Equal volume of a 0.02 M Zn+ solution and a 2.0M NH3 solution are mixed. Kf for [Zn(NH3)4] is 4.1*10^8. If enough sodium oxalate is added to make the solution 0.1 M in oxalate, will ZnC2O4 precipitate. Ksp ZnC2O4= 2.7*10^-8.
I set up an ICE table
which have me:
Kf= (Zn(NH3)4)/ (Zn^2+)(NH3^-)
Kf= (.02-x)/ (1.92+x)(x)
4.1*10^8 = .02/1.92x
x = 2.54*10^-11
but when I used the [] values using x, in the Q equation. I don't get the correct Q for comparing to the K so say whether it will precipitate or not
To determine whether ZnC2O4 will precipitate, we need to compare the value of Q (the reaction quotient) to the value of Ksp (the solubility product constant).
Since ZnC2O4 dissociates in water according to the equation:
ZnC2O4(s) ⇌ Zn2+(aq) + C2O42-(aq)
The expression for Q would be:
Q = [Zn2+][C2O42-]
Given that the concentration of Zn2+(aq) is equal to the concentration of [Zn(NH3)4] from the formation constant expression, we can substitute [Zn(NH3)4] with the value you calculated: 2.54*10^-11.
Now, let's calculate the concentration of C2O42-(aq):
Since enough sodium oxalate is added to make the solution 0.1 M in oxalate, [C2O42-] = 0.1 M.
Substituting the values into the Q expression, we have:
Q = (2.54*10^-11)(0.1) = 2.54*10^-12
Now, let's compare Q to Ksp to determine if ZnC2O4 will precipitate:
If Q > Ksp, then ZnC2O4 will precipitate.
If Q < Ksp, then ZnC2O4 will not precipitate.
Given that Ksp = 2.7*10^-8, we can conclude that Q < Ksp (2.54*10^-12 < 2.7*10^-8).
Therefore, based on the comparison of Q and Ksp, ZnC2O4 will not precipitate in the solution.
To determine if ZnC2O4 will precipitate when enough sodium oxalate is added, we need to compare the value of Q (the ion product) to the value of Ksp (the solubility product). If Q is greater than Ksp, the compound will precipitate.
Let's start with the initial equation for the formation of [Zn(NH3)4]:
Zn^2+ + 4NH3 ↔ [Zn(NH3)4]2+
From the given information, we can set up an ICE table as follows:
| | Zn^2+ | NH3 | [Zn(NH3)4]2+ |
|------------|---------|---------|-----------------|
| Initial | 0.02 M | 2.0 M | 0 M |
| Change | -x | -4x | +x |
| Equilibrium | 0.02-x | 2.0-4x | x |
Using the given Kf value, we can write the equation for Kf as follows:
Kf = [Zn(NH3)4]2+ / (Zn^2+)(NH3)^4
Substituting the equilibrium concentrations from the ICE table into the Kf expression:
4.1 * 10^8 = x / (0.02 - x) * (2.0 - 4x)^4
Now, we solve for x.
Note: This equation is quite complex, and solving it analytically might be challenging. You can use numerical methods or a calculator capable of numerical solving to find the value of x.
Once you have determined the value of x (which you have already calculated as 2.54 * 10^-11), you can proceed to find the concentration of Zn^2+ and NH3 in the solution.
Now, let's determine if ZnC2O4 will precipitate when enough sodium oxalate is added. We need the concentration of Zn^2+ and oxalate (C2O4^2-) in the solution.
Since the volume of the Zn^2+ solution and NH3 solution used is equal, and the initial concentration of Zn^2+ is 0.02 M, the concentration of Zn^2+ in the final solution is also 0.02 M.
Considering the stoichiometry of the reaction, the concentration of oxalate (C2O4^2-) in the final solution is also 0.02 M (assuming all the sodium oxalate dissociates to provide C2O4^2-).
Now, we calculate Q, the ion product:
Q = [Zn^2+][C2O4^2-] = (0.02)(0.1) = 0.002
Finally, we compare Q (0.002) to Ksp (2.7 * 10^-8) to determine if ZnC2O4 will precipitate.
Since Q < Ksp, ZnC2O4 will not precipitate when enough sodium oxalate is added to make the solution 0.1 M in oxalate.
Note: It's important to double-check the calculations and units to ensure accuracy.
I might try a little different approach.
Zn^+2 + 4NH3 ==> Zn(NH3)4^+2
First, since the problem states that equal volumes are mixed, that means Zn(II) is 0.02/2 = 0.01 initially and NH3 is 2/2 = 1 M initially.
Since the Kf for the Zn ammine is so LARGE, the final concn of the ammine will be essentially 0.01 M and the final concn of NH3 will be 1 to start less 4*Zn^+2 = 1-0.04 = 0.96 at the end. So plug 0.96 for NH3 (don't forget to raise it to the 4th power), 0.01 for the Zn ammine complex, and the value for Kf, and calculate Zn(II). Then substitute that into Qsp and compare with Ksp.