The solubility of insoluble substances changes depending on the nature of the solution. Below are two solutions
in which Cu(OH)2 is dissolved; in each case, the solubility of Cu(OH)2 in those solutions is not the same as it is
in pure water.
(a) For each Cu(OH)2 solution below, state whether the solubility is greater than or less than that of pure water
and briefly account for the change in solubility.
i. 1-molar HCl(aq)
ii. 1-molar Cu(NO3)2(aq)
A saturated solution of lead(II) iodide in pure water has a lead ion concentration of 1.3 × 10-3 M at 25°C.
(b) Calculate the value for the solubility-product constant of PbI2 at 25°C.
(c) Calculate the molar solubility of PbI2. in a 0.10 M Pb(NO3)2 solution at 25°C.
(d) To 333 milliliters of a 0.12 M Pb(NO3)2 solution, 667 milliliters of 0.015 M KI is added. Does a precipitate form?
Support your answer with calculations and a concluding statement from those calculations.
#1.
i. more soluble in HCl because
Cu(OH)2 + 2HCl ==> CuCl2 + 2H2O
ii. less soluble in Cu(NO3)2 because of the common ion effect of Cu^2+ from the Cu(NO3)2. In pure water,
.......Cu(OH)2 ==> Cu^2+ + 2OH^-
inital..solid.......0.......0
change..solid.......x.......2x
equil...solid.......x.......2x
If we add extra Cu^2+ from Cu(NO3)2, that disturbs the equilibrium and Le Chatelier's principle says the reaction will shift to the left which means less solubility.
#2b
.........PbI2 ==> Pb^2+ + 2I^-
I........solid.....0.......0
C........solid.....x.......2x
E........solid.....x.......2x
Substitute the E line into Ksp expression. The problem tells you the value of x, calculate value for 2x, substitute into Ksp expression and solve for Ksp.
c. Same process as b bu Pb is not x + 0.1. Substitute into Ksp and solve for x which gives you the molar solubility of PbI2.
d. Calculate M Pb(NO3)2 and M KI in the mixed solution. Then calculate Qsp and compare with Ksp.
Post your work if you get stuck.
(a) For each Cu(OH)2 solution below, state whether the solubility is greater than or less than that of pure water and briefly account for the change in solubility.
i. In 1-molar HCl(aq), the solubility of Cu(OH)2 is greater than that in pure water. This is because HCl is a strong acid, which ionizes completely in water to produce a high concentration of H+ ions. The high concentration of H+ ions suppresses the formation of hydroxide ions, shifting the equilibrium toward the dissolved Cu2+ ions.
ii. In 1-molar Cu(NO3)2(aq), the solubility of Cu(OH)2 is greater than that in pure water. Cu(NO3)2 dissociates in water to form Cu2+ ions, which can combine with hydroxide ions to form Cu(OH)2. The presence of additional Cu2+ ions increases the solubility of Cu(OH)2.
(b) To calculate the solubility-product constant of PbI2 at 25°C, we need to use the given lead ion concentration in the saturated solution:
[Pb2+][I-]^2 = Ksp
Using the given concentration of [Pb2+] = 1.3 × 10-3 M, and assuming the concentration of [I-] is equal to the solubility (S) of PbI2:
(1.3 × 10-3)(S)^2 = Ksp
Solving for Ksp:
Ksp = (1.3 × 10-3)^3 = 2.197 × 10-9
(c) To calculate the molar solubility of PbI2 in a 0.10 M Pb(NO3)2 solution, we can use the common ion effect. In this case, the Pb(NO3)2 provides additional Pb2+ ions, which can shift the equilibrium and decrease the solubility of PbI2.
Let the solubility of PbI2 be x.
The solubility product expression would be:
[Pb2+][I-]^2 = (0.10 + x)(x)^2 = Ksp
Simplifying and solving for x:
(0.10 + x)(x)^2 = 2.197 × 10-9
x^3 + 0.10x^2 - 2.197 × 10-9 = 0
Solving this equation would give the molar solubility of PbI2 in the 0.10 M Pb(NO3)2 solution.
(d) To determine if a precipitate forms when 333 milliliters of a 0.12 M Pb(NO3)2 solution is mixed with 667 milliliters of 0.015 M KI, we can calculate the ion concentrations after mixing and compare them to the solubility product constant of PbI2.
