# If I know, say, the mass of precipitated PbCrO4 when I mixed PbCl2 solution (100mL exact) and K2CrO4 (excess), how would I go about calculating the original concentrations of Pb2+ and Cl- in the original PbCl2 solution?

Asked by HeroicZach (195) May 9th, 2009

Again, all I know is the mass of precipitated PbCrO4…this isn’t good that I can’t do this, eh?

Here’s how I tried to solve it:

I used the molar mass of PbCrO4 to determine that at the end I had .0022mol of Pb2+ ions. I said this was the original concentration in solution, but I know that is wrong, because not all Pb2+ precipitated (this is one reaction that has a Ksp, that much I know, and that Ksp is not overwhelmingly product-favored).

Can you help me get to where I need to be? Thank you!

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Are you certain that your answer is incorrect? Ksp=[Pb][CrO4], and as you add K2CrO4, you are increasing the concentration of CrO4 in solution, and the equilibrium position must shift such that the amount of Pb and CrO4 decreases. This is accomplished by precipitating PbCrO4. It will continue to shift as you add more, and the fact that K2CrO4 is in excess implies that it is added until things stop changing; this occurs when essentially all of the Pb in solution has been precipitated. Thus, there should have been .0022mol Pb ions in solution, making for a concentration of .022M of Pb2+ and .044M Cl-. But of course I could very well be wrong. I would be very interested in knowing, given that I take my chemistry AP test on Tuesday.

Jayne (6751)

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