# Is titration suitable for sodium nitrate?

Sep 10, 2015

Acid-base titration would not be suitable for $N a N {O}_{3}$.

#### Explanation:

The nitrate ion, $N {O}_{3}^{-}$, is a very weak base that would be protonated only under strongly acidic conditions. Therefore, acid-base titration would be an unsuitable method to analyze $N a N {O}_{3}$ solutions.

Sep 17, 2015

Maybe a redox titration, you'll probably need to heat the solution before doing it though.

#### Explanation:

As the previous answerer said, an acid base titration wouldn't be suitable for $N a N {O}_{3}$ if water is the solvent, as $N {O}_{3}^{-}$ is too weak a base to change the pH to make a titration suitable. The same goes for $N {a}^{+}$ but being too weak an acid.

$N {O}_{3}^{-}$ has no insoluble salts, or at least no common insoluble salts, so we can't do a precipitation titration either. $N {a}^{+}$ has this same trouble.

$N {O}_{3}^{-}$ isn't a good coordinating agent so a complexation titration isn't viable either. There is a complexating agent for $N {a}^{+}$ and other alkali metals, but it's rarely used and expensive so it isn't worth considering.

$N {a}^{+}$ won't reduce to $N {a}^{0}$ easily, and if it does it'll react with water violently so it isn't a good idea to do it either.
$N {O}_{3}^{-}$ does have a redox reaction though,

$N {O}_{3}^{-} + 3 {H}^{+} + 2 {e}^{-} \leftrightarrow H N {O}_{2} + {H}_{2} O$ ${E}_{red}^{0} = 0 , 94$

Which could theoretically be coupled with ${I}^{-}$ to produce ${I}_{3}^{-}$ which can then be titrated with ${S}_{2} {O}_{3}^{-} 2$.

$3 {I}^{-} \leftrightarrow {I}_{3}^{-} + 2 {e}^{-}$ ${E}_{\odot x}^{0} = - 0 , 536$

That being said, I'm not sure if the $\Delta E$ would be big enough outside of standard conditions.

${E}^{0}$ values copied from Skoog's fundamentals of Analytic Chemistry