Question

To neutralize completely $20$ ml of $0.1$ M aqueous solution of phosphorous acid (${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}$), the volume of $0.1$ M aqueous solution required is:
A. $10$ ml
B. $20$ml
C. $40$ml
D. $60$ml

Answer 1

The solution of this question involves neutralization reaction and milliequivalent concept. Milliequivalents is one-thousandth of equivalent${\text{(1}}{{\text{0}}^{{\text{ - 3}}}}{\text{)}}$.The equivalent can be defined as how much quantity of substance will react with hydrogen ions or the atomic weight of the substance in gram divided by its valance.

Complete step by step answer:
Complete step-by-step answer:
We have studied that neutralization reaction involves a reaction between an acid and base and forms salt and water. These Reactions involve the complete use of both acid and base.Water is formed as a result of the reaction between ${{\text{H}}^{{\text{ + \;}}}}$and ${\text{O}}{{\text{H}}^{\text{-}}}$ ions. Below is the normal equation between acid and base
$Acid + Base \to Salt + water$
We will solve this by using the milliequivalents concept.
We know that neutralization reaction is occurring so the milliequivalents should be equal.
${\text{Milliequivalents of }}{{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}}{\text{ = Milliequivalents of KOH}}$
${{20 \times 0}}{{.1 \times 2 = V}} \times {\text{0}}{\text{.1}}$
Where$20$ ml is the solution used in the reaction.
$0.1$is the molarity of phosphorous acid.
Basicity of phosphorous acid is $2$(because phosphorous acid a dibasic acid) due to its decomposition and it is as shown in the reaction below;
${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{3}}} \to {\text{2}}{{\text{H}}^{\text{ + }}}{\text{ + HP}}{{\text{O}}_{\text{3}}}^{{\text{2 - }}}$
By solving the above equation, we get,
${\text{V}}$ = $40$ ml

So, the correct answer is Option C.

Note: We know that electrolytes concentration is mostly measured as milliequivalents. It is a chemical activity with respect to $1$ mg of hydrogen. So, we can say that $1$m equivalent is denoted by $1$ mg of hydrogen. Milliequivalents can be calculated by using the formula
${\text{mEq = }}\dfrac{{{{mg \times valence}}}}{{{\text{molar mass}}}}$
The concept of milliequivalent is helpful in titration also to calculate in mole because the volume will be in millilitre, molarity in mol per litre. It should be noted that neutralization reaction is a type of ionisation reaction because hydrogen ions and hydroxide ions are formed. The water formed in the reaction is a by-product.