# What concentration of NO_3^(-) results when 697 mL of 0.515 M NaNO_3 is mixed with 635 mL of 0.785 M Ca(NO_3)_2?

Dec 22, 2015

["NO"_3^(-)] = "1.02 M"

#### Explanation:

As you know, sodium nitrate, ${\text{NaNO}}_{3}$, and calcium nitrate, "Ca"("NO"_3)_2, are both soluble in aqueous solution.

This means that when dissolved in water, these ionic compounds will dissociate completely and exist as cations and anions.

More specifically, they will dissociate to form

${\text{NaNO"_text(3(aq]) -> "Na"_text((aq])^(+) + "NO}}_{\textrm{3 \left(a q\right]}}^{-}$

and

${\text{Ca"("NO"_3)_text(2(aq]) -> "Ca"_text((aq])^(2+) + color(red)(2)"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

Notice that sodium nitrate dissociates in a $1 : 1$ mole ratio with the nitrate anions, which means that every mole of the former will produce $1$ mole of nitrate anions in solution.

On the other hand, calcium nitrate dissociates in a $1 : \textcolor{red}{2}$ mole ratio with the nitrate anions, which means that every mole of the former will produce $\textcolor{red}{2}$ moles of nitrate anions in solution.

Use the molarities and volumes of the two solution to figure out how many moles of each salt you start with

$\textcolor{b l u e}{c = \frac{n}{V} \implies n = c \cdot V}$

${n}_{N a N {O}_{3}} = {\text{0.515 M" * 697 * 10^(-3)"L" = "0.3590 moles NanO}}_{3}$

n_(Ca(NO_3)_2) = "0.785 M" * 635 * 10^(-3)"L" = "0.4985 moles Ca"("NO"_3)_2

This corresponds to

0.3590 color(red)(cancel(color(black)("moles NaNO"_3))) * ("1 mole NO"_3^(-))/(1color(red)(cancel(color(black)("mole NaNO"_3)))) = "0.3590 moles NO"_3^(-)

and

0.4985 color(red)(cancel(color(black)("moles Ca"("NO"_3)_2))) * (color(red)(2)" moles NO"_3^(-))/(1color(red)(cancel(color(black)("mole Ca"("NO"_3)_2)))) = "0.9970 moles NO"_3^(-)

The total number of moles of nitrate anions will thus be

${n}_{\text{total" = 0.3590 + 0.9970 = "1.356 moles NO}} _ {3}^{-}$

Now, in order to get the concentration of the anions, you need to get the total volume of the solution in liters

${V}_{\text{total" = 697 * 10^(-3)"L" + 635 * 10^(-3)"L" = "1.332 L}}$

Therefore, the concentration of the nitrate anions will be

$\left[\text{NO"_3^(-)] = "1.356 moles"/"1.332 L" = color(green)("1.02 M}\right)$

The answer is rounded to three sig figs.