Question

If the Ksp of Zn (CN)2 (s) is 3×10^-16, what percentage of Zn2+ is precipitated when we know that
[Zn 2+]=0.36M
[CN-]=5×10^-3M
Ksp=3×10^-16

Answer 1

I looked at this two ways. If Ksp is
Zn(CN)2 ==> Zn^2+ + 2CN^- then the reverse of that is
Zn^2+ + 2CN^- ==> Zn(CN)2 \ 1/Ksp = 3.333E15
That is such a HUGE number that we can make an assumption that ALL of the CN^- will be used which would leave 0.36 - 0.0025 = about 0.357 or approx (0.357/0.36)*100 = about 99.3% pptd.
OR we cn do it this way.
............Zn^2+ + 2CN^- ==> Zn(CN)2
I..........0.36........0.005.............0
C...........-x..........2x.................x
E.......0.36-x.....0.005-2x..........x

Then K=3.333E15 = [Zn(CN)2]/(Zn^2+)(CN^-)^2
Plug in the E line and solve for x.It's a cubic equation but you can find solutions on Google. I didn't check my answer but i obtained 0.0012 for x which is (Zn^2+). Then 0.36-0.0012 = 0.359, then
(0.359/0.36)*100 = 99.7%
Take your pick. I didn't check my math; the answers may be even closer than that. It's hardly worth solving a cubic in my opinion.

Post your work if you have questions.

Answer 2

DrBob, I have a different approach...

Mixing 0.36M (Zn⁺²) + 0.005M (CNˉ) => 0.005M Zn(CN)₂ formed + (0.3600M - 0.005M) = 0.3575M Zn⁺² in excess => (b/c CNˉ is the limiting reactant.)

….. Zn⁺²……….+ ……… 2CNˉ…….=>..Zn(CN)₂ + *Unreacted excess Zn⁺²
½(0.005M)…. + …… (0.005M)…=>… (0.005M) + 0.3575M unreacted Zn⁺²
*The excess Zn⁺² doesn’t have anything to react with & then functions a common ion limiting the solubility of the Zn(CN)₂ that is formed during the original mix.

Solubility of Zn(CN)₂ in presence of 0.3575M Zn as common ion:
…….. Zn(CN)₂ … =>…. Zn⁺²….. +…. 2CNˉ; Ksp = 3 x 10ˉ¹⁶
Ceq: --------- …. => 0.3575M …….. 2x
Ksp = [Zn⁺²][CNˉ]² = (0.3575)(2x)² = 1.43x² = 3 x 10ˉ¹⁶ => x = 1.448 x 10ˉ⁸M = [Zn⁺²] remaining in solution from formation of Zn(CN)₂ from original mix.

The Zn⁺² precipitating is = 0.0025M(original Zn used) – 1.448 x 10 ˉ⁸M(Zn remaining in solution) = ~2.5 x 10ˉ³M(Zn pptg as Zn(CN)₂(s) …

=> ~2.5 x 10ˉ³ mole Zn⁺²(65.37 g/mole) = 0.1634 g Zn⁺² precipitated from original formation of Zn(CN)₂.
=> 0.3600 mole Zn⁺² (65.37 g/mole) =23.53 g Zn⁺² original mass of zinc in the 0.36M solution.

 %Zn precipitating as Zn(CN)₂(s) from original 0.36M Zn⁺² solution = (0.1634/23.53)100% = 0.694% of original Zn⁺² (w/w).

Answer 3

Doc48. A couple of comments.
1. I ignored the common ion from Zn excess because I didn't think it would make that much diference. I think your calculations show that.too.

2. Your values agree with mine; i.e., 0.0025 M fpr Zn(CN)2 formed and 0.36-0.0025 = 0.3575 (but I rounded mne to 0.3570) but here is where I went wrong. The (0.3575/0.36)*100 = 99.3% IS THE AMOUNT IN SOLUTION. It's the (0.0025/0.36)*100 = 0.694% pptd. Same answer if converted to grams but that's extra work I didn't want to do. The last step is where I went really wrong. I lost track of where the Zn was. Thanks for catching my error.

3. I'm inclined to think that the cubic solution (but I lost track of the Zn there too) would be 100%-99.7% = 0.3% pptd might be slightly better than the 0.694. answer but I don't want to go through the math again.

Answer 4

Not to be difficult, but the 0.3% indicates a much lower solubility for the Zn(CN)2 than the common ion calculation would support in the presence of 0.3575M Zn. The 1.448 x 10^-8M solubility is accurate with respect to the data given. A lower solubility for the 1:2 ionization ratio of Zn(CN)2 is just not supported. All other points are right on point. :-)