# Question 7c2f4

Apr 20, 2016

Yes, a precipitate will form.

#### Explanation:

Yes, the volumes are needed, and yes, you also need the solubility product constant, ${K}_{s p}$, of silver sulfate, ${\text{Ag"_2"SO}}_{4}$, which you can find listed here

${K}_{s p} = 1.2 \cdot {10}^{- 5}$

https://en.wikipedia.org/wiki/Silver_sulfate

The idea here is that you need to use the molarities and volumes of the two solutions to determine how many moles of each reactant you're mixing.

Once you know that, use the total volume of the resulting solution to find the concentration of the silver cations, ${\text{Ag}}^{+}$, and of the sulfate anions, ${\text{SO}}_{4}^{2 -}$, in the target solution.

As you know, the number of moles of solute can be calculated using molarity and volume

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The number of moles of silver nitrate, which will be equal to the number of moles of silver cations, will be

n_(Ag^(+)) = "0.120 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))

${n}_{A {g}^{+}} = {\text{0.0120 moles Ag}}^{+}$

Do the same to find the number of moles of potassium sulfate, which will be equal to the number of moles of sulfate anions

n_(SO_4^(2-)) = "0.010 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(150.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))

${n}_{S {O}_{4}^{2 -}} = {\text{0.0015 moles SO}}_{4}^{2 -}$

The total volume of the solution will be

${V}_{\text{total" = V_}} \left(A g N {O}_{3}\right) + {V}_{{K}_{2} S {O}_{4}}$

${V}_{\text{total" = "100.0 mL" + "150.0 mL" = "250.0 mL}}$

The concentration of the two ions in this solution will be

["Ag"^(+)] = "0.0120 moles"/(250.0 * 10^(-3)"L") = "0.0480 mol L"^(-1)

["SO"_4^(2-)] = "0.0015 moles"/(250.0 * 10^(-3)"L") = "0.0060 mol L"^(-1)#

Now, the net ionic equation that describes this double replacement reaction looks like this

$\textcolor{red}{2} {\text{Ag"_ ((aq))^(+) + "SO"_ (4(aq))^(2-) -> "Ag"_ 2"SO}}_{4 \left(s\right)} \downarrow$

By definition, the solubility product constant for this reaction will be

${K}_{s p} = \left[{\text{Ag"^(+)]^color(red)(2) * ["SO}}_{4}^{2 -}\right]$

All you have to do now is use the concentrations of the two ions in this solution to calculate the solubility quotient, ${Q}_{s p}$, and compare it with the value of the ${K}_{s p}$.

In order for a precipitate to form, you need

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{Q}_{s p} > {K}_{s p}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Plug in your values to find

${Q}_{s p} = {\left(0.0480\right)}^{\textcolor{red}{2}} \cdot 0.0060 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{1.38 \cdot {10}^{- 5}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Since you have ${Q}_{s p} > {K}_{s p}$, it follows that a precipitate will form.