Question

Question #1e971

Answer 1

In simple terms, and for lack of a better description, let's say it depends on the amount of "active groups"

Explanation:

Let's take NaOH: this readily (and completely) disintegrates in water, and produces `Na^+` and `OH^-` ions.

If you have 1 Mol of NaOH and dissolve it in 1 litre of `H_2O`, then you will get 1 mol of `Na^+`-ions and 1 mol of `OH^-`-ions, making it a 1M solution.

Now consider the same with 1 Mol `Mg(OH)_2`.

This will dissolve into `Mg2^+`-ions and once again `OH^-`-ions,

BUT:

You now have twice as many `OH^-`-ions as with `NaOH`.
Also, we are dealing here with `Mg^"2+"`, which can for instance bind with 2 `Cl^-`-ions.

Therefore, though you started with 1 Mol of `Mg(OH)_2`, you end up with a solution that could be seen as 2 Molar with regard to the amount of `OH^-`.

Could be a bit confusing, that's why Normality was introduced:

So:

a solution of 1 M `NaOH` = 1N(ormal), as it yields 1 mol `OH^-`-ions,
a solution of 1 M `Mg(OH)_2` = 2N(ormal), as it yields 2 mol `OH^-`-ions.

Same of course goes for acids, where you take `H_3O^+` as a yardstick...

Answer 2

Normality is similar to molarity but it uses "gram-equivalent weight" of a solute in stating amount of solute per litre (molarity uses "gram molecular weight".)

Explanation:

For example, if you have 1 M sulphuric acid, in an acid/base reaction it would be 2 N ("twice normal"), due to each mole of sulphuric acid yielding 2 moles of `H_3O^+` ions.

On the other hand, if you used 1 M sulphuric acid for something like precipitation of sulphates, then it would be 1 N ("normal") because each mole of sulphuric acid yields a mole of sulphate ions.