Which of the following compounds will be more soluble in acidic solution than in pure water?
PbCl2
FeS
Ca(ClO4)2
CuI
METHOD:
To determine the solubility in Acidic Solution for PbCl2 in variable form. No Ksp given. ASSUME: All the above are bases.
PbCl2–>Pb+2 + 2Cl- (generally show the dissociation of PbCl2 in an acid solution that’s capable)
use this 2Cl- since this is essentially the Anion(electron being donated by the base PbCl2 by Lewis Acid rule)
2Cl- + H+ –> HCl (use the H+ because this indicative of an acidic solution, if basic soln use OH-)
ICE Table in Variable Form
Find solubility product constant=ksp, then find actual values of ksp in text/online
Find Solubility of the compounds: s
WORK:
PbCl2: ksp=4s^3.
Looked in textbook ksp for PbCl2 to be 1.1710^-5
solve for s ====> s=[solubility of PbCl2]= 3.4210^-3
FeS: ksp=s^2
where ksp= 3.7210-19
==> s=[Solubility of FeS] = 4.5310^-7
CuI: ksp= s^2
where ksp = 1.2710^-12
s=[solubility of CuI] = 1.1310^-6
SUMMARY:
Solubilities: [PbCl2]= 3.4210^-3 [FeS]=4.5310^-7 [CuI]=1.13*10^-6 (only did these three)
Most soluble would the be PbCl2. Correct? Is this a possible method?
Answer:
B) Fes
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