# What reaction occurs when 25.0 mL of 0.600 M "NaOH" are mixed with 15.0 mL of 0.400 M "HNO"_3? What is the balanced equation and the net ionic equation?

Mar 31, 2016

A neutralization reaction.

#### Explanation:

You're mixing aqueous solutions containing sodium hydroxide, $\text{NaOH}$, a strong base, and nitric acid, ${\text{HNO}}_{3}$, a strong acid, which means that you're dealing with a neutralization reaction.

The strong base and the strong acid will ionize completely in aqueous solution to form cations and anions

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

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

Now, when these two solutions are mixed, the hydroxide anions, ${\text{OH}}^{-}$, and the hydronium ions, ${\text{H"_3"O}}^{+}$, will neutralize each other and form water, $\text{H"_2"O}$.

The reaction will also produce aqueous sodium nitrate, ${\text{NaNO}}_{3}$.

The balanced chemical equation for this reaction will look like this

${\text{NaOH"_ ((aq)) + "HNO"_ (3(aq)) -> "H"_ 2"O"_ ((l)) + "NaNO}}_{3 \left(a q\right)}$

The complete ionic equation will look like this

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

Eliminate spectator ions, which are ions present on both sides of the equation

$\textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{Na"_ ((aq))^(+)))) + "OH"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+) + color(red)(cancel(color(black)("NO"_ (3(aq))^(-)))) -> 2"H"_ 2"O"_ ((l)) + color(red)(cancel(color(black)("Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("NO}}_{3 \left(a q\right)}^{-}}}}$

to get the net ionic equation, which looks like this

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{OH"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+) -> 2"H"_ 2"O}}_{\left(l\right)}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, what's interesting here is that the problem provides you with the molarities and volumes of the two solutions.

This means that you can determine if the neutralization will be complete or incomplete.

You know that you have

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to get

${n}_{N a O H} = \text{0.600 M" * 25.0 * 10^(-3)"L" = "0.0150 moles NaOH}$

${n}_{H N {O}_{3}} = {\text{0.400 M" * 15.0 * 10^(-3)"L" = "0.00600 moles HNO}}_{3}$

Notice that you have more moles of sodium hydroxide than of nitric acid.

This means that the acid will be completely consumed before all the hydroxide ions will be neutralized $\to$ you're dealing with an incomplete neutralization.

The resulting solution will thus contain

${n}_{N a O H} = 0.0150 - 0.00600 = \text{0.00900 moles NaOH}$

${n}_{H N {O}_{3}} = \text{0 moles } \to$ completely consumed

${n}_{N a N {O}_{3}} = 0 + 0.00600 = {\text{0.00600 moles NaNO}}_{3}$

The chemical species present in the resulting solution will include

${\text{Na}}^{+}$, ${\text{OH}}^{-} \to$ from the ionization of the remaining sodium hydroxide

${\text{Na}}^{+}$, ${\text{NO}}_{3}^{-} \to$ the aqueous sodium nitrate produced by the reaction