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#?
1 Answer
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,
You will have
#"Ag"_ color(blue)(2)"SO"_ (4(s)) rightleftharpoons color(blue)(2)"Ag"_ ((aq))^(+) + "SO"_ (4(aq))^(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
#" ""Ag"_ color(blue)(2)"SO"_ (4(s)) " "rightleftharpoons" " color(blue)(2)"Ag"_ ((aq))^(+) " "+" " "SO"_ (4(aq))^(2-)#
By definition, the solubility product constant,
#K_(sp) = ["Ag"^(+)]^color(blue)(2) * ["SO"_4^(2-)]#
In your case, this will be equal to
#K_(sp) = (color(blue)(2)s)^color(blue)(2) * s = 4s^3#
Rearrange to solve for
#s = root(3) (K_(sp)/4)#
Plug in your value to find
#s = root(3)((1.2 * 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
#color(green)(|bar(ul(color(white)(a/a)color(black)(s = "0.0144 mol L"^(-1))color(white)(a/a)|)))#
- Part (b)
This time, you're interested in finding the molar solubility of silver sulfate in a solution that is
Unlike silver sulfate, silver nitrate is soluble in aqueous solution, which means that it dissociates completely to form silver cations and nitrate anions
#"AgNO"_ (3(aq)) -> "Ag"_ ((aq))^(+) + "NO" _(3(aq))^(-)#
As shown by the
#["Ag"^(+)] = ["NO"_3^(-)] = "0.10 M"#
This means that the ICE table for the dissociation of the silver sulfate will look like this
#" ""Ag"_ color(blue)(2)"SO"_ (4(s)) " "rightleftharpoons" " color(blue)(2)"Ag"_ ((aq))^(+) " "+" " "SO"_ (4(aq))^(2-)#
This time, the solubility product constant will take the form
#K_(sp) = (color(blue)(2)s + 0.10)^color(blue)(2) * s#
Now, because the value of the
#2s + 0.10 ~~ 0.10#
This means that you have
#K_(sp) = 0.10^color(blue)(2) * s#
which gets you
#s = K_(sp)/0.010 = (1.2 * 10^(-5))/0.010 = 1.2 * 10^(-3)#
Therefore, the moalr solubility of silver sulfate in a solution that is
#color(green)(|bar(ul(color(white)(a/a)color(black)(s = "0.0012 mol L"^(-1))color(white)(a/a)|)))#
As you can see, the molar solubility of the salt decreased as a result of the presence of the silver cations
You can sue the same approach to find the answer to part (c). This time, the common ion will be the sulfate anion,