Have you ever wondered how chemists dilute a stock solution to the desired concentration with such accuracy? Well, they are able to do so using a very important equation in the chemist’s toolkit, i.e., N1V1 = N2V2.
In this article, you will learn how to use N1V1 = N2V2 with lots of examples, so continue reading!
N1V1 = N2 V2 is formally referred to as the normality equation. It is actually an alternative way of writing the dilution formula; C1V1 = C2V2.
If the concentration (C) of a solution is recorded in terms of normality (N), then C1V1 = C2V2 transforms into N1V1 = N2V2; where,
N1 = The normality of the original, undiluted solution.
V1 = The initial volume of the given undiluted solution.
N2 = The final normality achieved after diluting the given solution.
V2 = Final volume of the solution prepared after dilution.
Normality (N) is the concentration of a solution measured in the number of equivalents of a solute per liter of the solution.
An equivalent is the amount of a substance that can either donate or accept one mole of a specific chemical species, such as H+ ions, OH– ions, and/or electrons.
The concept of normality is most applicable for measuring the concentration of acidic or alkaline solutions.
Normality can be simply determined from the molarity of a solution.
Molarity (M) is defined as the moles of solute dissolved per liter of the solution.
To convert molarity into normality, we can use the formula:
Normality (N) = M x Eq.
where;
M= molarity (moles/L)
Eq.= Number of equivalents per mole
For example, an HCl molecule has only 1 proton (H+ ion) available for donation. So, the number of equivalents per mole (Eq) = 1. The molar mass of HCl is 36.5 g/mol.
1 molar HCl solution contains 36.5 grams of hydrochloric acid.
So, 1 normal solution of HCl = 36.5 gram equivalents per litre.
Contrarily, 2 H+ ions are available per H2SO4 molecule for donation. Here, the number of equivalents per mole (Eq) = 2. The molar mass of H2SO4 is 98.08 g/mol.
1 molar H2SO4 solution contains 98.08 grams of sulfuric acid.
1 M H2SO4 = 2 N H2SO4 = 98.08 gram equivalents per L.
Thus, 1 normal H2SO4 = 98.08 / 2 = 49.0 gram equivalents per liter.
In short, normality (N) is usually measured in g equivalent per L.
In contrast, the volume (V) of a solution is calculated in liters (L) or milliliters (mL) where 1 L = 1000 mL.
What are the units used in the N1V1 = N2V2 equation?
As discussed above, in the equation N1V1 = N2V2, the symbols N1 and N2 represent normality which is one way of measuring the concentration of a solution, while V1 and V2 denote volume.
Normality (N) is measured in gram equivalents of solute per liter of the solution if the volume (V) of the solution is measured in liters (L).
However, you must keep in mind that consistency in units is integral for calculating an unknown value by using the normality equation.
Consistency in units implies that if N1 of the stock solution is given in g eq. /L, then we will determine N2 in g eq. /L too.
Similarly, both V1 and V2 must be measured in the same units (either L or mL).
Now let’s find out, through the help of the given examples, how to use N1 V1 = N2 V2 for preparing different concentration solutions in the chemistry laboratory.
How to use N1V1 = N2V2 equation?
Based on which quantity is unknown, the normality equation can be rearranged in one of the following forms.
Example # 1:Sarah is a BS chemistry student. She was given a stock solution of HCl prepared in distilled water. The initial concentration of the given solution was 0.05 N. She carefully added 25 mL of this 0.05 N HCl solution in a 100 mL volumetric flask and diluted it up to the mark with water. Can you help her find the final concentration of the diluted solution?
In the above example, the initial concentration (normality) of the stock solution of HCl is given, i.e., N1 = 0.05 N.
The initial volume (V1) is also given, i.e., V1 = 25 mL.
Sarah diluted it up to the mark to make the final volume V2 = 100 mL.
So we can easily apply the normality equation and find the final concentration, i.e., N2.
∴ N1V1 = N2V2
⇒ (0.05) (25) = N2 (100)
∴ N2 = 0.0125 N.
Result: Sarah diluted the HCl solution such that the normality of the solution obtained is 0.0125 N.
Example # 2:If Sam wants to prepare a 500 mL solution with a normality of 0.5 by diluting a 1 N stock solution, how much of the stock solution and how much water does he need to use?
From the information given in the question statement, we know;
⇒ N1 = 1.0 N and N2 = 0.5 N.
The final volume he needs to achieve V2 = 500 mL.
The initial volume (V1) required of the 1.0 N stock solution to prepare 0.5 N solution can be determined as follows:
∴ N1V1 = N2V2
⇒ (1.0) (V1) = (0.5)(500)
∴ V1 = 250 mL.
