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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.

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  • $\begingroup$ Lookup the chemical definition of normality. $\endgroup$ – MaxW Apr 25 at 20:14
  • $\begingroup$ No. Of gram equivalents/ volume of solution in litres $\endgroup$ – gucci Apr 25 at 20:15
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    $\begingroup$ 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 $\ce{HCl}$. $\endgroup$ – MaxW Apr 25 at 20:23
  • $\begingroup$ Yeah that's what i was thinking. Its N=M$\times$ n factore ==> 1.8 =0.9 $\times 2$. $\endgroup$ – gucci Apr 26 at 1:30
  • $\begingroup$ @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. $\endgroup$ – M. Farooq Apr 26 at 12:13
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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.

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  • $\begingroup$ Ok.....but i am eager to know as to what would happen if we compared molarity and normality @M.Farooq $\endgroup$ – gucci Apr 26 at 1:52
  • $\begingroup$ H2SO4: 1 M = 2 N // HCl: 1 M = 1 N $\endgroup$ – Poutnik Apr 26 at 3:51
  • $\begingroup$ 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. $\endgroup$ – M. Farooq Apr 26 at 12:11

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