How to find Normality (N)

Normality is one of the concentration units of solutions in stoichiometry. It is defined as the gram equivalents of solute, dissolved per liter of a solution. Gram equivalents (G.Eq) are the mass of a substance that can produce one mole of chemically active species in solutions.

For example, 1 normal (N) solution of sulphuric acid;

Sulphuric acid (H2SO4) has a molar mass of 98 g/mol. It means that if its 98g are diluted to make a liter of solution, its normality will be 2N (2H in one H2SO4 molecule). Therefore, in order to make a 1N solution,

49 grams of H2SO4 ⇒ One mole of hydrogen ions (active species)

Thus, one gram equivalents (required to make 1N solution) of sulphuric acid = 49 grams.

In order to calculate the normality of a given solution, chemists titrate it against a counter solution with a known concentration. Acid and bases produce hydrogen ions H+ and hydroxyl OH ions as active species. In salts, these species are electrons.

Normality = Number of Gram equivalents / Liter of solution  →  (i)

Gram equivalent= Molar wt / active species  →  (ii)

Units of Normality

When gram equivalent weight is divided per liter of solution, the units used are g.eq/ liter or just ‘N’.

 Equivalent weight in normality

Gram equivalent weight is the mass of a substance that produces one mole active species (like H+ or OH). The nature of active species depends on the type of reactions, taking place due to counter-active species present inside solutions.

For example, in acid base titration reactions, only hydrogen (H+) and hydroxyl (OH) ions react to neutralize each other so, they shall be called the active species.

Significance of Normality

Normality is a useful parameter to indicate the strength of the solutions based on the activity of their constituents. It is frequently used in standard lab procedures. For example, in wastewater analysis, normal (N) concentrations are calculated, as active species are the main concern there.

The normality calculations do not require a balanced chemical equation. Although, it helps in balancing chemical equations. The species that take part in the redox reaction are actually electrons. Hence, we can balance the chemical equations by balancing the number of active species on both sides of the chemical equations.

For example, phosphoric acid ionizes to produce 3 ions of H+,

H3PO4  →  3H+ + PO43-  (contain 3 active species)

Normality formulae

The normality of a solution can be found via several methods, the best and the most convenient of which is to treat the solution (with unknown concentration) against a standard-normal solution. For that, a primary standard solution is first prepared, then titrated against the secondary solution, until the endpoint.

The endpoint and equivalence points are of prime importance in titrimetry, but during titration, we are only concerned with visible endpoints with suitable chemical indicators.

Calculation of normality by acid-base titrations

It is really important to choose the solution of counteractive species (acid against a base). These chemical species are used as an indirect measure of the normality of a given solution. For this purpose, volumes of both counter-active species along with the normality of the standard solution are required.

N1V1 =N2V2

where,

  • Normality of first chemical (N1)
  • Volume of first chemical (V1)
  • Normality of second chemical (N2)
  • Volume of second chemical  (V2)

Molarity (M) to Normality (N)

To calculate normality from given molarity, the number of chemically active species is simply multiplied with the molarity value, in order to get, normality.

N= Number of active species x molarity

N = a.M

Percentage purity (%) to Normality (N)

Normality can be calculated from the percentage purity of a substance by the following formula;

Normality = [Percentage purity x density x 10] / Gram equivalent weight of the substance

\fn_phv N = \frac{%purity \times \rho \times 10}{Eq.wt}

For example, 37% HCL solution is available as a laboratory reagent. It has a density of 1.49 kg/m3 and a gram equivalent weight of 36.461 g. If its normality is calculated, it comes out to be 15.120 N.

Values of normality and molarity for some chemical reagents are shown below as examples;

Chemical reagents
Density (kg/m3)
Molarity (M) Normality (N)
Sulfuric acid (98%) 1.84 18.4 36.8
Nitric acid (70%) 1.42 15.8 15.8
Perchloric acid (70%) 1.67 11.6 11.6
Sodium hydroxide (47%) 1.5 17.6 17.6
Acetic acid (99.5%) 1.05 17.4 17.4

Explanation about Normality (M)

Normality of HCl

HCL hydrochloric acid is a very important lab chemical. During titration, a known concentration solution of this acid is usually required. Its gram equivalent weight is 36.5 g and it has 1 active species (H+) per molecule. Hence, one normal solution of HCL gas contains 36.5 grams of HCL gas dissolved per liter of solution.

