What is the molar solubility of copper(II) carbonate at 25 oC in a solution that is 0.055 M in copper(II) chloride?
I have done one and I am not sure I am doing it correct. If you can show your work for myself to understand. That would be awesome. Thank You!!
wow- fiesty drbob
To determine the molar solubility of copper(II) carbonate, we need to use the concept of solubility product constant (Ksp). The solubility product constant (Ksp) is an equilibrium constant that describes the extent to which a compound dissolves in a solution.
The chemical equation for the dissociation of copper(II) carbonate (CuCO3) in water is:
CuCO3(s) ⇌ Cu2+(aq) + CO3^2-(aq)
The balanced equation shows that one mole of copper(II) carbonate dissociates to produce one mole of copper(II) ions and one mole of carbonate ions.
First, we need to write the equilibrium expression using the equation coefficients as exponents:
Ksp = [Cu2+][CO3^2-]
Next, we need to determine the concentrations of Cu2+ and CO3^2- ions in the solution.
In this case, the solution is 0.055 M in copper(II) chloride (CuCl2). Copper(II) chloride dissociates completely to produce Cu2+ ions.
Hence, the concentration of Cu2+ ions can be considered as 0.055 M.
Since copper(II) carbonate is sparingly soluble, the concentration of CO3^2- ions formed from its dissociation can be considered negligible. Therefore, we can assume that the concentration of CO3^2- ions is zero initially.
Substituting the concentration values into the equilibrium expression:
Ksp = [0.055][0] = 0
As a result, the solubility product constant (Ksp) for copper(II) carbonate is zero, indicating that the compound is insoluble and has negligible solubility in the given solution.
Therefore, we can conclude that the molar solubility of copper(II) carbonate in a solution that is 0.055 M in copper(II) chloride at 25°C is effectively zero.
It would have been far easier for you to show your work and let us tell you where your weak/strong points are. As it is I do all the typing and you compare. Henceforth, you type and we will compare.
CuCO3(s)==> Cu^2+ + CO3^-
...x.........x.......x
.........CuCl2(aq) ==> Cu^2+ + 2Cl^-
initial..0.055...........0......0
final soln..0.........0.055..2*0.055
Ksp CuCO3 = (Cu^2+)(CO3^2-)
Ksp = (x+0.055)(x)
Solve for x = solubility CuCO3.
Note: This is a common ion problem with the Cu(II) from CuCl2 being the common ion. Therefore the concn Cu(II) is x from the CuCO3 and 0.055 from the CuCL2.