Let's explore the efficiency of chloride ion precipitation in seperating Ag^+ from Pb^2+ in a solution. Suppose that [Pb^2+]= 0.045 mol/L and [Ag^+]= 0.084 mol/L. Calculate the chloride ion concentraion required to just start precipitation of Pb^2+ ion and then calculate the concentration of Ag^+ left in solution at that chloride ion concentratio(Assume the volume of the solution does not change). Is this an efficient method of separating the two ions?
Ksp = (Pb^2+)(Cl^-)^2
Substitute 0.045M for Pn^2+ and solve for Cl^-.
Then Ksp = (Ag^+)(Cl^-)
Use that Cl^- and solve for Ag^+ remaining.
To calculate the chloride ion concentration required to start the precipitation of Pb^2+ ions, we need to determine the solubility product constant (Ksp) for lead chloride (PbCl2). The Ksp for PbCl2 is 1.6 x 10^-5 at 25°C.
The equation for the dissolution of PbCl2 in water is:
PbCl2 (s) ⇌ Pb^2+ (aq) + 2Cl^- (aq)
At equilibrium, the concentration of Pb^2+ (aq) will be equal to the saturation concentration of PbCl2, given by the product of the Pb^2+ and Cl^- ion concentrations:
[Pb^2+] = [Cl^-]^2
Given that [Pb^2+] = 0.045 M, we can rearrange the equation to solve for [Cl^-]:
[Cl^-] = √([Pb^2+])
= √(0.045)
≈ 0.212 M
Therefore, the chloride ion concentration required to just start the precipitation of Pb^2+ ions is approximately 0.212 M.
To calculate the concentration of Ag^+ ions remaining in solution, we need to consider the reaction between silver chloride (AgCl) and Ag^+ ions. The solubility product constant (Ksp) for silver chloride (AgCl) is 1.8 x 10^-10 at 25°C.
The equation for the dissolution of AgCl in water is:
AgCl (s) ⇌ Ag^+ (aq) + Cl^- (aq)
Since AgCl is a strong electrolyte, it completely dissociates, and the concentration of Cl^- ions will be equal to the [Cl^-] from before, which is 0.212 M.
Using the Ksp expression for AgCl, we can calculate the concentration of Ag^+ ions remaining in solution at the chloride ion concentration of 0.212 M:
[Ag^+][Cl^-] = Ksp
[Ag^+](0.212) = 1.8 x 10^-10
Rearranging the equation to solve for [Ag^+]:
[Ag^+] = (1.8 x 10^-10)/(0.212)
≈ 8.49 x 10^-10 M
Therefore, the concentration of Ag^+ ions left in solution at the chloride ion concentration of 0.212 M is approximately 8.49 x 10^-10 M.
Now, to determine the efficiency of this method, we compare the remaining concentration of Ag^+ ions to the initial concentration. The initial concentration of Ag^+ ions was given as 0.084 M, and the final concentration is approximately 8.49 x 10^-10 M.
The efficiency can be calculated as:
Efficiency = (Final concentration / Initial concentration) x 100
= (8.49 x 10^-10 M / 0.084 M) x 100
≈ 1.01 x 10^-6 %
This method of chloride ion precipitation is highly efficient in separating Ag^+ from Pb^2+ ions, as it results in an extremely low concentration of Ag^+ ions (approximately 1.01 x 10^-6 % of the initial concentration) remaining in solution.