What is the solubility ( in g/L) of lead(II) chromate, PbCrO4 in 0.13M potassium chromate, K2CrO4.

Expert Solution

PbCrO4 --------> Pb⁺² + CrO4^2-

0.13

s s

s 0.13+s

Ksp of PbCrO4 is 2.8*10^-13

Ksp = [Pb⁺²][CrO4^2-]

2.8*10^-13 = s * (0.13+s)

s = 2.15*10^-12 mole/L

solubility of PbCrO4 = 2.15*10^-12 moles/L

molar mass of PbCrO4 = 323g/mole

solubility of PbCrO4 = 2.15*10^-12 moles/L * 323g/moles

= 6.94*10^-10 g/L

queen_honey_blossom answered 10 months ago

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