Zn(CN)2 ==> Zn^2+ + 2CN^-
HCN ==> H^+ + CN^-
Note that as H^+ increases, H^+ combines with CN^- which shifts the solubility equilibrium to the right to make Zn(CN)2 more soluble.
If we let S = solubility of Zn(CN)2, then S = (Zn^2+) and 2S = (CN^-) + (HCN)
We use Ka = 6.2E-10 = (H^+)(CN^-)/(HCN).
For pH = 2.92, (H^+) = 0.0012. Substitute that into Ka and solve for (HCN) in terms of CN^-. Knowing HCN (in terms of CN), use the above solubility equation of
2S = (CN^-) + (HCN) and solve for CN^-.
Then (Zn^2+)(CN^-)^2 = Ksp
and solve for S.
I went through in a hurry an obtained about 0.065 M
Ignoring activities, determine the molar solubility (S) of Zn(CN)2 in a solution with a pH = 2.92. Ksp (Zn(CN)2) = 3.0 × 10-16; Ka (HCN) = 6.2 × 10-10.
Completely lost!
4 answers
I think there is an easier way to do this. Let me know if you want to go through it.
This problem requires you to solve for solubility(S) by using a mass balance equation.
Write out pertinent reactions to establish the mass balance equation...
1) Zn(CN)2 <--> Zn^2+ + 2CN^-
2) CN^- + H^+ <--> HCN (NOTE: this reaction is an inverse of the HCN dissociation! More on that later)
3) H2O <--> H^+ + OH^-
Mass balance: 2[Zn2+] = [CN-] + [HCN]
(If you don't see why we multiply the concentration of Zn2+ by 2, put it into words. "Zn(CN)2...The concentration of CN is twice the concentration of Zn" It may seem wrong, but that's how you set up the mass balance.)
Equations for Reactions-- Use the values given in the problem to set up as many equations as you can to solve for [Zn2+]
1) Ksp= 3.0E-16 = [Zn^2+][CN^-]^2
2) 1/Ka= 1.6E9 =[HCN] / [CN^-][H^+]
****Note that we take the inverse of the acid dissociation constant(6.2E-10) for the HCN reaction(=1.6E+9)****
3) Kw= 1.0E-14 = [H+][OH-]
[H+]=10^-2.92=1.20E-3
Last Step: Combine the reaction equations with the mass balance equation to solve for [Zn2+]
Ka=1.6E9=[HCN]/[CN]1.20E-3 === [CN]1.92E6=[HCN] (to mass balance eq)
2[Zn] = [CN]+[HCN]
[Zn]=[CN]+([CN]1.92E6) /2
[Zn] = 9.60E5[CN] (to Ksp eq)
Ksp=3.0E-16=9.60E5[CN] [CN]^2 and solve= 6.8E-8 M [CN-]
Since the problem asks to solve for Zn(CN)2, we know that in this equation the concentration of Zn ions in solution is equal to the molar solubility of Zn(CN)2 so-
[Zn] = 9.60E5[CN]= 9.60E5 (6.8E-8)= 0.065 M Zn(CN)2
Hope that helped!
Write out pertinent reactions to establish the mass balance equation...
1) Zn(CN)2 <--> Zn^2+ + 2CN^-
2) CN^- + H^+ <--> HCN (NOTE: this reaction is an inverse of the HCN dissociation! More on that later)
3) H2O <--> H^+ + OH^-
Mass balance: 2[Zn2+] = [CN-] + [HCN]
(If you don't see why we multiply the concentration of Zn2+ by 2, put it into words. "Zn(CN)2...The concentration of CN is twice the concentration of Zn" It may seem wrong, but that's how you set up the mass balance.)
Equations for Reactions-- Use the values given in the problem to set up as many equations as you can to solve for [Zn2+]
1) Ksp= 3.0E-16 = [Zn^2+][CN^-]^2
2) 1/Ka= 1.6E9 =[HCN] / [CN^-][H^+]
****Note that we take the inverse of the acid dissociation constant(6.2E-10) for the HCN reaction(=1.6E+9)****
3) Kw= 1.0E-14 = [H+][OH-]
[H+]=10^-2.92=1.20E-3
Last Step: Combine the reaction equations with the mass balance equation to solve for [Zn2+]
Ka=1.6E9=[HCN]/[CN]1.20E-3 === [CN]1.92E6=[HCN] (to mass balance eq)
2[Zn] = [CN]+[HCN]
[Zn]=[CN]+([CN]1.92E6) /2
[Zn] = 9.60E5[CN] (to Ksp eq)
Ksp=3.0E-16=9.60E5[CN] [CN]^2 and solve= 6.8E-8 M [CN-]
Since the problem asks to solve for Zn(CN)2, we know that in this equation the concentration of Zn ions in solution is equal to the molar solubility of Zn(CN)2 so-
[Zn] = 9.60E5[CN]= 9.60E5 (6.8E-8)= 0.065 M Zn(CN)2
Hope that helped!
Ashley, thank you. I'd been working on this problem for hours. God bless you
0 answers