Question

What is the `"normality"` of a `4.0*g` mass of sodium hydroxide dissolved in a `1.0*L` volume of water?

Answer 1

`"Normality"=1.0*N`

Explanation:

`"Normality"` is defined similarly to `"molarity"`.

`"Normality"=("Moles of "H_3O^+"or "HO^-)/"Volume of solution"`

Here, `"normality"=((4.0*g)/(40.0*g*mol^-1))/(0.100*L)=1.0*N`.

This will react quantitatively with one equiv of acid. One the other hand, a solution of diprotic `H_2SO_4(aq)` that is `0.50*mol*L^-1` with respect to `H_2SO_4`, but necessarily `1.0*mol*L^-1` with respect to `H_3O^+` (why?), would also be regarded as `1*N`.