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.
1 answer
I didn't go any further but if I multiply Kf*Ksp I get 1.2E8.
2 answers