# The Ksp for silver sulfate (Ag_2SO_4) is 1.2*10-5. How do you calculate the solubility of silver sulfate in each of the following: a). water b). 0.10 M AgNO_3 c). 0.43 M K_2SO_4?

Aug 8, 2016

Here's what I got.

#### Explanation:

I'll show you how to solve parts (a) and (b) and leave part (c) to you as practice.

• Part (a)

Silver sulfate, ${\text{Ag"_2"SO}}_{4}$, is considered insoluble in aqueous solution, which implies that a dissociation equilibrium between the dissociated ions and the undissolved solid is established when you dissolve the salt in water.

You will have

${\text{Ag"_ color(blue)(2)"SO"_ (4(s)) rightleftharpoons color(blue)(2)"Ag"_ ((aq))^(+) + "SO}}_{4 \left(a q\right)}^{2 -}$

Now, when you dissolve the salt in pure water, the initial concentration of the dissolved ions will be equal to zero. You can use an ICE table to find the equilibrium concentration of the two ions

${\text{ ""Ag"_ color(blue)(2)"SO"_ (4(s)) " "rightleftharpoons" " color(blue)(2)"Ag"_ ((aq))^(+) " "+" " "SO}}_{4 \left(a q\right)}^{2 -}$

color(purple)("I")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaacolor(black)((+color(blue)(2)s))aaaaaaacolor(black)((+s))
color(purple)("E")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(color(blue)(2)s)aaaaaaaaaaacolor(black)(s)

By definition, the solubility product constant, ${K}_{s p}$, is equal to

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

In your case, this will be equal to

${K}_{s p} = {\left(\textcolor{b l u e}{2} s\right)}^{\textcolor{b l u e}{2}} \cdot s = 4 {s}^{3}$

Rearrange to solve for $s$, the molar solubility of silver sulfate in pure water

$s = \sqrt{{K}_{s p} / 4}$

Plug in your value to find

$s = \sqrt{\frac{1.2 \cdot {10}^{- 5}}{4}} = 0.0144$

This means that in a saturated solution of silver sulfate, the concentration of the salt that will dissolve to produce ions is equal to

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{s = {\text{0.0144 mol L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$\textcolor{w h i t e}{\frac{a}{a}}$

• Part (b)

This time, you're interested in finding the molar solubility of silver sulfate in a solution that is $\text{0.10 M}$ silver nitrate, ${\text{AgNO}}_{3}$.

Unlike silver sulfate, silver nitrate is soluble in aqueous solution, which means that it dissociates completely to form silver cations and nitrate anions

${\text{AgNO"_ (3(aq)) -> "Ag"_ ((aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

As shown by the $1 : 1$ mole ratios that exist between the solid and the dissolved ions, a $\text{0.10 M}$ silver nitrate solution will have

["Ag"^(+)] = ["NO"_3^(-)] = "0.10 M"

This means that the ICE table for the dissociation of the silver sulfate will look like this

${\text{ ""Ag"_ color(blue)(2)"SO"_ (4(s)) " "rightleftharpoons" " color(blue)(2)"Ag"_ ((aq))^(+) " "+" " "SO}}_{4 \left(a q\right)}^{2 -}$

color(purple)("I")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaaacolor(black)(0.10)aaaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaaaaacolor(black)(-)aaaaaaaaaaacolor(black)((+color(blue)(2)s))aaaaaaacolor(black)((+s))
color(purple)("E")color(white)(aaaaaacolor(black)(-)aaaaaaaaaacolor(black)(color(blue)(2)s + 0.10)aaaaaaaacolor(black)(s)

This time, the solubility product constant will take the form

${K}_{s p} = {\left(\textcolor{b l u e}{2} s + 0.10\right)}^{\textcolor{b l u e}{2}} \cdot s$

Now, because the value of the ${K}_{s p}$ is so small compared with the initial concentration of the silver cations, you can use the following approximation

$2 s + 0.10 \approx 0.10$

This means that you have

${K}_{s p} = {0.10}^{\textcolor{b l u e}{2}} \cdot s$

which gets you

$s = {K}_{s p} / 0.010 = \frac{1.2 \cdot {10}^{- 5}}{0.010} = 1.2 \cdot {10}^{- 3}$

Therefore, the moalr solubility of silver sulfate in a solution that is $\text{0.10 M}$ silver nitrate is

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{s = {\text{0.0012 mol L}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As you can see, the molar solubility of the salt decreased as a result of the presence of the silver cations $\to$ think common-ion effect here.

You can sue the same approach to find the answer to part (c). This time, the common ion will be the sulfate anion, ${\text{SO}}_{4}^{2 -}$, delivered to the solution by the soluble potassium sulfate, ${\text{K"_2"SO}}_{4}$.