Normality of sodium thiosulphate

To find out the normality (N) of sodium thiosulfate, consider this redox equation;

2Na2S2O3+I2Na2S4O6+2NaI

In this equation, two moles of thiosulphate ions lose two electrons, on average (n=1), which means that its molarity and normality shall be equal. Since it has a gram equivalent weight of 248.18 g/eq.wt, its 248.18 grams are dissolved per liter to prepare 1N nad 1M solution.

Normality vs molarity: which one to prefer

The comparison between normality (N) and molarity (M) is much similar to that of molarity (M) and molality (m).

Normality (N) is the number of gram equivalents of solute dissolved per liter of solution, whereas, molarity is the number of moles of solute dissolved per liter of solution. Normality and molarity are two different entities. They are sometimes intermixed in such a way that they are the same for many compounds and their solutions.

For example, 1 molar sodium hydroxide is exactly the same amount as 1N NaOH. Similarly, HCl, HNO3, HClO4, NaHSO4, NaHCO3, etc have the same amounts for 1 normal and 1 molar solution.

Preference over one another

As an example, when potassium permanganate (KMnO4) dissolves in water, it loses five electrons. These are the active species of this compound.

Molar wt of KMnO4 (Mol. wt) = 158.04 (g/mol)

No of active species (n) = 5

Gram equivalent wt (G.eq) = Mol. wt / n

Hence, G.eq  =  31.6 (g eq.wt)

Thus, it can be concluded that molarity (M) is easier to calculate as compared to normality (N). The reason is that normality involves the number of active species (n), which can only be known after studying its behavior in a reaction.

Key Takeaway(s)

How to find normality- Normality formulae

Concepts Berg

When to use Normality?

It is used in lab standard procedures where active species are involved.

How is the molarity of a percentage solution calculated?

The Molarity of the percentage solution is given by,

M = (% purity x Densityx 10) / Mr

How do you calculate the normality of Na2CO3?

For the normality of Na2CO3, we require its molecular weight (MW) and its number of active species.

Na2CO+ H2O → 2Na+ + 2OH + CO

One mole of sodium carbonate flourishes two moles of (OH) ions, so it’s n = 2,

Molar mass (Mr) = 106 g/mol

Gram equivalent (G.eq) = Mr / n   ⇒   106/2   ⇒   53 g

Hence, 53 grams of sodium carbonate is required to make 1000ml of its solution. This will make 1 normal (N) Na2CO3 solution.

How do you calculate normality from mass?

Mass can be converted to moles as follows;

Moles = Mass / Molar mass

Gram equivalent = Moles / no of active species

Normality = Gram equivalent / liters of solution

How to find normality from molarity?

The normality of a substance can be calculated from its molarity by the simple formula;

Normality= Molarity x number of active species

What is the normality of potassium dichromate?

In a chemical reaction, potassium dichromate reacts as an oxidizing agent. It gains 3 electrons such that its oxidation state changes from “+6” to “+3”.

Cr+6 → Cr+3

Number of active species = 3

G.eq wt = Molar mass / 3 ⇒ 294.18/3 ⇒ 98.06 grams

Hence, its normality is 3 times its molarity.

How would you derive the normality equation N1V1=N2V2?

Normality is given by,

Normality (N) = Gram equivalents (G.Eq) / Volume in liters (V) → (i)

Let’s take an example,

HCL + NaOH → NaCl + H2O

From a stoichiometrically balanced chemical equation, 1 mole of HCl is required to completely neutralize one mole of sodium hydroxide, so,

1 Gram equivalent of acid = 1 Gram equivalent of base → (ii)

From eq (i),

G.Eq of acid = Noramilty of acid (N1) x volume of solution (V1)

G.Eq of base = Normality of base (N2) x Volume of solution (V2)

Hence,

N1V1 = N2V2

References

  • Chemistry, fifth edition, by Steven S. Zumdhal and Susan A. Zumdhal (University of Illinois, Urbana Champaign, IL, USA)
  • Fundamentals of Analytical Chemistry, second edition, by Douglas A. Skoog (Stanford University) and Donald M. West (San Jose State College)

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