Question #4a466
1 Answer
Here's what I got.
Explanation:
The idea here is that you can use the solubility product constant,
Zinc cyanide is insoluble in aqueous solution, so right from the start you know that the salt's dissolution will be an equilibrium reaction
#"Zn"("CN")_ (2(s)) rightleftharpoons "Zn"_ ((aq))^(2+) + color(Red)(2)"CN"_ ((aq))^(-)#
Notice that each mole of zinc cyanide that dissociates produces
If you take
#"Zn"("CN")_ (2(s)) " "rightleftharpoons" " "Zn"_ ((aq))^(2+) " "+" " color(Red)(2)"CN"_ ((aq))^(-)#
By definition, the solubility product constant is equal to
#K_(sp) = ["Zn"^(2+)] * ["CN"^(-)]^color(Red)(2)#
In your case, this is equivalent to
#8.0 * 10^(-12) = s * (color(red)(2)s)^color(red)(2)#
#8.0 * 10^(-12) = 4s^3#
Rearrange to solve for
#s = root(3)( (8.0 * 10^(-12))/4) = 1.3 * 10^(-4)#
Since
#"molar solubility in pure water" = color(green)(|bar(ul(color(white)(a/a)color(black)(1.3 * 10^(-4)"M")color(white)(a/a)|)))#
For the second part of the problem, you must find the molar solubility of the salt in a solution that contains
Zinc nitrate is a soluble salt, so expect it to dissociate completely in aqueous solution
#"Zn"("NO"_ 3)_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"NO"_ (3(aq))^(-)#
Notice that every mole of zinc nitrate added to the solution will produce
#["Zn"^(2+)]_0 = "0.10 M"#
Plug this into an ICE table to find the new molar solubility of zinc cyanide.
#"Zn"("CN")_ (2(s)) " "rightleftharpoons" " "Zn"_ ((aq))^(2+) " "+" " color(Red)(2)"CN"_ ((aq))^(-)#
This time, the solubility product will be
#8.0 * 10^(-12) = (0.10 + s) * (color(red)(2)s)^color(red)(2)#
#8.0 * 10^(-12) = 0.40 * s^2 + 4s^3#
Rearrange to get
#4s^3 + 0.40s^2 - 8.0 * 10^(-12) = 0#
This cubic equation will produce one positive solution and two negative solutions. Since you're looking for concentration, you can discard the negative solutions to find
#s = 4.5 * 10^(-6)#
This time, the molar solubility of the salt is
#"molar solubility in 0.10 M Zn"("NO"_3)_2 = color(green)(|bar(ul(color(white)(a/a)color(black)(4.5 * 10^(-6)"M")color(white)(a/a)|)))#
Both answers are rounded to two sig figs.
So, does this result make sense?
The solubility of the salt decreased because the solution contained one of its ions -- this is known as the common-ion effect.
The presence of the zinc cations caused the dissolution equilibrium to shift to the left, which implies that less solid dissociated to form ions.