Question

Use activity coefficients to calculate molar solubility of Zn(OH)₂ in

a) a mixture of 40.0mL of 0.250M KOH with 60.0mL de ZnCl₂ 0.0250M

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

c) K₂SO₄ 0.0300M M

Answer 1

activity coefficient Yi is represented as,

logYi = -0.51 x Zi^2 x sq.rt(I)/1+3.3 x Ri x sq.rt(I)

Where,

Yi = acitivity coefficient for the ion

Zi = charge of ion

I = ionic strength

Ri = effective ionic radius

Ionic strength I = 1/2 sum (CiZi^2)

where, Ci = concentration of ion and Zi = charge of ion

The relation of activity coefficient and Ksp is for say ions A and B,

Ksp = [Yi[A]]a . [YiB]]b

with, [A]a and [B]a be the effective concentration of ions in solution

a) moles of KOH = molarity x volume = 0.25 x 0.04 = 0.01 mole

moles of ZnCl2 = 0.025 x 0.06 = 0.0015 mole

So, 1 mole of ZnCl2 will react with 2 moles of NaOH to give 1 mole of Zn(OH)2

0.0015 moles of ZnCl2 reacts with 0.003 moles of NaOH to form 0.0015 moles of Zn(OH)2

Total volume of solution = 40 + 60 = 100 ml = 0.1 L

molarity of Zn(OH)2 = moles/L = 0.0015/0.1 = 0.015 M

[Zn2+] = 0.015 M with charge +2

[OH-] = 2 x 0.015 = 0.03 M with charge 1

So,

ionic strength of Zn(OH)2 I = 1/2[0.015x2^2+0.03x1^2] = 0.09

effective ionic radius in nm for ions is,

Zn2+ = 0.6 ; OH- = 0.35

acitivity coefficient

logY[Zn2+] = -0.51 x 2^2 x sq.rt(0.09)/1+3.3 x 0.6 x sq.rt(0.09)

Y[Zn2+] = 0.413

logY[OH-] = -0.51 x 1^2 x sq.rt(0.09)/1+3.3 x 0.35 x sq.rt(0.09)

Y[OH-] = 0.770

let x be the amount of Zn2+ in solution at equilibrium the, 2x will be the amount of Cl- in solution.

Ksp for Zn(OH)2 = 5 x 10^-17 = [Zn2+][Cl-]^2

5 x 10^-17 = 0.413(x).(0.770)^2(2x)^2

x^3 = 5.105 x 10^-17

x = 3.71 x 10^-6 M

So the molar solubility of Zn(OH)2 = 3.71 x 10^-6 M

b) moles of KOH = molarity x volume = 0.1 x 0.02 = 0.002 mole

moles of ZnCl2 = 0.025 x 0.05 = 0.00125 mole

So, 1 mole of ZnCl2 will react with 2 moles of NaOH to give 1 mole of Zn(OH)2

0.00125 moles of ZnCl2 reacts with 0.0025 moles of NaOH to form 0.00125 moles of Zn(OH)2

Total volume of solution = 20 + 50 = 70 ml = 0.07 L

molarity of Zn(OH)2 = moles/L = 0.00125/0.07 = 0.018 M

[Zn2+] = 0.018 M with charge +2

[OH-] = 2 x 0.018 = 0.036 M with charge 1

So,

ionic strength of Zn(OH)2 I = 1/2[0.018x2^2+0.036x1^2] = 0.108

effective ionic radius in nm for ions is,

Zn2+ = 0.6 ; OH- = 0.35

acitivity coefficient

logY[Zn2+] = -0.51 x 2^2 x sq.rt(0.108)/1+3.3 x 0.6 x sq.rt(0.108)

Y[Zn2+] = 0.393

logY[OH-] = -0.51 x 1^2 x sq.rt(0.108)/1+3.3 x 0.35 x sq.rt(0.108)

Y[OH-] = 0.755

let x be the amount of Zn2+ in solution at equilibrium the, 2x will be the amount of Cl- in solution.

Ksp for Zn(OH)2 = 5 x 10^-17 = [Zn2+][Cl-]^2

5 x 10^-17 = 0.393(x).(0.755)^2(2x)^2

x^3 = 5.58 x 10^-17

x = 3.82 x 10^-6 M

So the molar solubility of Zn(OH)2 = 3.82 x 10^-6 M

c) K2SO4 0.03 M

We don't have Zn(OH)2 here.

K2SO4 is quite soluble in aqueous solution.