Question

1. The Ksp of Bi(OH)3 is 3.0*10^-36 , and the Ksp of Zn(OH)2 is 3*10^-17 .

a) Can Bi3+ be separated from Zn2+ by the addition of an NaOH solution to an acidic solution that contains 0.190M Zn2+ and 0.190M Bi3+ ?

b) At what hydroxide ion concentration will the second cation precipitate?

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

2. Keq = 2.71*10^-35 for the following reaction at 298K:

A(s) + 3B(g) ⇄ C(g)

a) Find the equilibrium pressure of C(g) in a container at 298K containing excess of A(s) and initial pressure of 1.57*10^4 bar of B.

b) What volume of container is expected to contain a single molecule of compound C (1 atm = 1.013 bar)

Answer 1

(1)

(a)

Yes it is possible to separate these two cations because of the extremely large difference in the two Ksp values.

Bi(OH)₃ will ppt first because of its extremely lower Ksp value.

(b)

The second cation to ppt is Zn2+.

Using the Ksp relation for Zn(OH)2:

Ksp = [Zn²⁺]*[OH^-]² = [0.19]*[OH^-]² = 3*10^-17

Solving we get:

[OH^-] = 1.26*10^-8 M

(c)

Using the Ksp expression for Bi(OH)₃:

Ksp = [Bi³⁺]*[OH^-]³ = [Bi³⁺]*[1.26*10^-8]³ = 3*10^-36

Solving we get:

[Bi³⁺] = 1.49*10^-12 M

Hope this helps !