Question

Question #4a466

Answer 1

Here's what I got.

Explanation:

The idea here is that you can use the solubility product constant, `K_(sp)`, 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^@"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

`"Zn"("CN")_ (2(s)) rightleftharpoons "Zn"_ ((aq))^(2+) + color(Red)(2)"CN"_ ((aq))^(-)`

Notice that each mole of zinc cyanide that dissociates produces `1` mole of zinc cations and `color(red)(2)` moles of cyanide anions.

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

`"Zn"("CN")_ (2(s)) " "rightleftharpoons" " "Zn"_ ((aq))^(2+) " "+" " color(Red)(2)"CN"_ ((aq))^(-)`

`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_(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`

`s = root(3)( (8.0 * 10^(-12))/4) = 1.3 * 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 `"0.10 M"` zinc nitrate.

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 `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.

`"Zn"("CN")_ (2(s)) " "rightleftharpoons" " "Zn"_ ((aq))^(2+) " "+" " color(Red)(2)"CN"_ ((aq))^(-)`

`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 * 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.