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?

asked by Tanner

13 years ago

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1 answer

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.

answered by DrBob222

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Question

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?

1 answer

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.

Ask a New Question or answer this question.

DrBob222 · Answer

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.