[JohnCHM]

I'm new to this forum and have a question.I'm not looking for the answer but I need help in getting started with this problem. I just dont know where to start. Any way Here is the question.

A Pb2+ ion selective electrode is employed to deterimine lead concentration downstream from a mine. A 50mL sample is collected and returns a signal of -0.3200V. A second sample is spiked with 5mL of 6x10^-2M leadperchlorate and a signal of -0.2800V is observed. What is the concentration of Pb2+ in the sample?

The first thing I did was multiply (0.005L)(6x10^-2M leadperchlorate) = 3x10^-4 mole leadperchlorate than I dont know what to do?? Can some one give me some pointers

[JohnCHM]

I think I figured it out. Let me know if I'm doing this right.

pPb = - log[pb2+] = (E'cell - K) / (0.0592/2)

-log (50mL)[Pb2+] + (5mL)(6x10^-2M)  /  (50mL + 5 mL) = -(E"cell - K) / (0.0592/2)

-log 0.9091[Pb2+] + 0.0055 = --(E"cell - K) / (0.0592/2)

-log [Pb+]  / (0.9091[Pb2+] + 0.0055) = 2(E"cell - E'cell) / 0.0592

-log{ [Pb+]  / (0.9091[Pb2+] + 0.0055)} = 2(-0.2800V -(-.3200V)) / 0.0592 = 1.351

[Pb2+] / (0.9091[PB2+] + 0.0055) = 10^-1.351 = 0.0446

[Pb2+] = 0.0405[Pb2+] + 2.45x10^-4

[Pb2+] = 2.55x10^-4M

[Borek]

Please clarify what you are doing - what is K, why are you adding log of concentration to concentration and so on. Or did you eat some parentheses and you hope others will guess which ones?

Similar question was posted not so long ago:

http://www.chemicalforums.com/index.php?board=2;action=display;threadid=5945;start=msg26393#msg=26428

in 8^th message you will find equation that you should use. Perhaps you even used it, hard to tell

[JohnCHM]

Sorry, I've figured out the problem, so if you want Borek you can delete this post. I just threw something up really quick and didnt realize it was so messy. Thanks for responding though

[Borek]

Let it hang for the moment.

Post your final solution - it will be sort of guidance for others.