Question

Explain fully using equilibrium why the pH needs to be above 8 and an excess of NaOH added for the separate cation II Cu(II) and Cation III Zn2+

Answer 1

Cu2+ when reacts with NaOH it forms Cu(OH)2 which is later dehydrolysed to form black colored CuO.

Cu(OH)2(s) ---- > Cu2+(aq) + 2 OH-(aq), Ksp = 4.8x10^-20

Ksp = 4.8x10^-20 =  [Cu2+(aq)]x[OH-(aq)] = S x (2S)² = 4S³

=> S = cuberoot (4.8x10^-20 / 4) = 2.29x10^-7

=> S = [OH-(aq)] = 2.29x10^-7

For precipitation to occur, ionic product > solubility product(Ksp)

Hence for precipitation to occur, [OH-(aq)] > S

=> [OH-(aq)] >  2.29x10^-7

=> 10^-14 / [H+(aq)] >  2.29x10^-7 [ [OH-(aq)]x[H+(aq)] = 10^-14 ]

=> 10^-14 / 2.29x10^-7 >  [H+(aq)]

=> 4.367x10^-8 > [H+(aq)] or [H+(aq)] <  4.367x10^-8

=> log[H+(aq)] < log(4.367x10^-8 ) = -7.36

=> - log[H+(aq)] > 7.36

=> pH > 7.36

Hence for precipitation to start pH should be greater than 7.36 and for better result it should be greater than 8. When we add excess NaOH ionic product increases and precipitation occurs rapidly.

Same is the case for Zn2+