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Have you ever come across the term “normality” while studying titrations or chemical solutions?
Normality is a way to express how strong a solution is, especially when acids or bases are involved. It tells us how many reactive parts called equivalents are present in one litre of a solution. In simple terms, it helps us understand how much of a substance can actually take part in a chemical reaction. This concept is widely used in labs, especially during acid- base reactions and titration experiments.
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Normality is a way to measure the concentration of a solution, just like molarity. But instead of just counting the number of moles, normality focuses on the active part of the substance that reacts in a chemical reaction. This active part is called an equivalent. For example, in an acid base reaction, normality tells us how many hydrogen ions(H+) an acid can give or how many hydroxide ions (OH-) a base can take. In redox reactions, it shows how many electrons are gained or lost. In precipitation reactions, it helps measure the ions that combine to form a solid.
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The normality of a solution is the gram equivalent weight of a solute per litre of solution. It may also be called the equivalent concentration. In simple words, the number of gram equivalents of solute dissolved per litre of the solution. It is denoted by N. It is used for acid or base solution.
For units of concentration, the symbols N, eq/L, or meq/L (= 0.001 N) are used. For example, 0.1 N HCl could be used to describe the concentration of a hydrochloric acid solution.
A gram equivalent weight or equivalent is a measure of the reactive capacity of a given chemical species (ion, molecule, etc.). The equivalent value is determined using the molecular weight and valence of the chemical species. Normality is the only concentration unit that is reaction dependent.
\(N = \frac{\text{Gram equivalents of solute}}{\text{Volume of solution in litres}}\)
\({Gram equivalents} = \frac{\text{Weight}}{\text{Equivalent Weight}} \times \frac{1000}{V\ (\text{ml})}\)
Where VVV is the volume of solution in millilitres.
\({Equivalent Weight} = \frac{\text{Molar Mass}}{n}\)
Where n is the number of electrons exchanged (for redox), or \(H+^++/OH−^-−\) ions (for acid-base), or ionic charge in reactions.
where
$\mathrm{n}=$ number of $H^{+}$in Acid
OH− in base and for salt charge present in ionic forms equivalent weight is a new thing in the formula.
Carefully go through the below information on how to calculate normality and make a note of the normality equation to refer in for revision.
The below normality formula is important to calculate normality.
Normality can be calculated as:
\text { Normality formula }=\frac{\text { number of gram quivalents of solute }}{\text { volume of solution in litres }}
The term ‘gram equivalents’ has been used herein the normality definition. Basically, gram equivalent is the amount of a substance that will react or supply 1 mole of hydrogen ions (H+) or 1 mole of electrons (e–). For example, we have sulphuric acid H2SO4. When sulphuric acid dissociates completely, it ‘supplies’ 2 moles of H+ and 1 mole of sulfate ion.
\(\mathrm{H}_2 \mathrm{SO}_4(\mathrm{aq})=2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{SO}_4^{2-}(\mathrm{aq})\)
This means that 1mole of sulphuric acid (98.08 g/mol) gives 2 moles of hydrogen ions. Hence,\frac{98.08}{2}=49.04 \mathrm{~g} / \mathrm{mol} will give 1 mol of hydrogen ions. So, 49.04 g/eq is the gram equivalent weight of H2SO4.
In the case of NaOH, it dissociates into
\(\mathrm{NaOH}_{(\mathrm{s})} \xrightarrow[\text { complete dissociation }]{\text { Dissolves in water with }}\) \(\mathrm{Na}_{(\mathrm{aq})}^{+}+\mathrm{OH}_{(\mathrm{aq})}^{-}\)
That means, 1 mole of NaOH gives 1 mole of sodium ions and 1 mole of hydroxide ions. The molar mass of sodium hydroxide is 40 g/mol and it will ‘react’ with 1 mole of hydrogen ions (when hydroxide ions react with hydrogen ions, water is produced). So, the gram equivalent of NaOH will be 40g/eq.
In this way, you can find gram equivalents of all the other compounds as well.
\text { Gram equivalent of solute }=\frac{\text { mass of solute }}{\text { equivalent mass }}
The unit of normality is gm equivalent per liter.
Here are some solved examples on the normality formula.
Solved Example 1: We have 36.5 grams of 1 L HCl solution. Calculate the Normality of the HCl solution.
Solution:
Molecular Mass of HCl = 36.5
Number of H^+ ion = 1
So equivalent Weight = Molecular mass/1 = 36.5
No. of Gram Equivalent of Solute = Mass of Solute/Equivalent Weight = 36.5 /36.5 = 1
Volume here is given as 1 L
Normality will be calculated as
Normality = No. of Gram Equivalent of Solute/ Volume of Solution in L = 1 N
Solved Example 2: Calculate the normality of NaOH solution Formed by dissolving 0.2 gm NaOH to make 250 ml solution.
\begin{aligned}
& N=\frac{\text { Gram eq. of solute }}{\text { Volume of sol. In litre }} \\
& \text { No. of Gram Eq. of Solute }=\frac{\text { weight }}{\text { Equivalent weight }} \\
& \text { Now, Equivalent weight }=\frac{\text { Molar Mass }}{n}=\frac{23+16+1}{1}=40 \\
& \text { So, } N=\frac{\text { weight }}{\text { Equivalent weight }} \times \frac{1000}{V} \\
& =\frac{2}{40} \times \frac{1000}{250} \\
& =0.2 \mathrm{~N}
\end{aligned}
Feature |
Normality (N) |
Molarity (M) |
Meaning |
Measures active part (equivalents) in 1 litre of solution |
Measures number of moles in 1 litre of solution |
Focus |
Looks at how many parts actually take part in a reaction |
Just counts total number of moles, not specific action |
Used In |
Acid-base, redox, and precipitation reactions |
General chemistry and solution calculations |
Formula |
Normality = Equivalents / Volume (L) |
Molarity = Moles / Volume (L) |
Changes with Reaction Type? |
Yes, it depends on the type of reaction |
No, it's always the same |
Units |
Equivalents per litre (eq/L) |
Moles per litre (mol/L) |
Example |
1 M H₂SO₄ = 2 N (because it gives 2 H⁺ ions) |
1 M H₂SO₄ = 1 M (always based on total moles) |
Normality is used in many real life chemistry situations:
Always know what type of reaction you’re dealing with (acid-base, redox, etc.,)
We hope this article was helpful in clearing all your doubts and queries and helped in your exam preparation. Need more help with your studies? You can also check out other Chemistry topics on the Testbook website. Get help from the experts to prepare for the exams with selective study material, mock tests and valuable insights all designed to help you score better marks in the exam. Download the free Testbook App to grab some exclusive offers now.
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