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Question: The volume of \(0.025{\text{M Ca}}{({\text{OH}})_2}\) solution which can neutralize 100 ml of \({10^...

The volume of 0.025M Ca(OH)20.025{\text{M Ca}}{({\text{OH}})_2} solution which can neutralize 100 ml of 10−4M H3PO4{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

Explanation

Solution

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 dm−3mol{\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+{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 OH−O{H^ - } ions a base molecule may produce to get the acidity value.
Molarity of  H3PO4{\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}= M1{M_1}= 10−4M{10^{ - 4}}M
Volume of  H3PO4{\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}= V1{V_1}= 100 ml
Molarity of Ca(OH)2Ca(OH){_2}= M2{M_2}= 0.025 M
Basicity of  H3PO4{\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}= 3
Molarity x Basicity Equals Normality
Normality of  H3PO4{\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}= N1{N_1}= 10−4M{10^{ - 4}}Mx 3
Normality of  H3PO4{\text{ }}{{\text{H}}_3}{\text{P}}{{\text{O}}_4}= N1{N_1}= 3×10−4N3 \times {10^{ - 4}}N
Acidity of Ca(OH)2Ca(OH){_2}= 2
Molarity x Acidity = Normality
Normality of Ca(OH)2Ca(OH){_2}= N2{N_2}= 0.025 x 2
Normality of Ca(OH)2Ca(OH){_2}= N2{N_2}= 0.05 N
Now
2H3PO4+3Ca(OH)2→Ca3(PO4)2+6H2O2{{\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
V1N1=V2N2{V_1}{N_1} = {V_2}{N_2}
⇒V2=N1V1N2=3×10−4×100ml0.05\Rightarrow {V_2} = \dfrac{{{N_1}{V_1}}}{{{N_2}}} = \dfrac{{3 \times {{10}^{ - 4}} \times 100{\text{ml}}}}{{0.05}}
⇒V2=0.6ml\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.