Calculate the mass of Pb3(PO4)2 (Ksp = 1.00×10-54) which will dissolve in 100 ml of water.
As the reaction progresses, due to the stoichiometry of the
reaction, +3s of [Pb2^+] and +2s of [PO4^3-] will be made.
Ksp = 1.00x10^-54 = [Pb^2+]^3 [PO4^3-]^2 = [3s]^3 [2s]^2 =
s = 6.21x10^-12 M
b) (6.21x10^-12 mol Pb3(PO4)2 / 1 L) x (811.54 g Pb3(PO4)2/ 1
mol Pb3(PO4)2) = 5.04x10^-9 g Pb3(PO4)2 / 1 L
c) Same process as parts a) and b), but with a different
PbCrO4 <--> Pb^(2+) + CrO4^(2-)
Plumbous ion concentration is increased by s, chromate ion
concentration is increased by s.
2.00x10^-16 = s^2
s = 1.41x10^-8 mol/ L
Convert moles to grams to find solubility in terms of grams,
then convert liters to mL in order to find the mass of PbCrO4 that will
dissolve in 100 mL.
(1.41x10^-8 mol/ L) x (323.2 g / mol) x (1 L / 1000 mL) x
(100 mL) = 4.56x10^-7 g PbCrO4
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