Normality of substance 1 at Equivalence point Solution

STEP 0: Pre-Calculation Summary
Formula Used
Normality of Solution 1 = Normality of Solution 2*(Volume of Solution 2/Volume of Solution 1)
N1 = N2*(V2/V1)
This formula uses 4 Variables
Variables Used
Normality of Solution 1 - (Measured in Mole per Cubic Meter) - The Normality of Solution 1 is described as the number of gram or mole equivalents of solution 1 present in one liter of solution 1.
Normality of Solution 2 - (Measured in Mole per Cubic Meter) - The Normality of Solution 2 is described as the number of gram or mole equivalents of solution 2 present in one liter of solution 2.
Volume of Solution 2 - (Measured in Cubic Meter) - The volume of solution 2 gives the volume of the solution 2 in liters.
Volume of Solution 1 - (Measured in Cubic Meter) - The Volume of Solution 1 gives the volume of solution 1 in liters.
STEP 1: Convert Input(s) to Base Unit
Normality of Solution 2: 0.005 Equivalents per Liter --> 5 Mole per Cubic Meter (Check conversion ​here)
Volume of Solution 2: 1E-05 Liter --> 1E-08 Cubic Meter (Check conversion ​here)
Volume of Solution 1: 0.0002 Liter --> 2E-07 Cubic Meter (Check conversion ​here)
STEP 2: Evaluate Formula
Substituting Input Values in Formula
N1 = N2*(V2/V1) --> 5*(1E-08/2E-07)
Evaluating ... ...
N1 = 0.25
STEP 3: Convert Result to Output's Unit
0.25 Mole per Cubic Meter -->0.00025 Equivalents per Liter (Check conversion ​here)
FINAL ANSWER
0.00025 Equivalents per Liter <-- Normality of Solution 1
(Calculation completed in 00.020 seconds)

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National Institute Of Technology (NIT), Surathkal
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Number of Equivalents and Normality Calculators

Normality of substance 1 at Equivalence point
​ LaTeX ​ Go Normality of Solution 1 = Normality of Solution 2*(Volume of Solution 2/Volume of Solution 1)
Normality of substance 2 at Equivalence point
​ LaTeX ​ Go Normality of Solution 2 = Normality of Solution 1*(Volume of Solution 1/Volume of Solution 2)
Number of Equivalents of Solute
​ LaTeX ​ Go Number of Equivalents = Mass of Solute/Equivalent Weight
Normality given Molarity and Valency Factor
​ LaTeX ​ Go Normality = Molarity*N Factor

Normality of substance 1 at Equivalence point Formula

​LaTeX ​Go
Normality of Solution 1 = Normality of Solution 2*(Volume of Solution 2/Volume of Solution 1)
N1 = N2*(V2/V1)

What is normality?

Normality is described as the number of gram or mole equivalents of solute present in one liter of a solution. When we say equivalent, it is the number of moles of reactive units in a compound. Normality in chemistry is one of the expressions used to measure the concentration of a solution. It is abbreviated as ‘N’ and is sometimes referred to as the equivalent concentration of a solution. It is mainly used as a measure of reactive species in a solution and during titration reactions or particularly in situations involving acid-base chemistry.

What is equivalence point?

The equivalence point, or stoichiometric point, of a chemical reaction, is the point at which chemically equivalent quantities of reactants have been mixed. In other words, the moles of acid are equivalent to the moles of the base, according to the equation (this does not necessarily imply a 1:1 molar ratio of acid: base, merely that the ratio is the same as in the equation). It can be found by means of an indicator, for example, phenolphthalein or methyl orange.

How to Calculate Normality of substance 1 at Equivalence point?

Normality of substance 1 at Equivalence point calculator uses Normality of Solution 1 = Normality of Solution 2*(Volume of Solution 2/Volume of Solution 1) to calculate the Normality of Solution 1, The Normality of substance 1 at equivalence point formula is defined as the product of normality of substance 2 and the ratio of the volume of substance 2 to the volume of substance 1. Normality of Solution 1 is denoted by N1 symbol.

How to calculate Normality of substance 1 at Equivalence point using this online calculator? To use this online calculator for Normality of substance 1 at Equivalence point, enter Normality of Solution 2 (N2), Volume of Solution 2 (V2) & Volume of Solution 1 (V1) and hit the calculate button. Here is how the Normality of substance 1 at Equivalence point calculation can be explained with given input values -> 2.5E-7 = 5*(1E-08/2E-07).

FAQ

What is Normality of substance 1 at Equivalence point?
The Normality of substance 1 at equivalence point formula is defined as the product of normality of substance 2 and the ratio of the volume of substance 2 to the volume of substance 1 and is represented as N1 = N2*(V2/V1) or Normality of Solution 1 = Normality of Solution 2*(Volume of Solution 2/Volume of Solution 1). The Normality of Solution 2 is described as the number of gram or mole equivalents of solution 2 present in one liter of solution 2, The volume of solution 2 gives the volume of the solution 2 in liters & The Volume of Solution 1 gives the volume of solution 1 in liters.
How to calculate Normality of substance 1 at Equivalence point?
The Normality of substance 1 at equivalence point formula is defined as the product of normality of substance 2 and the ratio of the volume of substance 2 to the volume of substance 1 is calculated using Normality of Solution 1 = Normality of Solution 2*(Volume of Solution 2/Volume of Solution 1). To calculate Normality of substance 1 at Equivalence point, you need Normality of Solution 2 (N2), Volume of Solution 2 (V2) & Volume of Solution 1 (V1). With our tool, you need to enter the respective value for Normality of Solution 2, Volume of Solution 2 & Volume of Solution 1 and hit the calculate button. You can also select the units (if any) for Input(s) and the Output as well.
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