..........PbCl2(s) ==> Pb^2+ + 2Cl^-
x = solubility..x.......x.......2x
Ksp = (Pb^2+)(Cl^-)^2
Ksp = (x)(2x)^2
Solve for x which will give you the moles/L of PbCl2.
Mols/L x 0.48 L = moles in 480 mL.
Then moles x molar mass PbCl2 = grams PbCl2.
i need to find the; [Cl-] in PbCl2 solution,[Pb2+] in PbCl2 solution, and the ksp given: PbCl2(s)--> Pb2+(aq)+2Cl-(aq) Temp of solution =20.2 volume of Pbcl2 solution=25.00ml mass of dry AgCl2(g)=0.2543g #moles : AgCl(s)=1.77e-3
Would you expect the solubility of PbCl2 in water to change dramatically if it were to be dissolved into a solution of 3.091×10−4 M KCl? Yes, the solubility of PbCl2 would decrease dramatically. Yes, the solubility of PbCl2
What is the maximum mass(g) of KCL that can be added to 1.00L of a 0.0100 M lead(ii) chloride solution without causing any precipitation of PbCl2? Assume the additional KCl does not affect the volume of the solution. For, PbCl2,
1. What mass of lead chloride (PbCl2, MW= 278.2; Ksp= 1.8 x 10^-10) is dissolved in 250 mL of a saturated solution? 2. What mass lead chloride (PbCl2, MW=278.2; Ksp=1.8 x 10^-10) will dissolve in 150 mL of 0.050 M Pb(NO3)2?
In an experiment to determine the solubility of lead chloride (PbCl2), 5.6g of (NH4)2SO4 was added to a 250ml solution containing an unknown amount of dissolved lead chloride (PbCl2) resulting in the formation of Lead sulfate
Question One Lead chloride dissolves in water according to PbCl2(s) Pb^2+ + 2Cl^- (aq) The solubility in pure water has been measured to be 4.44g . L^-1 . Calculate the solubility product of lead chloride in pure water. Your