^ - }(aq)\)

\(HBr\)is a powerful acid, which will not react with water. The anion will react with the water since \(N{H_4}OH\) is a weak base.

The hydrolysis reactionis as shown below:

\(NH_4^ + (aq) + {H_2}O(l) \to N{H_3}(aq) + {H_3}{O^ + }(aq).\)

The final result of the hydrolysis reaction is \({H_3}{O^ + }\), indicating that the solution is acidic.

\((d) KCl{O_4}\)

The reaction of dissociation of \(KCl{O_4}\) is:

\(KCl{O_4}(aq) \to {K^ + }(aq) + ClO_4^ - (aq).\)

\({K^ + }\)is derived from \(KOH\), while \(ClO_4^ - \) is derived from \(HCl{O_4}\).

\(KOH\)is a strong base, which means it will not react with water. Because \(HCl{O_4}\) is a strong acid, it will not react with water either. The solution is neutral since there is no hydrolysis process here.

a) The solution is acidic.

b) The solution is basic.

c) The solution is acidic.

d) The solution is neutral.