Question

If a 0.100 M solution of NaOH is added to a solution containing 0.200 M Ni²⁺, 0.200 M Ce³⁺, and 0.200 M Cu²⁺, which metal hydroxides will precipitate first and last, respectively? K_sp for Ni(OH)₂ is 6.0x10^?16, for Ce (OH)₃ is 6.0 x 10^?22, and for Cu(OH)₂ is 4.8x10^?20.

A) Ni(OH)₂ first, Cu(OH)₂ last
B) Cu(OH)₂ first, Ni(OH)₂ last
C) Ni(OH)₂ first, Ce(OH)₃ last
D) Cu(OH)₂ first, Ce(OH)₃ last
E) Ce(OH)₃ first, Cu(OH)₂ last

Give detailed explanation

Answer 1

For Ni(OH)₂ <-----------------> Ni²⁺ +2OH^-

K_sp = 6.0x10^-16

[Ni²⁺] = 0.200M , [OH^-] = 0.100M

Hence ionic product(IP) = [Ni²⁺]x[OH^-]² = 0.200x(0.100)² = 0.00200 M²

Since IP>K_sp,  precipitation will occur.

Now [OH^-] required to precipitate Ni(OH)₂ can be calculated as

K_sp [Ni(OH)₂] =   [Ni²⁺]x[OH^-]² =  6.0x10^-16

=> [OH^-] = [(6.0x10^-16)/0.200]^1/2 = 5.48x10^-8 M

For Ce(OH)3 <-----------------> Ce³⁺ +3OH^-

K_sp = 6.0x10^-22

[Ce³⁺] = 0.200M , [OH^-] = 0.100M

Hence ionic product(IP) = [Ce³⁺]x[OH^-]³ = 0.200x(0.100)³ = 0.000200 M³

Since IP>K_sp,  precipitation will occur.

Now [OH^-] required to precipitate Ce(OH)₃ can be calculated as

K_sp [Ce(OH)₃] =   [Ce³⁺]x[OH^-]³ = 6.0x10^-22

=> [OH^-] = [(6.0x10^-22)/0.200]^1/3 = 1.44x10^-7 M

For Cu(OH)₂ <-----------------> Cu²⁺ +2OH^-

K_sp = 4.8x10^-20

[Cu²⁺] = 0.200M , [OH^-] = 0.100M

Hence ionic product(IP) = [Cu²⁺]x[OH^-]²  = 0.200x(0.100)² = 0.00200 M²

Since IP>K_sp,  precipitation will occur.

Now [OH^-] required to precipitate Cu(OH)₂ can be calculated as

K_sp [Cu(OH)₂] =   [Cu²⁺]x[OH^-]² = 4.8x10^-20

=> [OH^-] = [(4.8x10^-20)/0.200]^1/2 = 4.90x10^-10

Lowest the concentration of OH^- required fastest will be the precipitation.

Hence Cu(OH)₂ requires the lowest [OH-](4.90x10^-10 M) and will precipitate first.

and Ce(OH)₃ requires highest concentration(1.44x10^-7 M), it will precipitate last.

Hence D is the correct option.