Question
A solution with a volume of 1·00 dm3 is saturated with lead iodide, Pbl2. The concentration of iodide ions is 2·7 mol dm–3.Determine the solubility product of PbI2.
Answers
This is a common ion problem.
.....PbI2 ==> Pb^2+ + 2I^-
I....solid....0.......0
C....solid....x.......2x
E....solid....x.......2
Ksp = (Pb^2+)(I^-)^2
Ksp = ?
(I^-_ is 2.7 M.
(Pb^2+) = 1/2 (I^-)
Substitute and solve for Ksp.
.....PbI2 ==> Pb^2+ + 2I^-
I....solid....0.......0
C....solid....x.......2x
E....solid....x.......2
Ksp = (Pb^2+)(I^-)^2
Ksp = ?
(I^-_ is 2.7 M.
(Pb^2+) = 1/2 (I^-)
Substitute and solve for Ksp.
Related Questions
The concentration of iodide ions in a saturated solution of lead(II) iodide is __________ M. The sol...
20.00 ml of 0.100F Pb(No3)2 (FW 331.2) are mixed with 15.00 ml of 0.125 F Nal (FW 149.9). Lead (II)...
At 25C, the concentration of lead(II) sulfate in a saturated aqueous solution is 7.9 x 10^-M. Determ...
One way to remove lead ion from water is to add a source of iodide ion so that lead iodide will prec...