The n-factor of ${H_2}S{O_4}$ in the following reaction is:
${H_2}S{O_4} + KOH \to KHS{O_4} + {H_2}O$
A. \[1\]
B. $2$
C. $\dfrac{1}{2}$
D. $3$

Hint: The number of replaceable hydrogen atoms from an acid by a base is known as basicity of the acid and these number of replaceable hydrogen atoms denote the n-factor or the number of transferred electrons throughout the completion of the reaction.

Complete step by step answer:
We have been provided with the equation:
${H_2}S{O_4} + KOH \to KHS{O_4} + {H_2}O$
In this reaction, there is a neutralization taking place when a strong acid like the sulfuric acid reacts with a strong base like the potassium hydroxide to produce the potassium bisulfate salt along with the release of water from it. Let the oxidation number of sulfur in sulfuric acid be $x$ . Then,
$2( + 1) + x + 4( - 2) = 0$
Thus, the oxidation state of the sulfur atom in sulfuric acid is:
$x = + 6$
Now, with the replacement of a hydrogen atom with a potassium atom, we get a salt of potassium bisulfate. The oxidation state of the sulfur atom in this salt will be:
$1 + 1 + x + 4( - 2) = 0$
$x = + 6$
As we can see that, in both the cases there has been no change in the oxidation state of the sulfur atom. This means that with the transfer of one electron, the sulfuric acid releases a proton and in exchange of it, the potassium ion from the potassium hydroxide replaces this hydrogen atom in the weak conjugate base formed by the acid $(HSO_4^ - )$ .
Hence, the n-factor or the number of replaceable hydrogen atoms is equal to 1.

So, the correct answer is Option A .


Note:
The strong acids have a tendency of losing more protons as there is a complete dissociation of the acids in the aqueous solutions. Thus, a salt of a strong acid and a strong base has a higher value of ionization constant as their n-factor is high.