Question

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) when srso4(s) begins to precipitate

Answer 1

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