# How can normalities be interchangeably used?

So I was doing the following question on stoichiometry . The part $$\pu{1.80N}\ \ce{HCl}$$ can also be written as $$\pu{0.9 N}\ \ce{ H_2SO_4}$$ but I don't understand why .

$$\pu{10 g}$$ sample of 'gas liquor' ($$\ce{NH}^+_4$$ salt) is boiled with $$\ce{NaOH}$$ and the resulting $$\ce{NH3}$$ is passed into $$\pu{60 ml}$$ of $$\pu{1.8 N}\ \ce{HCl}$$ . Excess $$\ce{H_2SO4}$$ required $$\pu{10cm^3}$$ of $$\pu{0.40 N}\ \ce{NaOH}$$. What is the % of $$\ce{NH3}$$ in gas liquor?

I got the answer but just can't understand why the two can be used interchangeably.

• Lookup the chemical definition of normality. – MaxW Apr 25 '19 at 20:14
• No. Of gram equivalents/ volume of solution in litres – gucci Apr 25 '19 at 20:15
• Dah... I have read the question poorly again. // You are right, there is a mistake. There is half as much acid in 0.90 N $\ce{H2SO4}$ as there is in 1.80 N $\ce{HCl}$. // There is as much acid in 0.9 molar $\ce{H2SO4}$ as there is in 1.80 molar $\ce{HCl}$. – MaxW Apr 25 '19 at 20:23
• Yeah that's what i was thinking. Its N=M$\times$ n factore ==> 1.8 =0.9 $\times 2$. – gucci Apr 26 '19 at 1:30
• @gucci" "Its N=M× n factor ==> 1.8 =0.9 ×2"- You will never learn this gram equivalent concept if you memorize this relation. It does not work all the time, and fails for most redox reagents. – M. Farooq Apr 26 '19 at 12:13

The fundamental idea behind gram equivalents and normality is that one gram equivalent of an acid reacts with one gram equivalent of a base. This idea eliminates the need of using molarity and thinking about mole ratios for any acid base titration.

Equal volume 1 N HCl will react with 1 N NaOH. Similarly, 1 liter of 1 N $$\ce{H2SO4}$$ or 1 N $$\ce{H3PO4}$$ will consume 1 L of 1 N NaOH.

The key calculations is that you should know how to calculate normality from a given molecular weight and a balanced equation:

For example: $$\ce{HCl}$$ -> $$\ce{H+}$$ + $$\ce{Cl-}$$ gram equivalent weight= formula weight/ (No. of acidic protons) = 36/1 =36

$$\ce{H2SO4}$$ -> $$\ce{2H+}$$ + $$\ce{SO4^2-}$$ gram equivalent weight= formula weight/ (No. of acidic protons) = 98/2 = 49

Can you now make the connection that why "The part 1.80 N HCl can also be written as 0.9 N H2SO4" is quite wrong? because

1 N HCl = 1 N $$\ce{H2SO4}$$

and both solutions contain 1 gram equivalents of protons.

Yes if your teacher were comparing normality and molarity, then it is different.

Molarity calculations are different and far more easier. By definition, M of X= moles of A/ Total volume in which X is present.

A 1 M HCl just means there is one mole HCl (gas) dissolved in 1 L of solution Similarly, 1 M $$\ce{H2SO4}$$ is 1 mole $$\ce{H2SO4}$$ dissolved in 1 L of solution.

If you were to titrate 1 L of 1 M NaOH with 1 M HCl you would require 1 L of HCl. However, since sulfuric acid furnishes two moles of proton $$\ce{H+}$$ for each mole of sulfuric acid, only 500 mL would be required.

In short, 1 M HCl is equivalent to 0.5 M $$\ce{H2SO4}$$ in terms of titer for $$\ce{OH-}$$ ion.

Contrast this molarity unit with normality which is almost obsolete except in some South Asian colleges.

• Ok.....but i am eager to know as to what would happen if we compared molarity and normality @M.Farooq – gucci Apr 26 '19 at 1:52
• H2SO4: 1 M = 2 N // HCl: 1 M = 1 N – Poutnik Apr 26 '19 at 3:51
• The student needs to learn the concept of normalities, otherwise they will have no clue why 1 M H2SO4 = 2 N HCl. They may apply the same theme to H3BO3 as well. – M. Farooq Apr 26 '19 at 12:11