A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0L of solution at 25oC. The concentration of Pb+2 in the saturated solution is found to be 1.3*10-3 M.
Calculate the Ksp for PbI2(s) in water at 25 °C
R/ Kps=[Pb^2+][I^-1]^2, where, [I^-1]=2[Pb^2+]= 2.6*10^-3,
Kps=[Pb^2+][I^-]^2= 8.8*10^-9
B) Calculate the solubility of PbI2(s) in distilled water at 25oC. Answer in mg/L. Answer: 599.57 mg/L
PbI2-------->Pb^2+ + 2I^-
[]0 0 0 0
[]rxn 0 x 2x
[]equilibrio 0 x 2x
Kps=[x][2x]^2, where Kps= 8.8*10^-9
8.8*10^-9=(s)(4s^2)
s^3=8.8*10^-9/4
s=1.30054*10^-3 mol/L
(1.30057*10^-3 mol/L)(461 g/1 mol)(1000 mg/1g)=599.57 mg/L
C) Calculate the solubility of PbI2(s) in 0.1M NaI. Answer in mg/L.
is similar and the answer is 0.40568 mg/L
Is your prof picky about significant figures. You have too many in some cases.
To calculate the solubility of PbI2(s) in 0.1M NaI, we need to consider the common ion effect.
The balanced chemical equation for the dissociation of PbI2(s) in water is:
PbI2(s) ⇌ Pb^2+(aq) + 2I^-(aq)
Let's assume that x is the solubility of PbI2(s) in the presence of 0.1M NaI. Since NaI is a soluble salt, we can assume that the Na^+ ion concentration does not affect the solubility equilibrium.
The equilibrium expression for the solubility product constant (Ksp) is:
Ksp = [Pb^2+][I^-]^2
In the presence of 0.1M NaI, the concentration of I^- will be 0.1M + 2x. The concentration of Pb^2+ will be x. Therefore, the equilibrium expression becomes:
Ksp = (x)(0.1M + 2x)^2
Substituting the given value of Ksp (8.8*10^-9):
8.8*10^-9 = (x)(0.1M + 2x)^2
Now, we can solve this equation to find the value of x, which represents the solubility of PbI2(s) in 0.1M NaI.
After solving the equation, the value of x is found to be approximately 0.02028 M (mol/L).
To convert this to mg/L, we can use the molar mass of PbI2:
(0.02028 M)(461 g/mol)(1000 mg/g) = 9.33 mg/L
Therefore, the solubility of PbI2(s) in 0.1M NaI is approximately 9.33 mg/L.
To calculate the solubility of PbI2(s) in distilled water at 25oC, we can use the expression for the solubility product constant (Ksp) of PbI2:
Ksp = [Pb^2+][I^-]^2
Given that the concentration of Pb^2+ in the saturated solution is 1.3*10^-3 M, we can substitute this value into the equation:
Ksp = (1.3*10^-3)(2.6*10^-3)^2
= 8.8*10^-9
Therefore, the Ksp for PbI2(s) in water at 25oC is 8.8*10^-9.
To calculate the solubility of PbI2(s) in distilled water at 25oC, we need to consider the stoichiometry of the dissociation reaction:
PbI2(s) ⇌ Pb^2+ + 2I^-
Let the initial concentration of PbI2(s) be [PbI2]0 and the concentration of I^- ions be [I^-] = 2[Pb^2+] since the stoichiometric ratio is 1:2. At equilibrium, the concentration of Pb^2+ ions is x and the concentration of I^- ions is 2x.
Using the expression for Ksp, we can set up the following equation:
Ksp = [Pb^2+][I^-]^2
= x(2x)^2
= 4x^3
However, we know that the concentration of Pb^2+ ions in the saturated solution is 1.3*10^-3 M, so we can substitute this value into the equation:
8.8*10^-9 = 4x^3
Solving for x, the concentration of Pb^2+ ions at equilibrium:
x^3 = 8.8*10^-9 / 4
x^3 = 2.2*10^-9
x = (2.2*10^-9)^(1/3)
x = 1.30054*10^-3 M
Finally, to convert the solubility from moles per liter to milligrams per liter:
solubility = (1.30054*10^-3 mol/L) * (461 g/1 mol) * (1000 mg/1g)
= 599.57 mg/L
Therefore, the solubility of PbI2(s) in distilled water at 25oC is 599.57 mg/L.
For part C, calculating the solubility of PbI2(s) in a 0.1M NaI solution, the procedure is similar. The only difference is the initial concentration of I^- ions, which is now 0.1M due to the presence of NaI.
Following the same steps as before, the solubility is calculated to be 0.40568 mg/L in a 0.1M NaI solution.