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

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml

So what you would do is go through and add 3 teaspoons(tsp) 6 times because of the 6 tablespoons(tbsp). which will give you an answer of 18. another way is to multiply 3x6=18.