Assuming the volumes are additive, we can calculate the final concentrations of Pb2+ and I- ions:
[Pb2+] = (0.12 M) * (333 mL / (333 mL + 667 mL))
[I-] = (0.015 M) * (667 mL / (333 mL + 667 mL))
Comparing the calculated concentrations to the solubility product constant of PbI2, if [Pb2+][I-]^2 > Ksp, a precipitate would form. If [Pb2+][I-]^2 < Ksp, no precipitate would form.
Using the calculated concentrations, we can determine if a precipitate forms or not to support our answer.
(a) The solubility of Cu(OH)2 depends on the nature of the solution it is dissolved in.
i. In a 1-molar HCl(aq) solution, the solubility of Cu(OH)2 is greater than that in pure water. This is because the HCl solution is acidic, and the presence of excess H+ ions from the acid can react with the Cu(OH)2 and form soluble Cu2+ ions, resulting in a higher solubility.
ii. In a 1-molar Cu(NO3)2(aq) solution, the solubility of Cu(OH)2 is less than that in pure water. This is because the Cu(NO3)2 solution is a source of Cu2+ ions, which can react with OH- ions from the Cu(OH)2 and form insoluble Cu(OH)2, thereby reducing the solubility.
(b) To calculate the solubility product constant (Ksp) of PbI2 at 25°C:
The solubility product constant expression for PbI2 is:
Ksp = [Pb2+][I-]^2
Given that the lead ion concentration [Pb2+] is 1.3 × 10^-3 M, we need to find the iodide ion concentration [I-].
Since the solubility of PbI2 is given as saturated, it means that PbI2 has reached its maximum solubility and has fully dissociated into Pb2+ and I- ions.
Therefore, the concentration of iodide ions [I-] is equal to 2 times the concentration of lead ions [Pb2+] as per the balanced chemical equation for PbI2 dissociation.
[I-] = 2 × [Pb2+]
[I-] = 2 × 1.3 × 10^-3 M
[I-] = 2.6 × 10^-3 M
Substituting these values into the Ksp expression:
Ksp = (1.3 × 10^-3 M) × (2.6 × 10^-3 M)^2
Ksp = 8.12 × 10^-9
Therefore, the solubility product constant of PbI2 at 25°C is 8.12 × 10^-9.
(c) To calculate the molar solubility of PbI2 in a 0.10 M Pb(NO3)2 solution at 25°C:
Since the concentration of lead ions [Pb2+] in the solution is given as 0.10 M, we need to determine the concentration of iodide ions [I-] needed to reach equilibrium and satisfy the Ksp expression.
Using the balanced chemical equation for PbI2 dissociation:
PbI2 ⇌ Pb2+ + 2I-
We can assume that the solubility of PbI2 is x M. Therefore, the concentration of lead ions [Pb2+] in equilibrium will also be x M, and the concentration of iodide ions [I-] will be 2x M.
Substituting these values into the Ksp expression:
Ksp = (x)(2x)^2
8.12 × 10^-9 = 4x^3
x^3 = (8.12 × 10^-9) / 4
x^3 = 2.03 × 10^-9
x ≈ 1.26 × 10^-3 M
Therefore, the molar solubility of PbI2 in a 0.10 M Pb(NO3)2 solution at 25°C is approximately 1.26 × 10^-3 M.
(d) To determine if a precipitate forms when 333 milliliters of a 0.12 M Pb(NO3)2 solution is mixed with 667 milliliters of 0.015 M KI:
The balanced chemical equation for the reaction between Pb(NO3)2 and KI is:
Pb(NO3)2 + 2KI → PbI2 + 2KNO3
To check if a precipitate forms, we need to compare the Qsp (Ion Product) and Ksp values.
Calculating the initial concentrations of Pb2+ and I- ions:
[Pb2+] = 0.12 M
[I-] = 2 × [KI]
[I-] = 2 × 0.015 M
[I-] = 0.03 M
Substituting these concentrations into the Qsp expression:
Qsp = [Pb2+][I-]^2
Qsp = (0.12 M)(0.03 M)^2
Qsp = 1.08 × 10^-4
Comparing Qsp with the Ksp value obtained in part (b):
Qsp < Ksp
Since Qsp < Ksp, it indicates that the reaction has not reached equilibrium and will proceed further before a precipitate forms. Therefore, a precipitate of PbI2 will not form.
In conclusion, based on the calculations and comparison of Qsp and Ksp values, no precipitate of PbI2 will form when 333 milliliters of a 0.12 M Pb(NO3)2 solution is mixed with 667 milliliters of 0.015 M KI.