Question

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

1. Can Bi³⁺ be separated from Zn²⁺ by the addition of an NaOH solution to an acidic solution that contains 0.190 M Zn²⁺ and 0.190 M Bi³⁺? If can't leave answer fields in parts 2, 3 empty.

2. At what hydroxide ion concentration will the second cation precipitate? in mol/L

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

Answer 1

For Bi(OH)3 the solubility equilibrum is

Bi(OH)3 <-----> Bi⁺³  + 3 OH^-

- s 3s equilibrium concentrations where s mol/L is the soubility of salt.

similarlly for Zn(OH)2

Zn(OH)2 <-----> Zn⁺² + 2 OH^-

- s 2s   

Ksp of Bi(OH)3 = s.(3s)³ = 27s⁴= 3.0 x10^-36

since [Bi+3] = 0.190 M given

the [OH-] required to precipitate Bi+3 = Ksp /[Bi+3]

= 3.0 x10^-36 /0.190=1.579x 10^-35 M

similarly the [Oh-] required to precipitate Zn+2 = ksp/[Zn+2]

= 3.0x 10^-17 / 0.190

= 1.579x 10^-16 M

Q1) thus Bi+3 can be precipitated by the addition of NaOH solution much before the Zn+2 is precipitated.The concentration of OH- required for this precipitation is 1.579x 10^-35 M

Q2) The[OH-] concentration at which Zn+2 can be preicipated is 1.579x 10^-16 M

Q3) The concentration of BI+3 when the Zn+2 ion starts precipitation can be calulated as follows

[Bi+3] = Ksp of Bi[Oh]3 / Ksp of Zn(oh)2

= 3.0x 10^-36 /3.0x 10^-17

= 1.0 x10^-19 M

That is until almost all Bi+3 is precipitated, no precipitation of Zn+2 starts.