Question

The volume of $0.025{\text{M Ca}}{({\text{OH}})_2}$ solution which can neutralize 100 ml of ${10^{ - 4}}{\text{M }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$ is
(A) 10 ml
(B) 60 ml
(C) .6 ml
(D) 2.8 ml

Answer 1

Hint :Molar concentration (also known as molarity, quantity concentration, or substance concentration) is a measurement of a chemical species' concentration in a solution in terms of the amount of substance per unit volume of solution. The number of moles per litre, abbreviated as mol/L or \[mol{\text{ }}d{m^{ - 3}}\]in SI units, is the most often used unit for molarity in chemistry.

Complete Step By Step Answer:
The gram equivalent of solute dissolved in one cubic decimeter or one litre of solution is known as normality. The normalcy unit is N. It's the best option for titration calculations. The normality of a solution is equal to one. The product of equivalent mass and normality can be used to calculate the strength of a solution. It is measured in g per litre.
The normality of acids may be computed using the following formula:
Molarity x Basicity Equals Normality
Count the quantity of ${H^ + }$ions an acid molecule may produce to get the basicity value.
The normality of bases may be computed using the following formula:
Molarity x Acidity = Normality
Count the quantity of $O{H^ - }$ ions a base molecule may produce to get the acidity value.
Molarity of ${\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$= ${M_1}$= ${10^{ - 4}}M$
Volume of ${\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$= ${V_1}$= 100 ml
Molarity of $Ca(OH){_2}$= ${M_2}$= 0.025 M
Basicity of ${\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$= 3
Molarity x Basicity Equals Normality
Normality of ${\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$= ${N_1}$= ${10^{ - 4}}M$x 3
Normality of ${\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}$= ${N_1}$= $3 \times {10^{ - 4}}N$
Acidity of $Ca(OH){_2}$= 2
Molarity x Acidity = Normality
Normality of $Ca(OH){_2}$= ${N_2}$= 0.025 x 2
Normality of $Ca(OH){_2}$= ${N_2}$= 0.05 N
Now
$2{{\text{H}}_3}{\text{P}}{{\text{O}}_4} + 3{\text{Ca}}{({\text{OH}})_2} \to {\text{C}}{{\text{a}}_3}{\left( {{\text{P}}{{\text{O}}_4}} \right)_2} + {\text{6}}{{\text{H}}_2}{\text{O}}$
Using law of equivalence
${V_1}{N_1} = {V_2}{N_2}$
$ \Rightarrow {V_2} = \dfrac{{{N_1}{V_1}}}{{{N_2}}} = \dfrac{{3 \times {{10}^{ - 4}} \times 100{\text{ml}}}}{{0.05}}$
$ \Rightarrow {V_2} = 0.6ml$
Hecne option C is correct.

Note :
The Law of Chemical Equivalence asserts that all reactants and products in a chemical process must have the same equivalents. Volumetric analysis, also known as titrimetric analysis, is a method of quantitative chemical analysis in which the amount of a substance is determined by measuring the volume it occupies or, in broader use, the volume of a second substance that reacts with the first in established proportions.