Result: Sam requires 250 mL of 1.0 N stock solution, and he needs to add 500- 250 = 250 mL water to prepare the diluted solution.
More examples on N1V1 = N2V2 equation
Example # 3:A diluted solution (Y) is prepared from the 0.025 N stock solution (X). The normality of solution Y is 0.015 N in the final solution volume, 50 mL. Find the volume of solution X taken to prepare Y.
The following information is provided in the question statement:
⇒ N1 = 0.025 N
⇒ N2 = 0.015 N
⇒ V2 = 50 mL
The initial volume (V1) is unknown.
Now let’s apply the normality equation to find V1.
∴ N1V1 = N2V2
⇒ (0.025) (V1) = (0.015)(50)
∴ V1 = 30 mL.
Result: 30 mL solution X is taken to prepare solution Y.
Example # 4:1000 mL of 0.87 N solution P is prepared by diluting 0.75 L solution M with distilled water. What is the initial concentration of undiluted M in terms of normality?
The following information is provided in the question statement:
⇒ N2 = 0.87 N
⇒ V1 = 0.75 L
⇒ V2 = 1000 mL
The initial normality (N1) is unknown.
You must note that the units of V1 and V2 are inconsistent. So, we first need to convert either both volumes in litres or in millimeters (mL).
∴ 1 L = 1000 mL
⇒ V1 = 0.75 L = 0.75 x 1000 = 750 mL.
Now let’s apply the normality equation to find N1.
∴ N1V1 = N2V2
⇒ (N1) (750) = (0.87) (1000)
∴ N1 = 1.16 N.
Result: The initial concentration of undiluted solution M is 1.16 N.
Example # 5:A solution of HCl with unknown normality is titrated with 0.2 N NaOH solution. It is found that 25 mL of the NaOH solution is required to completely neutralize 10 mL of HCl solution. What is the normality of the HCl solution?
As per the acid-base neutralization reaction for HCl vs. NaOH;
1 mole HCl reacts completely with 1 mole NaOH. The titration formula is:
Here n1 = n2 = 1 while the concentration (C) is measured in normality (N), so the above equation becomes;
∴ N1V1 = N2V2, i.e., the normality equation.
As per the question statement, N1 = normality of HCl = unknown. N2 = normality of NaOH = 0.2 N.
⇒ V1 = volume of HCl = 10 mL
⇒ V2 = volume of NaOH used to titrate 10 mL HCl = 25 mL.
The equation N1V1 = N2 V2 is known as the normality equation. It is an alternative way of writing the dilution formula, i.e., C1 V1 = C2 V2, if the concentration (C) of a solution is measured in terms of normality (N).
What is normality?
Normality is the concentration of a solution measured in gram equivalents of solute dissolved per liter of the solution.
The symbol N is used to represent normality, calculated in g eq. / L.
What do the symbols N1, V1, N2 and V2 represent in the normality equation (N1V1 = N2V2)?
N denotes normality, while V stands for volume.
In the equation N1 V1 = N2 V2;
⇒ N1 = The normality of the original, undiluted solution.
⇒ V1 = The initial volume of the undiluted solution (N1) required to prepare N2.
⇒ N2 = The final concentration (in normality) achieved after diluting the given solution.
⇒ V2 = Final volume of the solution prepared after dilution.
How to use equation N1V1 = N2V2?
We can use equation N1V1 = N2V2 by finding an unknown volume or normality if the other three quantities are known.
The normality equation can be rearranged in one of the following ways to make the unknown quantity the subject of the formula.
Vishal Goyal is the founder of Topblogtenz, a comprehensive resource for students seeking guidance and support in their chemistry studies. He holds a degree in B.Tech (Chemical Engineering) and has four years of experience as a chemistry tutor. The team at Topblogtenz includes experts like experienced researchers, professors, and educators, with the goal of making complex subjects like chemistry accessible and understandable for all. A passion for sharing knowledge and a love for chemistry and science drives the team behind the website. Let's connect through LinkedIn: https://www.linkedin.com/in/vishal-goyal-2926a122b/
1 thought on “What is Normality Equation (N1V1=N2V2) in Chemistry? – Examples”
Ilayaraja
Sir please give me solution for following question
Eg.
V1- 200 ml N1- 269 ppm
This solution dissolved into
V2- 100 ml N2 – 57 ppm solution
What is the final solution of TDS in ppm
Topblogtenz is a website dedicated to providing informative and engaging content related to the field of chemistry and science. We aim to make complex subjects, like chemistry, approachable and enjoyable for everyone.
Sir please give me solution for following question
Eg.
V1- 200 ml N1- 269 ppm
This solution dissolved into
V2- 100 ml N2 – 57 ppm solution
What is the final solution of TDS in ppm