Question

If a 0.200 M solution of KOH is added to a solution containing 0.100 M Ni2+, 0.100 M Ce3+, and 0.100 M Cu2+, which metal hydroxide will precipitate first? Calculate the concentration of the first precipitated ion when the second soluble ion begins to precipitate. When the most soluble ion starts to precipitate, what is the concentration for the second soluble ion in the solution?

Ksp Ni(OH)2 = 6.0 X 10-16, Ksp Ce(OH)3 = 6.0 X 10-22, Ksp Cu(OH)2 = 4.8 X 10-20

Answer 1

Ni(OH)2 -------> Ni2+ + 2OH-

Ksp = 6 x 10^-16 = [Ni2+] [OH-]^2

=> 6 x 10^-16 = 0.1 x [OH-]^2

=> [OH-] = 7.75 x 10^-8 M = Minimum Conc. of OH- needed for Ni(OH)2 to start precipitation

Ce(OH)3 -------> Ce3+ + 3OH-

Ksp = 6 x 10^-22 = [Ce3+] [OH-]^3

=> 6 x 10^-22 = 0.1 x [OH-]^3

=> [OH-] = 1.82 x 10^-7 M = Minimum Conc. of OH- needed for Ce(OH)3 to start precipitation

Cu(OH)2 ------> Cu2+ + 2OH-

Ksp = 4.8 x 10^-20 = [Cu2+] [OH-]^2

=> 4.8 x 10^-20 = 0.1 x [OH-]^2

=> [OH-] = 6.93 x 10^-10 M = Minimum Conc. of OH- needed for Cu(OH)2 to start precipitation

Since the concentration of [OH-] required is minimum for Cu(OH)2, it begins to precipitate first

Ni(OH)2 will precipitate second

When Ni(OH)2 starts to precipitate [OH-] = 7.75 x 10^-8 M

Suppose X M of Cu(OH)2 precipitates

The reaction is

Cu2+ + 2OH- -------------------------> Cu(OH)2

0.1 - X .....7.75 x 10^-8 - 2X ............ X

Ksp = 4.8 x 10^-20 = (0.1 - X) (7.75 x 10^-8 - 2X)^2

=> X = 3.84 x 10^-8 M = Conc. of the first precipitated ion when the second soluble ion begins to ppt

Similarily,

Ni2+ + 2OH- ------------------> Ni(OH)2

0.1 - X... 1.82 x 10^-7 - 2X............X

Ksp = 6 x 10^-16 = (0.1 - X) (1.82 x 10^-7 - 2X)^2

=> X = 1.05 x 10^-7 M = the concentration for the second soluble ion in the solution