# When dissolved in equal volumes of water, which of the solutions will have a higher pH?

## If two equal numbers of moles of nitric acid and formic acid were dissolved in an equal volume of water, which one would have the higher pH?

Feb 28, 2017

The formic acid solution.

#### Explanation:

The idea here is that nitric acid, ${\text{HNO}}_{3}$, is a strong acid and formic acid, $\text{HCOOH}$, is a weak acid. This implies that nitric acid dissociates completely in aqueous solution to produce hydronium cations, ${\text{H"_3"O}}^{+}$

${\text{HNO"_ (3(aq)) + "H"_ 2"O"_ ((l)) -> "H"_ 3"O"_ ((aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

As you can see, every mole of nitric acid added to the solution will dissociate to produce $1$ mole of hydronium cations.

On the other hand, formic acid does not dissociate completely in aqueous solution to produce hydronium cations

${\text{HCOOH"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "HCOO}}_{\left(a q\right)}^{-}$

Formic acid has an acid dissociation constant, ${K}_{a}$, equal to

${K}_{a} = 1.77 \cdot {10}^{- 4}$

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

This means that for the above equilibrium reaction, you have

${K}_{a} = \left(\left[\text{H"_3"O"^(+)] * ["HCOO"^(-)])/(["HCOOH}\right]\right)$

Since ${K}_{a} < 1$, the equilibrium lies to the left, meaning that most of the moles of formic acid added to the solution will not ionize to produce hydronium cations.

For equal numbers of moles of nitric acid and formic acid dissolved in equal volumes of water, the concentration of hydronium cations will be higher for nitric acid and lower for formic acid.

["H"_ 3"O"^(+)]_ "nitric acid" > ["H"_ 3"O"^(+)]_ "formic acid"

Since

color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))

you can say that the higher the hydronium concentration in the resulting solution, the lower the pH. This implies that

$\text{pH"_ "formic acid" > "pH"_"nitric acid}$