Asked by Tara
the molar solubility of PbS, in a 0.150 M ammonium sulfide (NH4)2S solution is? The Ksp for PbS= 3.2E-28.
Answers
Answered by
DrBob222
...........PbS ==> Pb^+2 + S^-2
initial.....0.......0......0
change......x.......+x.....+x
final.......x........x......x
Ksp = (Pb^+2)(S^-2) = 3.2E-28
..............(NH4)2S ==> 2NH4^+ + S^-2
initial.......0.15M........0........0
change.........-0.15......0.30......0.15
final...........0..........0.30.....0.15
Substitute into Ksp and solve for PbS.
We want solubility PbS; therefore, we want to solve for x.
PbS = x
Pb^+2 = x
S^-2 = x+0.15
3.2E-28 = (x)(x+0.15).
Solve for x.
initial.....0.......0......0
change......x.......+x.....+x
final.......x........x......x
Ksp = (Pb^+2)(S^-2) = 3.2E-28
..............(NH4)2S ==> 2NH4^+ + S^-2
initial.......0.15M........0........0
change.........-0.15......0.30......0.15
final...........0..........0.30.....0.15
Substitute into Ksp and solve for PbS.
We want solubility PbS; therefore, we want to solve for x.
PbS = x
Pb^+2 = x
S^-2 = x+0.15
3.2E-28 = (x)(x+0.15).
Solve for x.
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