Question

$0.1\;M$ solution of which of the following substances is most acidic
a.$N{{H}_{4}}Cl$
b.$KCN\;$
c.$KCl\;$
d.$C{{H}_{3}}COONa$

Answer 1

Closely studying the given options, we can understand that all the given compounds are salts formed from the neutralization reaction of an acid and a base. Hence, we need to find the respective acid and base of the salt by taking its reaction with water. The salt of the strongest acid of all will be the most acidic.

Complete step by step answer:
The given compounds comprise the electron-rich component which acts as Lewis Base as well as an electron-deficient component which acts as Lewis acid. Hence, they act as neutral or near to neutral salts. We can also prove the statement by practically finding any of the given compounds which may come equal to or near to.
Now, we know that salt is formed by the neutralization reaction of acid and base with water as a co-product, which can be expressed as below
$Acid+Base\to Salt+Water$
Now, without considering the practical possibility of the reaction, we can invert the reaction so that acid and base can be formed by the reaction of salt and water, which can be expressed as
$Salt+Water\to Acid+Base$
Now, we will find the products of all the given salts by reaction with water
Reaction of Ammonium Chloride $N{{H}_{4}}Cl$ with water
$N{{H}_{4}}Cl+{{H}_{2}}O\to N{{H}_{4}}^{+}+C{{l}^{-}}+{{H}^{+}}+O{{H}^{-}}$
$N{{H}_{4}}^{+}+C{{l}^{-}}+{{H}^{+}}+O{{H}^{-}}\to N{{H}_{4}}OH+HCl$
Now, we should be able to understand that Ammonium Hydroxide $N{{H}_{4}}OH$ is a weak base and Hydrogen Chloride $HCl\;$ is a strong acid. Hence, the resultant salt obtained by neutralization will be acidic in nature.
Reaction of Potassium Cyanide $KCN\;$ with water
$KCN+{{H}_{2}}O\to {{K}^{+}}+C{{N}^{-}}+{{H}^{+}}+O{{H}^{-}}$
${{K}^{+}}+C{{N}^{-}}+{{H}^{+}}+O{{H}^{-}}\to KOH+HCN$
Now, we should be able to understand that Potassium Hydroxide $KOH\;$ is a strong base and Hydrogen Cyanide $HCN\;$ is a weak acid. Hence, the resultant salt obtained by neutralization will be basic in nature.
Reaction of Potassium Chloride $KCl\;$ with water
$KCl+{{H}_{2}}O\to {{K}^{+}}+C{{l}^{-}}+{{H}^{+}}+O{{H}^{-}}$
${{K}^{+}}+C{{l}^{-}}+{{H}^{+}}+O{{H}^{-}}\to KOH+HCl$
Now, we should be able to understand that Potassium Hydroxide $KOH\;$ is a strong base and Hydrogen Chloride $HCl\;$ is a strong acid. Hence, the resultant salt obtained by neutralization will be neutral in nature.
Reaction of Sodium Acetate $C{{H}_{3}}COONa$ with water
$C{{H}_{3}}COONa+{{H}_{2}}O\to N{{a}^{+}}+C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}+O{{H}^{-}}$
$N{{a}^{+}}+C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}+O{{H}^{-}}\to NaOH+C{{H}_{3}}COOH$
Now, we should be able to understand that Sodium Hydroxide $NaOH\$ is a strong base and Acetic acid $C{{H}_{3}}COOH$ is a weak acid. Hence, the resultant salt obtained by neutralization will be basic in nature.
Hence, from the above reactions, only $N{{H}_{4}}Cl$ is the acidic salt
Hence, the correct answer is Option $(A)$.
Note:
To obtain the respective acid and base of the salt, we must have the knowledge of the positive and the negative ion the salt will produce on dissociation when dissolved in water. The positive ion of the salt will react with hydroxyl ion and form base, and the negative ion of the salt will react with hydrogen ion and form acid. We must also have knowledge of the strong and weak acid and bases.