Question

The equation shows the neutralization reaction:-
${{H}_{2}}S{{O}_{4}}+2NaOH\xrightarrow{{}}N{{a}_{2}}S{{O}_{4}}+2{{H}_{2}}O$ . What is$N{{a}_{2}}S{{O}_{4}}$?

Answer 1

Hint:Neutralization reaction occurs when an acid reacts with a base and gives salt with water as products. This definition can be applied to the above reaction and we easily answer.

Complete step-by-step answer:First let us discuss the neutralization reaction.
A neutralization reaction is a reaction in which an acid and a base react quantitatively to produce salt and water. In this reaction, we can generally see the combination of ${{H}^{+}}$ and $O{{H}^{-}}$ions to form ${{H}_{2}}O$ as a product. When a strong acid reacts with the strong base, neutralization leads to exothermic reaction as heat is released.
Now, let us see what happens when sulfuric acid (${{H}_{2}}S{{O}_{4}}$ ) reacts with sodium hydroxide ($NaOH$).
-${{H}_{2}}S{{O}_{4}}$ in aqueous medium:-
 ${{H}_{2}}S{{O}_{4}}_{(aq)}\xrightarrow{{}}{{H}^{+}}_{(aq)}+S{{O}_{4}}{{^{2-}}_{(aq)}}\text{ ----------eq(A)}$
Since, ${{H}_{2}}S{{O}_{4}}$is a strong acid, therefore it completely dissociates in aqueous medium.
-$NaOH$ in aqueous medium:-
$NaO{{H}_{(aq)}}\xrightarrow{{}}N{{a}^{+}}_{(aq)}+O{{H}^{-}}_{(aq)}\text{ ----------eq(B)}$
Similarly, $NaOH$being a strong base, completely dissociates in aqueous medium.
 We know that, each ${{H}^{+}}$ion react with each $O{{H}^{-}}$ion to produce ${{H}_{2}}O$ but ${{H}_{2}}S{{O}_{4}}$produces 2${{H}^{+}}$ions therefore we require 2 $O{{H}^{-}}$ions for them. Similarly, each $S{{O}_{4}}^{2-}$ ion require 2$N{{a}^{+}}$ ions to produce$N{{a}_{2}}S{{O}_{4}}$ . Therefore, we multiply equation (B) with 2.
$2NaO{{H}_{(aq)}}\xrightarrow{{}}2N{{a}^{+}}_{(aq)}+2O{{H}^{-}}_{(aq)}\text{ ----------eq(C)}$
On equating the equation (A) and (B), we get the following reaction:-
${{H}_{2}}S{{O}_{4}}_{(aq)}+2NaO{{H}_{_{(aq)}}}\xrightarrow{{}}N{{a}_{2}}S{{O}_{{{4}_{(aq)}}}}+2{{H}_{2}}{{O}_{_{(aq)}}}$ \[
    \text{ }\!\![\!\!\text{ since }{{H}^{+}}+O{{H}^{-}}\xrightarrow{{}}{{H}_{2}}O \\
   \text{ }S{{O}_{4}}^{2-}+N{{a}^{+}}\xrightarrow{{}}N{{a}_{2}}S{{O}_{4}}] \\
 \]
Let’s check the reaction according to the definition of neutralization reaction,
When ${{H}_{2}}S{{O}_{4}}$(a strong acid) reacts with $NaOH$(a strong base) quantitatively, $N{{a}_{2}}S{{O}_{4}}$(salt) is produced along with${{H}_{2}}O$.
By this we conclude that, $N{{a}_{2}}S{{O}_{4}}$is a salt produced in this neutralization reaction.

Note:When you come across chemical reactions, try to balance it and solve further. Also, these neutralization reactions are very helpful while performing acid-base titrations.