# Question 23f8d

Dec 21, 2015

(I) ${\text{Na}}^{+}$, (II) ${\text{Ag}}^{+}$, and (III) ${\text{NO}}_{3}^{-}$.

#### Explanation:

So, you're mixing two solutions that contain soluble ionic compounds and want to determine which ions will be present in the resulting solution.

As you know, sodium chloride, $\text{NaCl}$, and silver nitrate, ${\text{AgNO}}_{3}$, are soluble in aqueous solution, which means that they exist as cations and anions

${\text{NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

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

Now, when you mix these solutions, an insoluble solid will actually precipitate out of solution - this is characteristic of a double replacement reaction.

More specifically, the silver cations, ${\text{Ag}}^{+}$, and the chloride anions, ${\text{Cl}}^{-}$, will form silver chloride, $\text{AgCl}$, which will precipitate out of solution.

The sodium cations and the nitrate anions are spectator ions, which means that they exist as ions both on the reactants' and on the products' side.

This tells you that you can expect the resulting solution to contain at least the sodium cations and nitrate anions.

In order to determine whether or not other ions will be present as well, you need to use the balanced chemical equation and the molarities of the two solutions.

"NaCl"_text((aq]) + "AgNO"_text(3(aq]) -> overbrace("AgCl"_text((s]) darr)^(color(blue)("insoluble")) + overbrace("NaNO"_text(3(aq]))^(color(red)("soluble"))#

Notice that you have a $1 : 1$ mole ratio between sodium chloride and silver nitrate. This tells you that the reaction consumes equal numbers of moles of each reactant.

The net ionic equation, for which spectator ions are excluded, will look like this

${\text{Cl"_text((aq])^(-) + "Ag"_text((aq])^(+) -> "AgCl}}_{\textrm{\left(s\right]}} \downarrow$

Use the molarities and volumes of the two solutions to determine how many moles of each reactant you're mixing

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

${c}_{N a C l} = 0.5 \text{mol"/color(red)(cancel(color(black)("dm"^3))) * 20 * 10^(-3)color(red)(cancel(color(black)("dm"^3))) = "0.010 moles NaCl}$

${c}_{A g N {O}_{3}} = 1.0 {\text{mol"/color(red)(cancel(color(black)("dm"^3))) * 20 * 10^(-3)color(red)(cancel(color(black)("dm"^3))) = "0.020 moles AgNO}}_{3}$

Since you have twice as many moles of silver nitrate, the reaction will completely consume the sodium chloride, i.e. the chloride anions, ${\text{Cl}}^{-}$, and leave you with excess silver nitrate, i.e. silver cations and nitrate anions.

The only ions that are completely consumed by the reaction are the chloride anions. The other ions will remain present in the resulting solution.

Therefore, the answers are (I) ${\text{Na}}^{+}$, (II) ${\text{Ag}}^{+}$, and (III) ${\text{NO}}_{3}^{-}$.