Question

The reaction of copper hydroxide and dil.${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$ Produces:
(A) Normal salt
(B) Mixed salt
(C) Complex salt
(D) Double salt

Answer 1

The acid generates the ${{\text{H}}^{\text{+}}}$ and with the hydroxide ion $\text{O}{{\text{H}}^{-}}$ from the base and produces the salt and water is called the neutralization reaction. The salt is formed by the replacement of the hydrogen ion by the copper and forms a normal copper sulfate salt.

Complete step by step solution:
According to the Arrhenius concept, an acid is a substance that dissociates to give hydrogen ion ${{\text{H}}^{\text{+}}}$ when dissolved in water. On the other hand, the base is a substance that dissociates to give hydroxyl ion $\text{O}{{\text{H}}^{-}}$ in the water when dissolved in the water.
All the metals other than the alkali metals form the hydroxide with the general formula $\text{M(OH}{{\text{)}}_{\text{2}}}$. The metal hydroxides release the hydroxide ion and thus increases the $\text{pH}$ of the solution. Therefore, the metal hydroxides are basic in solution.
The copper hydroxide $\text{Cu (OH}{{\text{)}}_{\text{2}}}$ is a base. But it reacts with the acid.
The neutralization reaction is a chemical reaction that is defined as the reaction in which the acid reacts with the base and forms the salt and water. The general reaction is as depicted follows,

\text{Acid} & \text{+} & \text{Base} & \to & \text{Salt} & \text{+} & \text{Water} \\\ \end{matrix}$$ In a neutralisation reaction, the hydrogen ion ${{\text{H}}^{\text{+}}}$ combines with the hydroxide ion $\text{O}{{\text{H}}^{-}}$ to form the water. This is generally an acid-base neutralization reaction. Let’s consider the reaction of copper hydroxide with the sulphuric acid $${{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}}$$. The copper hydroxide reacts with the sulphuric acid to form copper sulfate as salt and the water. The general reaction is as depicted below: $$\begin{matrix} \text{Cu(OH}{{\text{)}}_{\text{2}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} & \to & \text{CuS}{{\text{O}}_{\text{4}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{O} \\\ \text{(Copper hydroxide)} & {} & \text{(Sulfuric acid)} & {} & \text{(Copper sulphate)} & {} & \text{(water)} \\\ \end{matrix}$$ The salt generated is normal. Normal salt is formed by the replacement of the replaceable hydrogen from the acid by the metal or group elements that act as the metal. Here, the two hydrogens of sulphuric acid are replaced by the one copper ion which is acting as the metal and forms the copper sulfate which is a normal salt. **Hence, (A) is the correct option.** **Note:** There are different types of salts such as: 1) Acid salt: This salt contains the one or more replaceable hydrogen in the molecule and it is formed by the partial replacement of H atoms by metals or positive radicals. For example, $$\text{NaHS}{{\text{O}}_{\text{4}}}$$ ,$$\text{NaHC}{{\text{O}}_{\text{3}}}$$, etc. 2) Base salt: It is a salt that contains the ${{\text{O}}_{\text{2}}}-$ or $\text{OH}-$ in its molecules. Some of the examples are, $$\text{Mg }\left( \text{OH} \right)\text{ Cl}$$etc.