Now given that the molar solubility of PbCrO4 (Pb2+ and CrO4 2-) is 1.40 x 10 -8 mol/L, find the Ksp
Write the expression for Ksp
Subtitute the numbers and solve.
Post your work if you gt stuck.
How many kilograms of chalcopyrite needs to be mined to get 325g of pure Cu?
So my ksp expression is
Ksp=(Pb2+)^1(CrO4)^1
Ksp=(1.40 x 10^-8)^1(1.40 x 10^-8)^1
Ksp=1.96 x 10^-16
Is this right?
Yes, that's ight. You know, of course, that the exponents 1 are not needed; however, there is nothing wrong with putting them there AND it serves to remind you each time to place an exponent. Many of my students through the years forgot the exponent. Good work!
Don't piggy back your question on another question. Usually those get ignored. When you repost as a separate question, show the equation you want to use to go from chalcopyrite to Cu.
To find the Ksp (solubility product constant) for PbCrO4, we need to set up an equilibrium expression based on the solubility. The balanced equation for the dissociation of PbCrO4 is:
PbCrO4(s) ⇌ Pb2+(aq) + CrO4 2-(aq)
The molar solubility of PbCrO4 is given as 1.40 x 10^-8 mol/L. This means that at equilibrium, the concentration of Pb2+ and CrO4 2- ions in the solution will both be equal to this molar solubility value.
The equilibrium expression for PbCrO4 is:
Ksp = [Pb2+][CrO4 2-]
We can substitute the molar solubility into the expression:
Ksp = (1.40 x 10^-8)(1.40 x 10^-8)
Ksp = 1.96 x 10^-16
Therefore, the Ksp for PbCrO4 is 1.96 x 10^-16.