Asked by Jematormal91
A solution contains 0.0200 M Pb2 (aq) and 0.0200 M Sr2 (aq). If we add SO42–(aq), what will be the concentration of Pb2 (aq) when SrSO4(s) begins to precipitate?
Answers
Answered by
DrBob222
Ksp = (Pb^2+)(SO4^2-) = ?
Ksp = (Sr^2+)(SO4^2-) = ?
Substitute 0.02M for Pb^2+ and solve for SO4^2-. That will be the sulfate concn when PbSO4 starts to ppt since the Ksp for PbSO4 is smaller than that of SrSO4. The problem doesn't ask for that but it's nice to know.
Substitute 0.02M for Sr^2+ into that Ksp and solve for SO4^2-. That will be the sulfate concn when SrSO4 just begins to ppt. Take that sulfate concn, plug it into Ksp for PbSO4 and solve for Pb. That is the Pb^2+ concn when SrSO4 begins to ppt.
Ksp = (Sr^2+)(SO4^2-) = ?
Substitute 0.02M for Pb^2+ and solve for SO4^2-. That will be the sulfate concn when PbSO4 starts to ppt since the Ksp for PbSO4 is smaller than that of SrSO4. The problem doesn't ask for that but it's nice to know.
Substitute 0.02M for Sr^2+ into that Ksp and solve for SO4^2-. That will be the sulfate concn when SrSO4 just begins to ppt. Take that sulfate concn, plug it into Ksp for PbSO4 and solve for Pb. That is the Pb^2+ concn when SrSO4 begins to ppt.
Answered by
Jematormal91
Thank you so much!
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