When 1.0 x 10-3 moles of CuCl2(s) are added to 1.0 L of 1.0 x 10-3 M IO3
-
, the
A Trial Ksp > Ksp and a precipitate forms
B Trial Ksp < Ksp and a precipitate forms
C Trial Ksp > Ksp and no precipitate forms
D Trial Ksp < Ksp and no precipitate forms
show your work
To determine whether a precipitate will form, we need to compare the trial Ksp (the ion product) with the Ksp (the solubility product constant) for CuCl2.
The balanced equation for the dissociation of CuCl2 is:
CuCl2(s) -> Cu2+(aq) + 2Cl-(aq)
The Ksp expression for CuCl2 can be written as:
Ksp = [Cu2+][Cl-]^2
To find the concentrations of Cu2+ and Cl- ions in the solution, we need to calculate the moles of CuCl2 that dissociate.
Given:
Initial moles of CuCl2 = 1.0 x 10^-3 moles
Initial volume of solution = 1.0 L
Initial concentration of IO3- = 1.0 x 10^-3 M
The reaction between CuCl2 and IO3- can be represented as follows:
CuCl2(s) + 2IO3-(aq) -> Cu(IO3)4(aq) + 2Cl-(aq)
Since Cu(IO3)4 has a much smaller Ksp than CuCl2, we can assume it remains negligible compared to CuCl2.
The moles of Cu2+ ions formed will be equal to twice the moles of CuCl2 dissolved:
Moles of Cu2+ = 2 * (1.0 x 10^-3 moles) = 2.0 x 10^-3 moles
The moles of Cl- ions formed will be equal to twice the moles of Cu2+ ions:
Moles of Cl- = 2 * (2.0 x 10^-3 moles) = 4.0 x 10^-3 moles
Since the volume of the solution is 1.0 L, the concentrations of Cu2+ and Cl- ions can be calculated:
[Cu2+] = Moles of Cu2+ / Volume of solution = 2.0 x 10^-3 moles / 1.0 L = 2.0 x 10^-3 M
[Cl-] = Moles of Cl- / Volume of solution = 4.0 x 10^-3 moles / 1.0 L = 4.0 x 10^-3 M
Substituting these values in the Ksp expression:
Trial Ksp = [Cu2+][Cl-]^2 = (2.0 x 10^-3 M)(4.0 x 10^-3 M)^2 = 3.2 x 10^-8
To compare this trial Ksp with the actual Ksp for CuCl2:
Given Ksp for CuCl2 = 1.0 x 10^-6
Comparing the trial Ksp with the Ksp:
Trial Ksp < Ksp
Therefore, the answer is (D) Trial Ksp < Ksp and no precipitate forms.