Question

Using activity coefficients (find them with debye-huckel equation), calculate molar solubility of Zn(OH)₂ in

c) a mixture of 40.0mL of 0.250M KOH with 60.0mL de ZnCl₂ 0.0250M (answer should be 2.8*10^-13)

d) a mixture of 20mL 0.100M KOH and 50.0mL of 0.0250M ZnCl₂

Note: if you attempt this question please solve both parts (a and b) and show your work or else your attempt will be rated as wrong/reported

Answer 1

c) Equation for reaction:
2 KOH + ZnCl2 → 2KCl + Zn(OH)2
Mol KOH in 40ml 0.250M KOH = 0.250 x 0.04 L = 0.01mol KOH This will require 0.005mol ZnCl2
Mol ZnCl2 in 60ml of 0.025M ZnCl2 = 0.025 x 0.06L = 0.0015 mol ZnCl2 This is the limiting reactant.

The reaction: 0.003 mol KOH will react with 0.0015 mol ZnCl2, to produce 0.003 mol Zn(OH)2 and leaving 0.007mol KOH unreacted.This is in excess of the solubility of Zn(OH)2, which precipitates.

The problem is to find the molar concentration of the soluble portion of Zn(OH)2.
You have 0.007 mol KOH dissolved in 100ml solution: molarity = 0.007 / 0.1 L= 0.07M KOH solution.

[OH-] = 0.07M
We know that, Ksp Zn(OH)2 = 3.75*10^-15 ,
Zn(OH)2 ↔ Zn 2+ + 2OH-
Ksp = [Zn2+] *[ 2OH-]²
Because OH- is a common ion, you must use 0.07M for this concentration:
3.75*10^-15 = [Zn] * [4*0.0049] =   [Zn] * [0.0196]
[Zn2+] = 1.91 *10^-13

Molar solubility of Zn(OH)2 = 1.91 *10^-13 Exact result depends on the value of Ksp, but this is very close

d) a mixture of 20mL 0.100M KOH and 50.0mL of 0.0250M ZnCl₂

Mol KOH in 20ml 0.100M KOH = 0.1 x 0.02 L = 0.02 mol KOH This will require 0.01mol ZnCl2
Mol ZnCl2 in 50ml of 0.025M ZnCl2 = 0.025 x 0.05L = 0.00125 mol ZnCl2 This is the limiting reactant.

The reaction: 0.0025 mol KOH will react with 0.00125 mol ZnCl2, to produce 0.0025 mol Zn(OH)2 and leaving 0.007mol KOH unreacted.This is in excess of the solubility of Zn(OH)2, which precipitates.

The problem is to find the molar concentration of the soluble portion of Zn(OH)2.
You have 0.0075 mol KOH dissolved in 70ml solution: molarity = 0.0075 / 0.07 L= 0.107M KOH solution.

[OH-] = 0.107M
We know that, Ksp Zn(OH)2 = 3.75*10^-15 ,
Zn(OH)2 ↔ Zn 2+ + 2OH-
Ksp = [Zn2+] *[ 2OH-]²
Because OH- is a common ion, you must use 0.07M for this concentration:
3.75*10^-15 = [Zn] * [0.045796] =   [Zn] * [0.0196]
[Zn2+] = 8.18 *10^-14

Molar solubility of Zn(OH)2 = 8.18 *10^-14