Zn(OH)_2(s) is an important ingredient of sunscreen and surgical dressing. In th
ID: 1071195 • Letter: Z
Question
Zn(OH)_2(s) is an important ingredient of sunscreen and surgical dressing. In this question, we investigate different solution conditions that may change its solubility. a) The K_sp for Zn(OH)_2(s) is 5.1 timed 10^-17 at 25 degree C. Determine the molar solubility of Zn(OH)_2(s) in a buffer solution with pH = 11.50? b) Which of the following processes will increase the solubility of Zn(OH)_2 (s) by assuming enough Zn(OH)_2(s) is present in the solution? Note that one or more than one answers are in this question. Explain your choice(s). (i) changing the pH of the buffer solution from 11.5 to 12.0 (ii) changing the pH of buffer solution from 11.50 to 10.0 (iii) adding water (iv) adding more 1.0 M NaOH.Explanation / Answer
For Zn(OH)2
(a) Ksp = [Zn2+][OH-]^2
pOH = 14 - pH = 14 - 11.5 = 2.5
pOH = -log[OH-]
[OH-]= 0.00316 M
So,
molar solubility of Zn(OH)2 would be,
= Ksp/[OH-]^2
= 5.1 x 10^-17/(0.00316)^2
= 5.11 x 10^-12 M
(b) solubility of Zn(OH)2 would increase by,
(ii) changing the pH from 11.5 to 10.
This is according to LeChatellier's principle, as we decrease the pH, the solution becomes more acidic. To maintain the same pH that is same amount of OH- in solution, more Zn(OH)2 would dissolve to release OH- in solution.
(iv) adding water
As dilution goes up, dissociation goes up. More salt would dissolve in solution.
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