What is the molar concentration of I– in 1 liter of a saturated water solution of PbI2 if the Ksp of lead iodide is 1.4 × 10–8?
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Ksp = 1.4x10^-8 = [Pb 2+][I-]^2
if we look at an ICE chart, we see the following:
[PbI2] .....[Pb 2+].....[2I-]
some.........0..............0
less...........+x.........+2x
less.........x..............2x
1.4x10^-8 = (x)(2x)^2 = 4x^3
3.5x10^-9 = x^3
x = 0.00152M = [Pb 2+] [I-] = 0.00304M
if we look at an ICE chart, we see the following:
[PbI2] .....[Pb 2+].....[2I-]
some.........0..............0
less...........+x.........+2x
less.........x..............2x
1.4x10^-8 = (x)(2x)^2 = 4x^3
3.5x10^-9 = x^3
x = 0.00152M = [Pb 2+] [I-] = 0.00304M