Question

# The enthalpy of Neutralization of $HCl$ and $NaOH$ is x KJ. The heat evolved when $500$ ml of $2{\text{N}}$ of $HCl$ is mixed with $250$ ml of $4{\text{N}}$ $NaOH$ will be-A. xB. $\dfrac{x}{2}$ C.$\dfrac{x}{4}$ D.$\dfrac{{2x}}{3}$

Hint: Use formula of number of gram equivalent of a molecule,
$\Rightarrow {\text{No}}{\text{. of gram equivalent = N}} \times {\text{V}}$
Where N is the normality and V is the volume of solution in litres.

Step-by-Step Solution: Given that enthalpy of neutralization of $HCl$ and $NaOH$ is x KJ.
Normality of $HCl$=$2$ and Volume = $500$ ml$= \dfrac{{500}}{{1000}}$ litres.
And Normality of $NaOH$=$4$ and volume =$250$ml= $\dfrac{{250}}{{1000}}$ litres.
We have to find the heat of neutralization when these two solutions are mixed.
We know that the heat of neutralization is equal to the enthalpy change of reaction when one gram equivalent of acid is neutralized by one gram equivalent of base.
Here $HCl$ is acid and $NaOH$ is base. So we have to find the number of gram equivalents of both.
We know that
$\Rightarrow {\text{No}}{\text{. of gram equivalent = N}} \times {\text{V}}$
Where N is the normality and V is the volume of solution in litres.
Then No. of gram equivalents of $HCl$= ${\text{Normality of HCl}} \times {\text{Volume of acid}}$
On putting the given values we get,
$\Rightarrow {\text{No}}{\text{. of gram equivalents of HCl}} = \;2 \times \dfrac{{500}}{{1000}}$
On simplifying we get,
$\Rightarrow {\text{No}}{\text{. of gram equivalents of HCl}} = \;\dfrac{{1000}}{{1000}} = 1$
Now, No. of gram equivalents of $NaOH$= ${\text{Normality of NaOH}} \times {\text{Volume of base}}$
On putting the given values we get,
$\Rightarrow$ No. of gram equivalents of $NaOH = 4 \times \dfrac{{250}}{{1000}}$
On simplifying we get,
$\Rightarrow$ No. of gram equivalents of $NaOH = \dfrac{{1000}}{{1000}} = 1$
Since here one gram equivalent of acid is neutralized by one gram equivalent of base then,
Heat evolved= Enthalpy of neutralization
And we know that enthalpy of neutralization of $HCl$ and $NaOH$ is x KJ
∴Heat evolved= x KJ

Hence the correct answer is A.

Note: Gram equivalent weight gives the measure of reactive capacity of a molecule. Normality is the number of gram or mole equivalents of solute dissolved in one litre of a solution.
Number of gram equivalent can also be expressed by this formula-
$\Rightarrow {\text{No}}{\text{. of gram equivalent = }}\dfrac{{{\text{Weight of solute}}}}{{{\text{Equivalent weight of solute}}}}$
Equivalent weight is calculated by dividing the molecular weight by the charge number. Charge number is the number of protons or hydroxide which the compound contains.