
What is the principle of volumetric analysis? Derive the relationship $ {N_1}{V_1}\, = \,{N_2}{V_2} $.
Answer
547.8k+ views
Hint: Volumetric analysis is a widely used quantitative analytical method, it involves the measurement of the volume of a solution with known concentration, which is used to find the concentration of the analyte.
Complete step by step answer:
The basic principle of Volumetric analysis:
The solution which we want to analyze contains a chemical of unknown amount then the reagent reacts with that chemical of unknown amount in the presence of an indicator to show the end-point. End-point shows that the reaction is complete.
Then we measure the volumes by the method of titration which completes the reaction between the reagent and solution, then the amount of unknown chemical in the solution is calculated by using the mole fraction of the equation. After the end-point of reaction is reached, volumetric analysis calculations of the analyte are done by the formula - $ {C_a}\, = \,{C_t}{V_t}M/{V_a} $
Where, $ {C_a} $ is the concentration of the analyte.
$ {C_t} $ is the concentration of titrant. $ V $ is the volume of the titrant.
$ M $ is the mole ratio of the analyte and the reactant. $ V $ is the volume of the analyte.
$ {N_1}{V_1}\, = \,{N_2}{V_2} $ is the normality equation which can be simply derived by using the Law of equivalence, which states that the number of gram equivalents of reactant is equal to gram equivalents of product. Let’s take the example of an acid-base titration reaction to derive this equation.
As we know
$ Normality\,\,\, = \,\,\,\dfrac{{no.\,of\,gram\,equivalent}}{{Volume(in\,L)}} $
Let the $ Normality $ and $ Volume $ of the Acid be $ {N_1} $ and $ {V_1} $ and $ Normality $ and $ Volume $ of Base be $ {N_2} $ and $ {V_2} $ .
The no of gram equivalent will be ;
$ No.\,of\,gram{{ }}equivalent = Normality(N)\, \times \,Volume(V) $ $ $
And as we know,
$ No.\,of\,gram\,equivalent\,of\,Acid\, = \,No.\,of\,gram\,equivalent\,of\,Base $
Hence after putting the variables, the equation will be
$ {N_1}{V_1}\, = \,{N_2}{V_2} $
This is our desired equation.
Note:
Normality is described as the number of grams equivalent of solute that are present in one liter of solution. It is denoted by the letter $ N $ . after molarity, normality is the most used parameter to analyze the concentration of the solutes.
Complete step by step answer:
The basic principle of Volumetric analysis:
The solution which we want to analyze contains a chemical of unknown amount then the reagent reacts with that chemical of unknown amount in the presence of an indicator to show the end-point. End-point shows that the reaction is complete.
Then we measure the volumes by the method of titration which completes the reaction between the reagent and solution, then the amount of unknown chemical in the solution is calculated by using the mole fraction of the equation. After the end-point of reaction is reached, volumetric analysis calculations of the analyte are done by the formula - $ {C_a}\, = \,{C_t}{V_t}M/{V_a} $
Where, $ {C_a} $ is the concentration of the analyte.
$ {C_t} $ is the concentration of titrant. $ V $ is the volume of the titrant.
$ M $ is the mole ratio of the analyte and the reactant. $ V $ is the volume of the analyte.
$ {N_1}{V_1}\, = \,{N_2}{V_2} $ is the normality equation which can be simply derived by using the Law of equivalence, which states that the number of gram equivalents of reactant is equal to gram equivalents of product. Let’s take the example of an acid-base titration reaction to derive this equation.
As we know
$ Normality\,\,\, = \,\,\,\dfrac{{no.\,of\,gram\,equivalent}}{{Volume(in\,L)}} $
Let the $ Normality $ and $ Volume $ of the Acid be $ {N_1} $ and $ {V_1} $ and $ Normality $ and $ Volume $ of Base be $ {N_2} $ and $ {V_2} $ .
The no of gram equivalent will be ;
$ No.\,of\,gram{{ }}equivalent = Normality(N)\, \times \,Volume(V) $ $ $
And as we know,
$ No.\,of\,gram\,equivalent\,of\,Acid\, = \,No.\,of\,gram\,equivalent\,of\,Base $
Hence after putting the variables, the equation will be
$ {N_1}{V_1}\, = \,{N_2}{V_2} $
This is our desired equation.
Note:
Normality is described as the number of grams equivalent of solute that are present in one liter of solution. It is denoted by the letter $ N $ . after molarity, normality is the most used parameter to analyze the concentration of the solutes.
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