Problem: Given that the ΔG f ° for Pb2+(aq) and Cl −(aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, Ksp, for PbCl2(s).

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FREE Expert Solution

We’re being asked to determine thedetermine the solubility product, Ksp, for PbCl2(s).


PbCl2(s)  Pb2+(aq) + 2 Cl (aq)


Recall that ΔG˚rxn and K are related to each other:


ΔG°rxn=-RTlnK


We can use the following equation to solve for ΔG˚rxn:

 ΔG˚rxn =  ΔG˚f, products - ΔG˚f, reactants



Step 1: We can easily solve for ΔG˚rxn


Given: 

ΔG f ° for Pb2+(aq) =  −24.3 kJ/mol

ΔG f ° for Cl (aq) = −131.2 kJ/mol

ΔG f ° for Cl (aq) = −314.2 kJ/mol


 ΔG˚rxn =  ΔG˚f, products - ΔG˚f, reactants ΔG˚rxn =  [ΔG˚f, Pb2+ +2 ΔG˚f, Cl- ] - [ΔG˚f, PbCl2]ΔG˚rxn =  [(1 mol)(24.3 kJ/mol) + (2 mol)(131.2 kJ/mol) ]                                 - [ (1 mol)(314.2 kJ/mol)]

ΔG˚rxn = 27.5 kJ


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Problem Details

Given that the ΔG f ° for Pb2+(aq) and Cl (aq) is −24.3 kJ/mole and −131.2 kJ/mole respectively, determine the solubility product, Ksp, for PbCl2(s).

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Our data indicates that this problem or a close variation was asked in Chemistry - OpenStax 2015th Edition. You can also practice Chemistry - OpenStax 2015th Edition practice problems.