# Question e42f1

Mar 27, 2016

Here's what I got.

#### Explanation:

Silver nitrate, ${\text{AgNO}}_{3}$, and potassium sulfide, $\text{K"_2"S}$, are soluble ionic compounds, which means that they dissociate completely in aqueous solution to form cations and anions.

The two reactants can thus be represented as

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

${\text{K"_2"S"_text((aq]) -> 2"K"_text((aq])^(+) + "S}}_{\textrm{\left(a q\right]}}^{2 -}$

When these two aqueous solutions are mixed, silver sulfide, $\text{Ag"_2"S}$, an insoluble ionic compound that precipitates out of solution, will be formed.

In addition to silver sulfide, the reaction will also produce aqueous potassium nitrate, ${\text{KNO}}_{3}$, which will exist as dissociated cations and anions in solution.

The complete ionic equation looks like this

$\textcolor{red}{2} {\text{Ag"_text((aq])^(+) + color(red)(2)"NO"_text(3(aq])^(-) + 2"K"_text((aq])^(+) + "S"_text((aq])^(2-) -> "Ag"_color(red)(2)"S"_text((s]) darr + 2"K"_text((aq])^(+) + color(red)(2)"NO}}_{\textrm{3 \left(a q\right]}}^{-}$

In order to get the net ionic equation, you must remove the spectator ions, i.e. the ions that are present on both sides of the equation.

You will have

color(red)(2)"Ag"_text((aq])^(+) + color(red)(cancel(color(black)(color(red)(2)"NO"_text(3(aq])^(-)))) + color(blue)(cancel(color(black)(2"K"_text((aq])^(+)))) + "S"_text((aq])^(2-) -> "Ag"_color(red)(2)"S"_text((s]) darr + color(blue)(cancel(color(black)(2"K"_text((aq])^(+)))) + color(red)(cancel(color(black)(color(red)(2)"NO"_text(3(aq])^(-))))

This will get you

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\textcolor{red}{2} {\text{Ag"_text((aq])^(+) + "S"_text((aq])^(2-) -> "Ag"_color(red)(2)"S}}_{\textrm{\left(s\right]}} \downarrow} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

It's worth noting that silver sulfide is a black precipitate.

When aqueous solutions of calcium hydroxide, "Ca"("OH")_2, and sodium phosphate, ${\text{Na"_3"PO}}_{3}$, are mixed, the insoluble solid calcium phosphate, "Ca"_2("PO"_4)_3, and aqueous sodium hydroxide, $\text{NaOH}$, are formed.

The complete ionic equation looks like this - keep in mind that you have color(red)(3 calcium cations and $\textcolor{b l u e}{2}$ phosphate anions in calcium phosphate, "Ca"_color(red)(3)("PO"_4)_color(blue)(2)

$\textcolor{red}{3} {\text{Ca"_text((aq])^(2+) + (color(red)(3) xx 2)"OH"_text((aq])^(-) + (color(blue)(2) xx 3)"Na"_text((aq])^(+) + color(blue)(2)"PO"_text(4(aq])^(3-) -> "Ca"_3("PO"_4)_text(2(s]) darr + (color(blue)(2) xx 3)"Na"_text((aq])^(+) + (color(red)(3) xx 2)"OH}}_{\textrm{\left(a q\right]}}^{-}$

The net ionic equation will be

color(red)(3)"Ca"_text((aq])^(2+) + color(purple)(cancel(color(black)((color(red)(3) xx 2)"OH"_text((aq])^(-)))) + color(brown)(cancel(color(black)((color(blue)(2) xx 3)"Na"_text((aq])^(+)))) + color(blue)(2)"PO"_text(4(aq])^(3-) -> "Ca"_3("PO"_4)_text(2(s]) darr + color(brown)(cancel(color(black)((color(blue)(2) xx 3)"Na"_text((aq])^(+)))) + color(purple)(cancel(color(black)((color(red)(3) xx 2)"OH"_text((aq])^(-))))

which is equivalent to

color(green)(|bar(ul(color(white)(a/a)color(black)(color(red)(3)"Ca"_text((aq])^(2+) + color(blue)(2)"PO"_text(4(aq])^(3-) -> "Ca"_3("PO"_4)_text(2(s]) darr)color(white)(a/a)|)))#

Calcium phosphate is a white precipitate.