Asked by Anonymous
How many grams ofPbCl2will dissolve in 2.00 L of O.650M aqueous Pb(N03)2 solution?
Answers
Answered by
DrBob222
........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.
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.
Answered by
DrBob222
The question asked for grams PbCl2.
x is in M, then mols = M x L and mols x molar mass = grams
x is in M, then mols = M x L and mols x molar mass = grams
Answered by
Anonymous
@DrBob222 I understood up to the substitution part, can you explain that please?
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