# Question 1c86b

May 17, 2017

We recall that:

$\text{Molarity" = "mols solute"/"L solution}$

Define a general polyprotic acid, $\text{H"_n"A}$, whose complete dissociation would be represented as:

${\text{H"_color(red)(n)"A" + color(red)(n)"H"_2"O"(l) -> color(red)(n)"H"_3"O"^(+)(aq) + "A}}^{\textcolor{red}{n} -} \left(a q\right)$

where $n$ is the number of protons in the acid.

Then, its normality is:

" "ul(" "" "" "" "" "" "" "" "" "" "" "" "" "" ")
" "ulbb(|" ""Normality" = color(red)(n) xx "Molarity"" "| )

In a context that you might actually use, consider a polyprotic acid, such as ${\text{H"_3"PO}}_{4}$.

For example, if we have a $\text{1 M}$ molarity, then we actually have a $\text{3 N}$ normality, because it takes $\text{3 mol}$s of ${\text{OH}}^{-}$ to neutralize $\text{1 mol}$ of ${\text{H"_3"PO}}_{4}$.

Or, define a general hydroxide-containing strong base, "B"("OH")_n (where $\text{B}$ is the base, not boron). Then, its complete dissociation would be represented as:

${\text{B"("OH")_(color(red)(n))(aq) -> "B"^(color(red)(n)+)(aq) + color(red)(n)"OH}}^{-} \left(a q\right)$

Then, its normality is also as defined above.

For example, if we have a $\text{2 M}$ molarity of "Ba"("OH")_2, then we actually have a $\text{4 N}$ normality, because it takes $\text{4 mol}$s of ${\text{H}}^{+}$ to neutralize $\text{2 mols}$ of "Ba"("OH")_2#.

As a note, the normality unit is discouraged by IUPAC and NIST due to its ambiguity and dependence on context.