Question #1e971

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...

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.