Consider the reaction
Zn(OH)2(s) + 2CN-(aq) <--> Zn(CN)2(s) + 2OH-(aq)
a) Calculate K for the reaction. (Ksp Zn(CN)2= 8.0 x 10^-12)
b) Will Zn(CN)2 precipitate if NaCN is added to a saturated Zn(OH)2 solution?
I tried splitting this into two equations:
Zn(OH)2(s)<--> Zn(CN)2(s) + 2OH-(aq)
2CN-(aq) <--> Zn(CN)2(s) + 2OH-(aq)
But that really doesn't work stoichiometrically. I think I know how to do the problem if water is a "product", but I'm not sure how to work it out this way.
Rxn 1:
Zn(OH)2 ==> Zn^+2 + 2 OH^- Ksp = ??
Rxn 2:
Zn(CN)2 ==> Zn^+2 + 2CN^- Ksp = xx
Reverse Rxn 2 for which K'sp = 1/Ksp.
Add Rxn 2 for which Ksp = ??
This will give you the reaction you want and Krxn = Ksp(rxn 1) x K'sp(reverse of rxn 2).
Post your work if you get stuck.
I have a Ksp value in my book for Zn(OH)2 that equals 4 x 10^-17. Is this fine to use or am I supposed to find that value in the problem? And would this number have (1) sig figs or is it a "known" value?
Also, would I be correct in thinking that Zn(CN)2 will precipitate when
NaCN is added to the saturated Zn(OH)2 solution, because the left side of the equation increased so the right must as well?
I have an VERY old text from the 70s that lists Ksp for Zn(OH)2 as 3.3E-17. Your value, then, probably is the more up to date one. And if it lists 4E-17 that is one significant figure. It is an experimentally determined (in some cases calculated) value. It is not a "known" value.
Yes, I would have thought that Ksp for Zn(OH)2 would have been quoted in the problem since Ksp for Zn(CN)2 is listed.
No, I don't think your thinking is correct on the precipitation part. Frankly, I'm not sure that question can be answered since no concentration of CN^- is given. Check your problem to make sure you didn't just skip that part when copying to this forum. You might have one ion and you might have 25 grams. But you can get some handle on it by calculating the concentration of Zn^+2 and the hydroxide ion in a saturated solution of Zn(OH)2, then calculating the concentration of CN^- required to ppt Zn(CN)2.
To calculate the value of K for the given reaction, we need to use the Ksp value of Zn(CN)2, which is provided as 8.0 x 10^-12.
First, let's write the equations for the dissociation of Zn(OH)2 and Zn(CN)2:
Rxn 1: Zn(OH)2(s) ⇌ Zn^2+(aq) + 2OH^-(aq)
Rxn 2: Zn(CN)2(s) ⇌ Zn^2+(aq) + 2CN^-(aq)
Since we have the Ksp value for Zn(CN)2, we can use the reverse of Rxn 2 to get the K value for it: Ksp(Zn(CN)2) = 1/Ksp(reverse of Rxn 2).
Now, let's add the two reactions together in order to obtain the desired reaction:
Zn(OH)2(s) + 2CN^-(aq) ⇌ Zn(CN)2(s) + 2OH^-(aq)
For the overall reaction, we multiply the Ksp value of Zn(OH)2 with the Ksp value of the reverse reaction of Zn(CN)2:
K = Ksp(Zn(OH)2) * Ksp(reverse of Rxn 2)
Now, substitute the given Ksp values:
K = (4 x 10^-17) * (1 / (8.0 x 10^-12))
Simplify the expression:
K = (4 x 10^-17) / (8.0 x 10^-12)
K = 0.5 x 10^-5
Therefore, the value of K for the reaction is 0.5 x 10^-5.
Now, let's move on to part b of the question. We are asked if Zn(CN)2 will precipitate when NaCN is added to a saturated Zn(OH)2 solution.
To determine this, we need to consider the concentration of CN^- in the solution. However, the concentration of CN^- is not given in the question. Without this information, we cannot determine if Zn(CN)2 will precipitate or not.
To get some insight into this, you can calculate the concentration of Zn^2+ and OH^- in a saturated solution of Zn(OH)2. Then, you can calculate the concentration of CN^- required to precipitate Zn(CN)2. However, without the concentration of CN^- provided, we can't answer this question definitively.
To calculate the equilibrium constant (K) for the reaction, we can use the relationship between Ksp values.
The given Ksp values are:
Ksp(Zn(CN)2) = 8.0 x 10^-12
Ksp(Zn(OH)2) = 4 x 10^-17
First, we need to write the dissociation equations for Zn(OH)2 and Zn(CN)2:
Zn(OH)2(s) ⇌ Zn^2+(aq) + 2OH^-(aq) (Equation 1)
Zn(CN)2(s) ⇌ Zn^2+(aq) + 2CN^-(aq) (Equation 2)
Now, let's calculate the solubility product constant Ksp for Zn(OH)2 using Equation 1:
Ksp(Zn(OH)2) = [Zn^2+][OH^-]^2
We know that [OH^-] is equal to 2 times the concentration of Zn(OH)2 because two moles of OH^- ions are produced per mole of Zn(OH)2:
Ksp(Zn(OH)2) = [Zn^2+](2[Zn(OH)2])^2
= 4[Zn^2+][Zn(OH)2]^2
= 4 x 10^-17
Now, let's calculate the equilibrium constant K for the reaction using the relationship between Ksp values:
K = Ksp(Zn(CN)2) / (Ksp(Zn(OH)2))^2
= (8.0 x 10^-12) / (4 x 10^-17)^2
= 5 x 10^13
Therefore, the equilibrium constant (K) for the reaction is 5 x 10^13.
Now, let's move on to part b of the question.
To determine if Zn(CN)2 will precipitate when NaCN is added to a saturated Zn(OH)2 solution, we need to compare the reaction quotient (Q) with the equilibrium constant (K). If Q is greater than K, precipitation occurs.
The reaction quotient (Q) is calculated the same way as the equilibrium constant (K), but using the reactant and product concentrations at a particular point in the reaction.
In this case, we need the concentrations of Zn^2+, CN^-, and OH^- ions to calculate Q. However, no concentration values are given in the question. Therefore, it is not possible to determine if Zn(CN)2 will precipitate without the specific concentrations of the ions in the solution.