Question

If $0.1mol$ of salt is added to $1L$ water, which of these salts is expected to produce the most acidic solution?

• A. $Na{C}_2{H}_3{O}_2$
• B. ${NH}_4{NO}_3$
• C. $Cu{SO}_4$
• D. $Al{Cl}_3$

Answer 1

The correct option is D $Al{Cl}_3$

A) ${NaC}_2{H}_3{O}_2$

${CH}_{3}COONa$ is salt formed from weak acid ${CH}_{3}COOH$ and strong base $NaOH$. So, $0.1M$ of salt produce alkaline solution.

B) ${NH}_4{NO}_3$ is salt formed from ${NH}_{4}OH$ and ${HNO}_3$. Solution of ${NH}_4{NO}_3$ will be slightly acidic.

C) ${CuSO}_4$ is a salt formed from ${Cu\left(OH\right)}_2$ and $H_2{SO}_4$. Solution of ${CuSO}_4$ will be slightly acidic.

D) ${AlCl}_3$ is dissolved to form ${\left[{Al\left(H_{2}O\right)}_6\right]}^{3+}$ which is very much prone to release $H^{+}$ ions. So its solution will be most acidic.

${AlCl}_3+6H_{2}O⟶{\left[{Al\left(H_{2}O\right)}_6\right]}^{3+}+3{Cl}^{-}$

Due to higher electron with drawing effect of ${Al}^{3+}$ hydrogen of complex is more acidic.

${\left[{Al\left(H_{2}O\right)}_6\right]}^{3+}⟶{\left[{Al\left(OH\right)\left(H_{2}O\right)}_5\right]}^{2+}+H^{+}$