Homework Help: chemistry

Posted by nahang on Thursday, April 14, 2011 at 5:55am.

A solution is 2.0×10−2M in both chromate and sulfate anions and . To this solution, 0.55M Pb(NO)3 is slowly added.

What is [Pb2+] at the point at which the second anion begins to precipitate (in M)? The second anion being the, sulfate anion.

chemistry - DrBob222, Thursday, April 14, 2011 at 4:50pm
We need to think our way through this problem.
............Pb^2+ + CrO42- ==>PbCrO4
initial......0........0.02.........
So when (Pb2+)(CrO42-) = 1.8E-14, PbCrO4 will start to ppt. Therefore, (Pb2+) = 1.8E-14/(0.02) = 9.0E-14 when PbCrO4 FIRST begins to ppt. When all of the PbCrO4 has pptd, the Pb2+ = 1.8E-14. We continue adding Pb(NO3)2 and the PbSO4 begins to ppt when (Pb2+)(SO42-) = 1.6E-8. Since sulfate = 0.02M, then (Pb2+) = (Ksp/0.02) = ?? when the first molecule of PbSO4 forms.
Check my thinking.
chemistry - nahang, Thursday, April 14, 2011 at 9:54pm
Yup that's correct! I was just the wrong Ksp value! Wasted an hour on this question just because of a silly mistake (N)

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