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Vesna [10]
3 years ago
13

A solution contains 0.0150 m pb2 (aq) and 0.0150 m sr2 (aq). if we add so42–(aq), what will be the concentration of pb2 (aq) whe

n srso4(s) begins to precipitate
Chemistry
1 answer:
polet [3.4K]3 years ago
3 0
Ksp of SrSO4 = 3.2 x 10-7, so it will precipitate when Ksp > 3.2 x 10-7  
Same way, calculating for 0.0150 M Sr2 SrSO4 precipitates,
 when [ SO4 2-] > (3.2 x 10-7 / 0.0150) = 2.133 x 10^-05.
 Thus SrSO4 precipitates when SO4 2- > 2.133 x 10^-05 
 Now the Pb2+ at this point = 2.133 x 10^-05 x 0.0150 =3.2 x 10^-07 
 Hence Ksp for PbSO4 = 1.6 x 10-8 
 the solubility product for PbSO4 is more and the PbSO4 precipitates as
follows [Pb2+] = 1.6 x 10-8 / 2.133 x 10^-05 = 7.50 x 10^-04

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9. If the concentration of H2O (a reactant) is decreased and that of CO (a product) is increased, both actions lead to the equilibrium being shifted to the left.

10. Addition of catalyst to the system will only speed up the rate at which the system reach the equilibrium.

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