Question

Question #620d2

Answer 1

`"190 mL"`

Explanation:

Start by writing the balanced chemical equation that describes this double replacement reaction

`color(blue)(3)"AgNO"_ (3(aq)) + "K"_ 3"PO"_ (4(aq)) -> "Ag"_ 3"PO"_ (4(s)) darr + 3"KNO"_ (3(aq))`

Notice that the reaction consumes `color(blue)(3)` moles of silver nitrate for every `1` mole of potassium phosphate that takes part in the reaction.

This tells you that in order for all the moles of potassium phosphate to react, you need to ensure that you get at least `color(blue)(3)` times more moles of silver nitrate.

Now, you know that molarity represents the number of moles of solute present in `"1 L"` of a given solution.

Since

`25.0 color(red)(cancel(color(black)("mL"))) * "1 L"/(10^3color(red)(cancel(color(black)("mL")))) = 1/40` `"L"`

you can say that the potassium phosphate solution contains `1/40` of the number of moles it would contain in `"1 L"`, i.e. it will contain

`overbrace("0.50 moles K"_3"PO"_4)^(color(purple)("what you have in 1 L of solution")) * 1/40 = 0.05/4` `"moles K"_3"PO"_4`

This means that the silver nitrate solution must contain

`0.05/4 color(red)(cancel(color(black)("moles K"_3"PO"_4))) * (color(blue)(3)"moles AgNO"_3)/(1color(red)(cancel(color(black)("mole K"_3"PO"_4)))) = 0.15/4` `"moles AgNO"_3`

The silver nitrate solution has a molarity of `"0.20 mol L"^(-1)`, which means that you get `0.20` moles of silver nitrate for every `"1 L"` of solution.

Since

`(0.15/4color(red)(cancel(color(black)("moles AgNO"_3))))/(0.20color(red)(cancel(color(black)("moles AgNO"_3)))) = 3/16`

it follows that you will need a volume that is `3/16` times smaller than `"1 L"`.

Therefore, you can say that you will need

`3/16 * 1 color(red)(cancel(color(black)("L"))) * (10^3color(white)(.)"mL")/(1color(red)(cancel(color(black)("L")))) = "187.5 mL"`

of `"0.20 mol L"^(-1)` silver nitrate solution in order to ensure that all the moles of potassium phosphate take part in the reaction.

Rounded to two sig figs, the number of sig figs you have for the molarities of the two solutions, the answer will be

`color(darkgreen)(ul(color(black)("volume of AgNO"_3 = "190 mL")))`