A solution consists of 0.27 M MgCl2 and 0.7 M CuCl2. Calculate the concentration of hydroxide ion needed to separate the metal ions. The Ksp of Mg(OH)2 is 6.3x10-10, and the Ksp of Cu(OH)2 is 2.2x10-20. Multiply your answer by 106 and enter that number to 2 decimal places.

The concentration of hydroxide ion needed to precipitate Cu²⁺ whereas Mg²⁺ still in solution is 1.77x10⁻¹⁰

Explanation:

Based on the different Ksp of the hydroxides of the metal ions it is possible to precipitate 1 metal as hydroxide whereas the other still in solution.

Ksp of both hydroxides is:

Mg(OH)₂ ⇄ Mg²⁺ + 2OH⁻

Ksp = [Mg²⁺] [OH⁻]² = 6.3x10⁻¹⁰

Cu(OH)₂ ⇄ Cu²⁺ + 2OH⁻

Ksp = [Cu²⁺] [OH⁻]² = 2.2x10⁻²⁰

Replacing with the concentration of the metals:

[0.27] [OH⁻]² = 6.3x10⁻¹⁰

[OH⁻] = 4.83x10⁻⁵M

That means the precipitation of Mg²⁺ begins when [OH⁻] is 4.83x10⁻⁵M

[0.7] [OH⁻]² = 2.2x10⁻²⁰

[OH⁻] = 1.77x10⁻¹⁰

As the Cu²⁺ needs a low concentration of OH⁻ to begin precipitation,

the answer is c) gamma rays

hope i : )

it's filtration, but this is an assumption so my apologies if i got this