Question

The ${{V}_{1}} ml$ of $NaOH$ of normality $X$ and ${{V}_{2}} ml$ of $KOH$ of normality $Y$ are together sufficient to neutralize completely $100ml$ of $0.1N HCl$ . If ${{V}_{2}}:{{V}_{1}}=1:2$ and $X:Y=1:2$ , what fraction of the acid is neutralised by $KOH$ ?
(A) $0.25$
(B) $0.33$
(C) $0.50$
(D) $0.67$

Answer 1

A neutralization reaction can be defined as a chemical reaction in which an acid and a base react together quantitatively to form a product of salt and water. The above reaction in the question is also a neutralization reaction. We assume the values of the normality and volume according to the question and use it to calculate the moles of KOH and thus, the fraction of acid neutralised by it.

Formula Used:
For complete neutralization to occur, we have the following formula
${{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}={{N}_{3}}{{V}_{3}}$
Where
${{N}_{1}}$ is the normality of $NaOH$
${{V}_{1}}$ is the volume of $NaOH$
${{N}_{2}}$ is the normality of $KOH$
${{V}_{2}}$ is the volume of $KOH$
${{N}_{3}}$ is the normality of $HCl$
${{V}_{3}}$ is the volume of $HCl$ .

Complete Step-by-Step Solution
According to the question, the following information is provided to us:
The volume of $NaOH$ is ${{V}_{1}}$
The normality of $NaOH$ is $X$
The volume of $KOH$ is ${{V}_{2}}$
The normality of $KOH$ is $Y$
The volume of $HCl$ is $100ml={{10}^{-1}} Litre$
The normality of $HCl$ is $0.1$
The ratios of volumes of $KOH$ and $NaOH$ used is given as $2:1$
The ratios of normality of $NaOH$ and $KOH$ given as $1:2$
Now,
Let us suppose the normality of $NaOH$ as $X$
Then the normality of $KOH$ will be $2X$
Similarly,
Let us suppose the volume of $KOH$ be $V$
Then the volume of $NaOH$ will be $2V$
Now,
$[{{H}^{+}}]={{10}^{-1}}$
For complete neutralization to occur, we have the following formula
${{N}_{1}}{{V}_{1}}+{{N}_{2}}{{V}_{2}}={{N}_{3}}{{V}_{3}}$
Now, we will substitute the values of assumed and known variables in the given equation
That is,
$2VX+2VX={{10}^{-1}}\times {{10}^{-1}}$
$\Rightarrow 4VX={{10}^{-2}}$
Upon solving the above equation, we get
$VX=0.0025$
Now,
For $KOH=2\times VX=2\times 0.0025=0.005$
The fraction of acid neutralized by $KOH$ is given by the formula
$\dfrac{no. of moles of KOH}{total no. of moles of HCl} =\dfrac{0.005}{0.01}=0.5$
Hence, the correct option is (C).

Note
Reactions of ionic equations to net neutralization include solid bases, solid salts, water, and solid acids. Neutralisation is the reaction between an acid and a base that forms water and salt. Net ionic equations for neutralisation reactions can be provided by solid acids, solid bases, solid salts, and water.