We are given that the solubility product of platinum (II) sulfide is 9.9x10^-74
We are told to write a balanced equation for dissolving platinum (II) sulfide.
I put : PtS(S) <> Pt^2+(aq) + S^2-(aq)
and then we had to write the Ksp expression: Ksp=[Pt^2+][S2-]
The the final question says to: Calculate the volume of water required to dissolve 0.0010mg of platinum (ii) sulfide.
I do not know where to start for the problem, please help.
4 answers
Why are you posting this again?
because i forgot to mention the solubility product given
OK, I'm going to remove the duplicate below. Now you need patience until one of our chem tutors comes online.
.....PtS ==> Pt^2+ + S^2-
I....solid....0......0
C....solid....x......x
E....solid....x......s
Ksp = (Pt^2+)(S^2-)
You know Ksp, substitutethe E line into the Ksp expression and solve for x = solubility in mols/L.
Convert mols/L to grams/L with grams = mols x molar mass = ?
That give you grams PtS in 1000 mL. You want to dissolve 0.001 mg or 1E-6 grams.
1000 mL x ( 1E-6/solubility) = mL required to dissolve 0.001 mg.
I....solid....0......0
C....solid....x......x
E....solid....x......s
Ksp = (Pt^2+)(S^2-)
You know Ksp, substitutethe E line into the Ksp expression and solve for x = solubility in mols/L.
Convert mols/L to grams/L with grams = mols x molar mass = ?
That give you grams PtS in 1000 mL. You want to dissolve 0.001 mg or 1E-6 grams.
1000 mL x ( 1E-6/solubility) = mL required to dissolve 0.001 mg.