Ksp of Zn(OH)2 = [Zn2+][OH-]^2 = 5.2 x 10^-17 (a) pH of buffer = 11.5 pOH = 14 -

Question

Ksp of Zn(OH)2 = [Zn2+][OH-]^2 = 5.2 x 10^-17

(a) pH of buffer = 11.5

pOH = 14 - pH = 2.5

pOH = -log[OH-]

[OH-] = 3.16 x 10^-3 M

feed in Ksp equation,

molar solubility [Zn2+] = Ksp/[OH-]^2

                                    = 5.2 x 10^-17/(3.16 x 10^-3)^2 = 5.21 x 10^-12 M

(b) increase in solubility of Zn(OH)2 upon,

(i) addition of 1.0 M NaOH

[OH-] = 1.0 M

molar solubility [Zn2+] = 5.2 x 10^-17/(1)^2 = 5.2 x 10^-17

So molar solubility dcreased upon addition of 1 M NaOH

(ii) pH changed from 11.5 to 10.0

pOH = 14 - pH = 4

[OH-] = 1 x 10^-4 M

molar solubility [Zn2+] = Ksp/[OH-]^2

                                    = 5.2 x 10^-17/(1.0 x 10^-4)^2 = 5.2 x 10^-9 M

So molar solubility increased upon addition of buffer of pH 10.0

(iii) pH changed from 11.5 to 12.0

pOH = 14 - pH = 2

[OH-] = 1 x 10^-2 M

molar solubility [Zn2+] = Ksp/[OH-]^2

                                    = 5.2 x 10^-17/(1.0 x 10^-2)^2 = 5.2 x 10^-13 M

So molar solubility decreased upon changing pH of buffer from 11.5 to 12.0.

Explanation / Answer

3. Zn(0H)4s) is an important ingredient of sunscreen and surgical dressing. In this solution conditions that may change its solubility (a) (8pt) The K., for Zn(OH (s) is .210at 25 °C. Determine the molar solubility of Zn(OtH)s) in a buffer solution with pH-11.50 (b) (4pt) Which of the following processes will increase the solubility of Zn(OIH)(s) by assuming enough Zn(OH)s) is present in the solution? Note that one or more than one answers are in this question. Explain your choicels). (i) adding more 1.0 M NaOH (ii) adding water (il) changing the pH of buffer solution from 11.50 to 10.0 (iv) changing the pH of the buffer solution from 11.5 to 12.0

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