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A solution contains 0.0100 M Pb2 (aq) and 0.0100 M Sr2 (aq). If we add SO42–(aq),...

A solution contains 0.0100 M Pb2 (aq) and 0.0100 M Sr2 (aq). If we add SO42–(aq), what will be the concentration of Pb2 (aq) when SrSO4(s) begins to precipitate?

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Answer #1

Ksp of SrSO4 = 2.8*10^-7

Ksp of PbSO4   = 1.8*10^-8

SrSO4(s) ----------> Sr^2+ (aq) + SO4^2- (aq)

Ksp   = [Sr^2+][SO4^2-]

2.8*10^-7   = 0.01[SO4^2-]

[SO4^2-]    = 2.8*10^-7/0.01   = 2.8*10^-5 The value is just slightly greater than this then SrSO4 will precipitate
Ksp   = [Pb^2+][SO4^2-]

1.8*10^-8   = [Pb^2+]*2.8*10^-5

[Pb^2+]       = 1.8*10^-8/2.8*10^-5   = 6.43*10^-4M





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