Question

Question #e42f1

Answer 1

Here's what I got.

Explanation:

Silver nitrate, `"AgNO"_3`, and potassium sulfide, `"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

`"AgNO"_text(3(aq]) -> "Ag"_text((aq])^(+) + "NO"_text(3(aq])^(-)`

`"K"_2"S"_text((aq]) -> 2"K"_text((aq])^(+) + "S"_text((aq])^(2-)`

When these two aqueous solutions are mixed, silver sulfide, `"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, `"KNO"_3`, which will exist as dissociated cations and anions in solution.

The complete ionic equation looks like this

`color(red)(2)"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"_text(3(aq])^(-)`

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

`color(green)(|bar(ul(color(white)(a/a)color(black)(color(red)(2)"Ag"_text((aq])^(+) + "S"_text((aq])^(2-) -> "Ag"_color(red)(2)"S"_text((s]) darr)color(white)(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, `"Na"_3"PO"_3`, are mixed, the insoluble solid calcium phosphate, `"Ca"_2("PO"_4)_3`, and aqueous sodium hydroxide, `"NaOH"`, are formed.

The complete ionic equation looks like this - keep in mind that you have `color(red)(3` calcium cations and `color(blue)(2)` phosphate anions in calcium phosphate, `"Ca"_color(red)(3)("PO"_4)_color(blue)(2)`

`color(red)(3)"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"_text((aq])^(-)`

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.