Asked by Heather
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!
Completely lost!
Answers
Answered by
DrBob222
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
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
Answered by
DrBob222
I think there is an easier way to do this. Let me know if you want to go through it.
Answered by
Ashley
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!
Answered by
mikey
Ashley, thank you. I'd been working on this problem for hours. God bless you