When excess NaOH is added to Zn(OH)2, the complex ion Zn(OH)4 2- is formed. Determine the molar solubility of Zn(OH)2 in .10 M NaOH. Compare with the molar solubility of Zn(OH)2 in pure water.
I thought I knew how to do this problem, but I'm not getting the right answer. If someone could maybe tell me what I'm doing wrong I'd really appreciate it.
To begin with, I found K by multiplying Ksp and Kf. Ksp for Zn(OH)2 is 4X10-7 and Kf for Zn(OH)4 2- is 3X10+14. So K would be 1.2X10-2.
K=[Zn(OH)4 2-][Na+]2/[NaOH]2
I set up an ICE table
[NaOH] [Zn(OH)4 2-] [Na+]
I .10 0 0
C -2x +x +2x
E .10-2x x 2x
So then K would equal...
K=(x)(2x)2/(.10-2x)2
assuming x is much smaller than .10...
K=4x3/(.10)2
When I work that out, x is 3.1X10-2, but the answer is apparently 1X10-4. I'm not sure what I'm doing wrong.
I didn't go any further but if I multiply Kf*Ksp I get 1.2E8.
To determine the molar solubility of Zn(OH)2 in 0.10 M NaOH and compare it with the molar solubility in pure water, we need to consider the complexation reaction between Zn(OH)2 and NaOH.
First, we need to write the balanced equation for the formation of the complex ion Zn(OH)4 2-:
Zn(OH)2 + 4NaOH ⇌ Na2[Zn(OH)4] + 2NaOH
The equilibrium constant for the complexation reaction, Kc, can be expressed as:
Kc = [Zn(OH)4 2-] / ([Zn(OH)2] * [NaOH]^4)
Since Kc depends on the concentration of the species involved, we need to express [Zn(OH)4 2-] and [NaOH] in terms of x (the molar solubility of Zn(OH)2 in NaOH) and [NaOH].
Let's set up the ICE table for the reaction:
[Zn(OH)2] + 4[NaOH] ⇌ [Zn(OH)4 2-] + 2[NaOH]
Initial: x(M) 0.10 M 0 M 0.10 M
Change: -x(M) -4x(M) +x(M) +2x(M)
Equilibrium: x(M) 0.10 - 4x(M) x(M) 0.10 + 2x(M)
Now, plug the equilibrium concentrations into the equation for the equilibrium constant, Kc:
Kc = [Zn(OH)4 2-] / ([Zn(OH)2] * [NaOH]^4)
Kc = x / (x * (0.10 - 4x)^4)
However, we can simplify the expression by noting that the concentration of NaOH is much greater than the concentration of Zn(OH)2 (assuming x is much smaller than 0.10). Therefore, we can approximate 0.10 - 4x as 0.10.
So, Kc = x / (x * (0.10)^4)
Kc = 1 / (0.10^4) = 1.0 x 10^-4
Therefore, the equilibrium constant Kc is 1.0 x 10^-4.
Now, you need to find the molar solubility of Zn(OH)2 in 0.10 M NaOH. To do this, you'll need to solve for x in the equation:
Kc = x / (x * (0.10)^4)
Simplifying the equation, we get:
1.0 x 10^-4 = x / (0.10^4)
Rearranging the equation and solving for x, we get:
x = (1.0 x 10^-4) * (0.10^4) = 1.0 x 10^-8 M
Therefore, the molar solubility of Zn(OH)2 in 0.10 M NaOH is 1.0 x 10^-8 M.
Comparing this with the molar solubility of Zn(OH)2 in pure water, you would need to go through the same steps but set the concentration of NaOH to 0 M. This would lead to a different equilibrium constant Kc, and subsequently a different molar solubility value for Zn(OH)2 in pure water.