A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO?3?)?2? with 50.0 mL of

Question

A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO?3?)?2? with 50.0 mL of 0.50 MCo(NO?3?)?2?. NaOH is added to the mixture. Which of the two metal hydroxides precipitatesfirst? K?sp?(Cu(OH)?2?) = 2.2 × 10??20?, K?sp?(Co(OH)?2?) = 1.3 × 10??15

a. Co(OH)2

b. Co(NO3)2

c. Cu(No3)2

d. Cu(OH)2

e. no precipitate is formed.

When 0.015 mol of KOH is added to 1.00 L of 0.0010 M Ca(NO?3?)?2?. Which of the followingstatements is correct? [K?sp?(Ca(OH)?2?) = 6.5 × 10?–6?]

A) Calcium hydroxide precipitates until the solution is saturated.

B) The solution is unsaturated and no precipitate forms.

C) The concentration of calcium ions is reduced by the addition of the hydroxide ions.

D) One must know Ksp for calcium nitrate to make meaningful predictions on this system.

E) The presence of KOH will raise the solubility of Ca(NO3)2

Explanation / Answer

Cu(OH)2 will precipitate first because Ksp of Cu(OH)2 << Ksp of Co(OH)2.

so answer d is correct

mixing 50.0 mL of 0.50 M Cu(NO3)2 with 50.0 mL of 0.50 M Co(NO3)2
dilutes both Molarities in half to...
0.25 M Cu(NO3)2 & 0.25 M Co(NO3)2



you will likely need to re-do the calcs using your text's Ksp's
Ksp Co(OH)2 = 1.3×10^–15
Ksp Cu(OH)2 = 2.2×10^–20


Ksp Cu(OH)2 = 2.2×10^–20
Ksp Cu(OH)2 = [Cu+2] [OH-]^2
2.2×10^–20 = [0.25M] [OH-]^2
[OH-]^2 = 8.8 X 10-20
[OH-] = 2.97 X 10^-10 Molar


Ksp Co(OH)2 = 1.3×10^–15
Ksp Co(OH)2 = [Co+2] [OH-]^2
1.3×10^–15 = [0.25M] [OH-]^2
[OH-]^2 = 5.2 X 10-15
[OH-] = 7.2 X 10^-8 Molar

Cu(OH)2 will precipitate first...
because it needs the least amout of [OH-] to match its Ksp
Cu(OH)2 is saturated at [OH-] = 2.97 X 10^-10 Molar
add any more {OH-] & it will begin to precipitate

Co(OH)2 precipitates second
it requires a greater concentration of hydroxide to match its Ksp
[OH-] = 7.2 X 10^-8 Molar
has enough [OH-[ been added to just saturate the solution,
add any more hydroxide & it wil ppt

you have to add more than [OH-] = 2.97 X 10^-10 Molar to ppt Cu(OH)2
but you have to keep it less that [OH-] = 8.0 X 10^-8 , or else Co(OH)2 will ppt as well

Ksp Ca(OH)2 = [Ca+2] [OH-]^2
6.5×10^–6 = [0.001M] [OH-]^2
[OH-]^2 = 6.5 X 10-3 = 65 X 10-4
[OH-] = 8.06 X 10-2 Molar

so A statment is false

statment B is correct

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