Jump to content

Normality computations


Recommended Posts

10 hours ago, Dhamnekar Win,odd said:

image.png.78e63e148c9a021a02d7baafa1607875.png

Are these above normality computations correct? If yes how? I don't understand these computations.

Any chemistry help will be accepted.

Consider the following

NaOH

Ca(OH)2

HCl

H2SO4

One molecule of sodium hydroxide will neutralise one molecule of hydrochloric acid in an acid - base reaction.

But it takes two molecules of sodium hydroxide to neutralise one molecule of sulphuric acid.

So in some sense the sulphuric acid has twice the neutralising power of hydrochloric acid.

The equivalent mass or equivalent weight is an old fashioned method of recognising this fact and we say that the equivalent weight of sulphuric acid is half its molecular weight.
By the same token we say that the equivalent weight of hydrochloric acid is equal to its molecular weight.
Or we say that the molecular weight of hydrochloric acid contains one 'equivalent' and the molecular weight of sulphuric acid contains two 'equivalents'.

This method of 'equivalents' tells us how much of a given acid we need to neutralise a given quantity of sodium hydroxide.

 

The same thing goes the other way when we consider calcium hydroxide, as compared to sodium hydroxide.

It takes twice as much acid to neutralise the calcium hydroxide as for the sodium hydroxide.

So the molecular weight of sodium hydroxide contains one equivalent and the molecular weight of calcium hydroxide contains two equivalents.

 

 

Does this help ?

 

Link to comment
Share on other sites

Create an account or sign in to comment

You need to be a member in order to leave a comment

Create an account

Sign up for a new account in our community. It's easy!

Register a new account

Sign in

Already have an account? Sign in here.

Sign In Now
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.