Question

Question #32649

Answer 1

`["S"^(2-)] = "0.022 M"`

Explanation:

The idea here is that some of the sulfide anions, `"S"^(2-)`, will combine with the silver cations, `"Ag"^(+)`, to form the insoluble solid, and the rest will remain in solution.

Your goal here is to figure out the concentration of these sulfide anions that remain floating around uncombined with the silver cations.

The first thing to do here is to write the balanced chemical equation that describes this double replacement reaction

`color(blue)(2)"AgNO"_ (3(aq)) + "Na"_ 2"S"_ ((aq)) -> "Ag"_ 2"S"_ ((s)) darr + 2"NaNO"_ (3(aq))`

We can get rid of the spectator ions and write the net ionic equation

`color(blue)(2)"Ag"_ ((aq))^(2+) + "S"_ ((aq))^(2-) -> "Ag"_ 2"S"_ ((s)) darr`

Notice that the reaction consumes `color(blue)(2)` moles of silver cations for every mole of sulfide anions present in solution.

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

`20.0 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.10 moles Ag"^(+)/(1color(red)(cancel(color(black)("L")))) = "0.0020 moles Ag"^(+)`

`25.0 color(red)(cancel(color(black)("mL"))) * (1 color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * ("0.080 moles S"^(2-))/(1color(red)(cancel(color(black)("L")))) = "0.0020 moles S"^(2-)`

Now, you're mixing equal numbers of moles of silver nitrate and of sodium sulfide; however, because you need twice as many moles of silver nitrate in order to make sure that all the moles of sodium sulfide react, the silver nitrate will be your limiting reagent.

In other words, the silver cations will be completely consumed by the reaction, along with

`0.0020 color(red)(cancel(color(black)("moles Ag"^(+)))) * "1 mole S"^(2-)/(color(blue)(2)color(red)(cancel(color(black)("moles Ag"^(+))))) = "0.0010 moles S"^(2-)`

This implies that the resulting solution will contain

`overbrace("0.0020 moles S"^(2-))^(color(blue)("what you start with")) - overbrace("0.0010 moles S"^(2-))^(color(purple)("consumed")) = "0.0010 moles S"^(2-)`

The total volume of the resulting solution will be

`V_"total" = "20.0 mL" + "25.0 mL" = "45.0 mL"`

Therefore, you can say that the concentration of the sulfide anions in the resulting solution will be

`["S"^(2-)] = "0.0010 moles"/(45.0 * 10^(-3)"L") = color(green)(bar(ul(|color(white)(a/a)color(black)("0.022 M")color(white)(a/a)|)))`

The answer is rounded to two sig figs.

Here's a cool image showing how the precipitation of silver sulfide looks like