Question

A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2. Sodium hydroxide is then added to the mixture. Assume no change in volume.

Ksp = 2.2 × 10-20 for Cu(OH)2, Ksp = 1.3 × 10-15 for Co(OH)2

The hydroxide concentration at which the first metal hydroxide will just begin to precipitate is __________M

Using the integer designators, the metal hydroxide that precipitates first is {(1) Cu(OH)2 or (2) Co(OH)2 } _____________M

The hydroxide concentration at which the maximum amount of the first cation is precipitated while leaving the second cation completely in the solution is ___________M

Answer 1

Here, The Mixing of 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2 Dilutes Both Molarity in Half To, 0.25 M Cu(NO3)2 & 0.25 M Co(NO3)2.

Ksp = 2.2 × 10^-20 for Cu(OH)2

Ksp = 1.3 × 10^-15 for Co(OH)2

NOW,

Ksp Cu(OH)2 = 2.2 × 10^-20

Ksp Cu(OH)2 = [Cu+2] [OH-]^2

2.2 × 10^-20 = [0.25 M] [OH-]^2

[OH-]^2 = 8.8 × 10^-20

[OH-] = 2.97 × 10^-10 M

pOH = 9.53 & pH = 4.47

NOw,

Ksp CO(OH)2 = 1.3 × 10^-15

Ksp CO(OH)2 = [CO+2] [OH-]^2

1.3 × 10^-15 = [0.25 M] [OH-]^2

[OH-]^2 = 5.2 × 10^-15

[OH-] = 7.21 × 10^-8 M

Now, here Cu(OH)2 will precipitate first because it needs the least amt of [OH-] to match it's Ksp Cu(OH2) is saturated at [OH-] to molar which is at Ph = 4.47 . if any more [OH-] add to it will begin to precipitate.

Co(OH)2 precipitate 2nd because it requires greater concentration of hydroxide to match its Ksp [OH-] = 7.21 × 10^-8 not until you get a Ph = 7 has enough [OH-] if more [OH-] add to than it will begin to precipitate.

SO, finally

you have to add more than[OH-] = 2.97 × 10^-10 to precipitate Cu(OH2) & have to keep it lees that [OH-] = 7.21 × 10^-8, or else Co(OH)2 will precipitate first.