For the following equilibrium, if the concentration of lead ion is 5.3×10−7 M, what is Ksp for lead (II) chromate:
PbCrO4(s)↽−−⇀Pb2+(aq)+CrO2−4(aq)
The expression of Ksp is:
Ksp = [Pb + 2] * [CrO4-2]
It has to:
[Pb + 2] = [CrO4-2]
Is calculated:
Ksp = (5.3 × 10-7) ^ 2 = 2.8x10 ^ -13
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For the following equilibrium, if the concentration of lead ion is 5.3×10−7 M, what is Ksp...