Question

$NaOH+{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow { Na }_{ 2 }{ SO }_{ 4 }+{ H }_{ 2 }O$. Balance the given chemical equation.

Answer 1

Hint: In order to balance any equation, determine the n-factor or the valence factor of the reacting species. Since this is a neutralization reaction, determine the n-factor of the acid and the base.

Complete step by step solution:
The reaction given in the above question is a neutralization reaction. In a neutralization reaction, the acid and the base quantitatively react with one another to produce salt and water molecules.
$acid+base\longrightarrow salt+water$
Given below is an example for neutralization reaction:
$NaOH(aq)+HCl(aq)\longrightarrow NaCl(aq)+{ H }_{ 2 }O(l)$
If a strong acid and strong base react with each other such that there is complete neutralization, the pH of the resulting solution will be equal to 7. On the other hand, the neutralization of a strong acid and weak base will give a pH of less than 7. Similarly the neutralization of a strong base and a weak acid will give a pH of greater than 7.
Since the given reaction is between a strong acid and a strong base, the pH will be 7.
For balancing a neutralization reaction, we should follow the following steps:
(a). Determine the n-factor for the acid and the base. N-factor is the number of the protons and the hydroxide that are reacting in order to give the salt. Since ${NaOH}$ is reacting with $ { H }_{ 2 }S{ O } $ to give $ { Na }_{ 2 }{ SO }_{ 4 } $, therefore the n-factor for the acid is 2 and the n-factor for the base is 1.
(b). Now cross-multiply the simplest ratio of the n-factors for the acid and the base. Now balance all the atoms on both sides except for oxygen and hydrogen atoms.
(c). Now balance the oxygen atoms followed by hydrogen atoms on both sides.

Hence the balanced equation is:
$ 2NaOH+{ H }_{ 2 }S{ O }_{ 4 }\longrightarrow { Na }_{ 2 }{ SO }_{ 4 }+{ 2H }_{ 2 }O $

Note: The n-factor or the valence factor for a species is not a fixed quantity. It changes with the type of reaction, the products of the reaction so be careful while calculating the n-factor for a particular species.