Information Set 4:
Consider the reaction between 500 mL of 0.100 mol L–1 Pb(NO3)2(aq) and 500 mL of 0.200 mol L–1Na2CrO4(aq). 

Ksp for PbCrO4 (s) is 2.8 x 10–13.

What volume of 0.100 mol L–1 lead nitrate solution would give the necessary concentration of lead ion to start the formation of a precipitate?

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  1. I think the question is worded wrong because 500 mL of Pb(NO3)2 and 500 mL Na2CrO4 will ppt PbCrO4 big time. I think the problem is asking for volume of 0.1M Pb(NO3)2 needed to start pptn of PbCrO4 in 500 mL of 0.2M Na2CrO4.l
    If you start with 0.2M Na2CrO4, then (CrO4^2-) = 0.2 M so
    (Pb^2+)(CrO4^2-) = Ksp
    Solve for (Pb^2+) = about 1E-12M.
    Then (0.1M Pb^2+ x ?L = approx 1E-12 so ?L is a very small number. Technically one would need to add the volumes in because the volume added of Pb(NO3)2 would dilute that 0.2M Na2CrO4; however, since the volume of Pb ion needed is so small we can forget about the added volume acting as a diluent.

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