Question #1e971

Nov 10, 2017

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 $N {a}^{+}$ and $O {H}^{-}$ ions.

If you have 1 Mol of NaOH and dissolve it in 1 litre of ${H}_{2} O$, then you will get 1 mol of $N {a}^{+}$-ions and 1 mol of $O {H}^{-}$-ions, making it a 1M solution.

Now consider the same with 1 Mol $M g {\left(O H\right)}_{2}$.

This will dissolve into $M g {2}^{+}$-ions and once again $O {H}^{-}$-ions,

BUT:

You now have twice as many $O {H}^{-}$-ions as with $N a O H$.
Also, we are dealing here with $M {g}^{\text{2+}}$, which can for instance bind with 2 $C {l}^{-}$-ions.

Therefore, though you started with 1 Mol of $M g {\left(O H\right)}_{2}$, you end up with a solution that could be seen as 2 Molar with regard to the amount of $O {H}^{-}$.

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

So:

a solution of 1 M $N a O H$ = 1N(ormal), as it yields 1 mol $O {H}^{-}$-ions,
a solution of 1 M $M g {\left(O H\right)}_{2}$ = 2N(ormal), as it yields 2 mol $O {H}^{-}$-ions.

Same of course goes for acids, where you take ${H}_{3} {O}^{+}$ as a yardstick...

Nov 10, 2017

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 ${H}_{3} {O}^{+}$ 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.