Question 4a466

Jun 20, 2016

Here's what I got.

Explanation:

The idea here is that you can use the solubility product constant, ${K}_{s p}$, to find the molar solubility of zinc cyanide, "Zn"("CN")_2, in pure water and in water that contains zinc nitrate, "Zn"("NO"_3)_2, at ${25}^{\circ} \text{C}$.

Zinc cyanide is insoluble in aqueous solution, so right from the start you know that the salt's dissolution will be an equilibrium reaction

${\text{Zn"("CN")_ (2(s)) rightleftharpoons "Zn"_ ((aq))^(2+) + color(Red)(2)"CN}}_{\left(a q\right)}^{-}$

Notice that each mole of zinc cyanide that dissociates produces $1$ mole of zinc cations and $\textcolor{red}{2}$ moles of cyanide anions.

If you take $s$ to be the equilibrium concentration of the zinc cations, you will have

${\text{Zn"("CN")_ (2(s)) " "rightleftharpoons" " "Zn"_ ((aq))^(2+) " "+" " color(Red)(2)"CN}}_{\left(a q\right)}^{-}$

color(purple)("I")color(white)(aaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaaaacolor(black)(-)aaaaaaaaaaacolor(black)((+s))aaaaaaacolor(black)((+color(red)(2)s))
color(purple)("E")color(white)(aaaaacolor(black)(-)aaaaaaaaaaaaacolor(black)(s)aaaaaaaaaacolor(black)(color(red)(2)s)

By definition, the solubility product constant is equal to

${K}_{s p} = {\left[{\text{Zn"^(2+)] * ["CN}}^{-}\right]}^{\textcolor{R e d}{2}}$

In your case, this is equivalent to

$8.0 \cdot {10}^{- 12} = s \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}}$

$8.0 \cdot {10}^{- 12} = 4 {s}^{3}$

Rearrange to solve for $s$

$s = \sqrt{\frac{8.0 \cdot {10}^{- 12}}{4}} = 1.3 \cdot {10}^{- 4}$

Since $s$ represents the molar solubility of the salt, you will have

"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 $\text{0.10 M}$ zinc nitrate.

Zinc nitrate is a soluble salt, so expect it to dissociate completely in aqueous solution

${\text{Zn"("NO"_ 3)_ (2(aq)) -> "Zn"_ ((aq))^(2+) + 2"NO}}_{3 \left(a q\right)}^{-}$

Notice that every mole of zinc nitrate added to the solution will produce $1$mole of zinc cations. The initial concentration of the zinc cations in solution will thus be

["Zn"^(2+)]_0 = "0.10 M"

Plug this into an ICE table to find the new molar solubility of zinc cyanide.

${\text{Zn"("CN")_ (2(s)) " "rightleftharpoons" " "Zn"_ ((aq))^(2+) " "+" " color(Red)(2)"CN}}_{\left(a q\right)}^{-}$

color(purple)("I")color(white)(aaaaacolor(black)(-)aaaaaaaaaaaacolor(black)(0.10)aaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaaacolor(black)(-)aaaaaaaaaacolor(black)((0.10+s))aaaaacolor(black)((+color(red)(2)s))
color(purple)("E")color(white)(aaaaacolor(black)(-)aaaaaaaaaacolor(black)(0.10 + s)aaaaaaacolor(black)(color(red)(2)s)

This time, the solubility product will be

$8.0 \cdot {10}^{- 12} = \left(0.10 + s\right) \cdot {\left(\textcolor{red}{2} s\right)}^{\textcolor{red}{2}}$

$8.0 \cdot {10}^{- 12} = 0.40 \cdot {s}^{2} + 4 {s}^{3}$

Rearrange to get

$4 {s}^{3} + 0.40 {s}^{2} - 8.0 \cdot {10}^{- 12} = 0$

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

$s = 4.5 \cdot {10}^{- 6}$

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.