a solution contains .25M Ni(NO3)2 and .25 M Cu(NO3)2. Can the metal ions be separated by slowly adding Na2CO3?Assume that for successful separation 99% of the metal ion must be precipitated before the other metal ion begins to precipitate, and assume no vlume change on addition of Na2CO3.
What are the Ksp values for NiCO3 and CuCO3?
I looked up some Ksp values for CuCO3 and NiCO3 and found the following:
Ksp NiCO3 = 1.4 x 10^-7
Ksp CuCO3 = 2.5 x 10^-10
You should confirm those numbers.
(Cu^+2)(CO3^-2)...(2.5 x 10^-10
--------------- = ------------- =0.00179
(Ni^+)(CO3^-2)....(1.4 x 10^-7)
Therefore,(Cu^+2)/(Ni^+2) = 0.00179 and
(Cu^+2) = 0.00179*(Ni^+2).
If all of the Cu^+ is pptd, then before the next molecule of Na2CO3 is added, the (Ni^+2) = 0.25 and (Cu^+2) = 0.00179 x (0.25) = 0.000448 in solution. Thus that is the unpptd Cu^+2. [Note: It is better to be using moles at this point; however, since Na2CO3 is a solid we are not diluting the solution, and molarity works ok.]
Thus percent Cu^+2 pptd before the first molecule NiCO3 ppts is
0.000448/0.25)*100 = 0.18% which means that 99.8% pptd. Thus, the two can be separated. Check my work. Check my thinking.
To determine if the metal ions can be separated by slowly adding Na2CO3, we need to consider the solubility of the metal carbonates.
First, let's look at the solubility product constants (Ksp) for NiCO3 and CuCO3. The Ksp values at 25°C are as follows:
- Ksp(NiCO3) = 6.0 x 10^-10
- Ksp(CuCO3) = 2.5 x 10^-10
Based on the given information, we need to determine if 99% of one metal ion can be precipitated before the other metal ion begins to precipitate when Na2CO3 is slowly added to the solution.
Let's consider the precipitation of NiCO3. To precipitate 99% of the Ni(2+) ions, the concentration of Ni(2+) remaining in the solution should be less than or equal to 1% of its initial concentration.
Using the given initial concentration of 0.25M Ni(NO3)2, the remaining concentration of Ni(2+) should be less than or equal to 0.0025M.
To determine the concentration of carbonate ion that will give a remaining concentration of Ni(2+) less than or equal to 0.0025M, we can use the Ksp expression for NiCO3:
NiCO3(s) ⇌ Ni(2+)(aq) + CO3(2-)(aq)
Ksp = [Ni(2+)]eq [CO3(2-)]eq
Let's assume that x is the concentration of carbonate ion formed by the dissociation of Na2CO3. Since it is a 1:1 mole ratio between Ni(2+) and CO3(2-), the concentration of Ni(2+) remaining should be around (0.25 - x).
Using the Ksp expression and neglecting x in comparison with 0.25, we have:
Ksp = (0.25 - x)(x)
0.25 x 10^-9 = x^2
Taking the square root of both sides, we have:
x ≈ 5 x 10^-5 M
Therefore, when the carbonate ion concentration is around 5 x 10^-5 M, 99% of the Ni(2+) ions will be precipitated.
Now let's check the precipitation of CuCO3. If we slowly add Na2CO3 until the carbonate ion concentration reaches 5 x 10^-5 M, the concentration of Cu(2+) remaining in the solution should be less than or equal to 0.0025M based on the 99% precipitation criteria.
Using the Ksp expression for CuCO3:
CuCO3(s) ⇌ Cu(2+)(aq) + CO3(2-)(aq)
Ksp = [Cu(2+)]eq [CO3(2-)]eq
Using the same approach as before:
Ksp = (0.25 - y)(y)
0.25 x 10^-9 = y^2
Taking the square root of both sides, we have:
y ≈ 5 x 10^-5 M
Therefore, when the carbonate ion concentration is around 5 x 10^-5 M, 99% of the Cu(2+) ions will be precipitated.
Based on these calculations, it seems that both Ni(2+) and Cu(2+) ions will start precipitating at around the same carbonate ion concentration. Therefore, it is unlikely that the metal ions can be separated by slowly adding Na2CO3 while maintaining the 99% precipitation criteria.
To determine if the metal ions can be separated by slowly adding Na2CO3, we need to consider the solubility of their respective metal carbonates.
1. Ni(NO3)2 + Na2CO3 → NiCO3 + 2NaNO3
2. Cu(NO3)2 + Na2CO3 → CuCO3 + 2NaNO3
Based on the given information, the requirement for successful separation is that 99% of one metal ion must precipitate before the other metal ion begins to precipitate.
First, let's calculate the concentration of the metal carbonates after precipitation.
For NiCO3:
Initial concentration of Ni(NO3)2 = 0.25 M
0.99 * 0.25 M = 0.2475 M Ni(NO3)2 remaining
Assuming no volume change, let's assume the total volume after adding Na2CO3 remains at 1 liter.
Now, let's use stoichiometry to determine the concentration of NiCO3 formed:
Ni(NO3)2 : NiCO3 → 1 : 1 (according to the balanced equation)
The concentration of NiCO3 formed will also be 0.2475 M.
For CuCO3:
Initial concentration of Cu(NO3)2 = 0.25 M
0.99 * 0.25 M = 0.2475 M Cu(NO3)2 remaining
Using stoichiometry:
Cu(NO3)2 : CuCO3 → 1 : 1 (according to the balanced equation)
The concentration of CuCO3 formed will also be 0.2475 M.
Since both NiCO3 and CuCO3 have the same concentration after the required precipitation, it indicates that both metal ions will precipitate simultaneously. Therefore, the metal ions cannot be separated by slowly adding Na2CO3 under the given conditions.