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Question #e42f1

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Answer 1 · 2016-03-27T01:25:16

Here's what I got.

Explanation:

Silver nitrate, ${AgNO}_{3}$ $"AgNO"_3$ , and potassium sulfide, $K_{2} S$ $"K"_2"S"$ , 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

${AgNO}_{3(aq]} \to {Ag}_{(aq]}^{+} + {NO}_{3(aq]}^{-}$ $"AgNO"_text(3(aq]) -> "Ag"_text((aq])^(+) + "NO"_text(3(aq])^(-)$

$K_{2} S_{(aq]} \to 2 K_{(aq]}^{+} + S_{(aq]}^{2 -}$ $"K"_2"S"_text((aq]) -> 2"K"_text((aq])^(+) + "S"_text((aq])^(2-)$

When these two aqueous solutions are mixed, silver sulfide, ${Ag}_{2} S$ $"Ag"_2"S"$ , 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, ${KNO}_{3}$ $"KNO"_3$ , which will exist as dissociated cations and anions in solution.

The complete ionic equation looks like this

$2 {Ag}_{(aq]}^{+} + 2 {NO}_{3(aq]}^{-} + 2 K_{(aq]}^{+} + S_{(aq]}^{2 -} \to {Ag}_{2} S_{(s]} ⏐ ⏐ ↓ + 2 K_{(aq]}^{+} + 2 {NO}_{3(aq]}^{-}$ $color(red)(2)"Ag"_text((aq])^(+) + color(red)(2)"NO"_text(3(aq])^(-) + 2"K"_text((aq])^(+) + "S"_text((aq])^(2-) -> "Ag"_color(red)(2)"S"_text((s]) darr + 2"K"_text((aq])^(+) + color(red)(2)"NO"_text(3(aq])^(-)$

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.

![archives.library.illinois.edu)

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.

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