How many grams ofPbCl2will dissolve in 2.00 L of O.650M aqueous Pb(N03)2 solution?
........PbCl2 ==> Pb^2+ + 2Cl^-
I.......solid......0........0
C.......solid......x........2x
E.......solid......x........2x
.........Pb(NO3)2 --> Pb^2+ + 2NO3^-
I........0.650M........0.......0
C.......-0.650.......0.650..0.650
E........0...........0.650..0.650
Ksp = (Pb^2+)(Cl^-)^2
Substitute Ksp value.
(Pb^2+) = x from PbCl2 + 0.650 from Pb(NO3)2
(Cl^-) = x
Solve for x.
The question asked for grams PbCl2.
x is in M, then mols = M x L and mols x molar mass = grams
@DrBob222 I understood up to the substitution part, can you explain that please?
To find the number of grams of PbCl2 that will dissolve in the given solution, we need to understand the concept of solubility.
Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature and pressure. It is usually expressed in grams of solute per liter of solvent (g/L).
To determine the solubility of PbCl2, we need to refer to solubility data or solubility product constant (Ksp) values. The Ksp is an equilibrium constant for the dissolution of a sparingly soluble compound.
However, the given question doesn't provide the solubility data or the Ksp value. Therefore, it is not possible to calculate the exact number of grams of PbCl2 that will dissolve in the solution without this information.
If you have access to the solubility data or Ksp value of PbCl2, I can assist you in calculating the solubility and the number of grams that will dissolve in the given solution.