Question

The K_sp of Cr(OH)₃ is 6.7*10^-33, and the K_sp of Zn(OH)₂ is 3*10^-17

1) Can Cr³⁺ be seperated from Zn²⁺ by the addition of an NaOH solution to an acidic solution that contiains 0.250 M Zn²⁺ and 0.120 M Cr³⁺?

2) If yes for one, at what hydroxide ion concentration will the second cation precipitate?

3) What is the concentration of the other cation at that hydroxide ion concentration?

Answer 1

a)

Ksp = [Cr+3][OH-]^3

[OH-] = ((6.7*10^-33)/(0.12))^(1/3) = 3.822*10^-11

for

Ksp = [Zn+2][OH-]^2

[OH-] = Ksp/[Zn+2] ^0.5 = ((3*10^-17)/(0.25))^0.5 = 1.095*10^-8

clearly

we can separate them, since overall difference:

[OH]zinc / [OH]cr = (1.095*10^-8) / (3.822*10^-11) = 286 times

Q2

if yes --> find OH- when Zn precipitates

as stated -- > Zn precipitates as Zn(OH)2 = (1.095*10^-8) M

Q3

find [Cr+3] at this concentration

Ksp = [Cr+3][OH-]^3

6.7*10^-33 = [Cr+3] ( (1.095*10^-8) )^3

[Cr+3] = (6.7*10^-33) / ((1.095*10^-8) )^3

[Cr+3] = 5.103 *10^-9 M