I enjoyed chemistry in college and have always wanted an excuse to try it at home. A relative is currently very worried about lead from various sources. I think this is a great excuse to get back into chemistry. I was looking for a procedure that could determine the quantity of lead in a water sample. I was not able to find much on the topic but I did find this. https://chemistry-dictionary.yallascience.com/2017/05/determination-of-lead-by-edta-titration.html

My only concern with this procedure is I'm not sure what concentrations are required for a positive result, and I do not know if the test is invalidated by the presence of other metals in the solution.

Obviously "ship it off to a lab" is not the answer here, as testing for lead is only an excuse to buy some glassware. Procedures that require expensive equipment, or require a high quality fume extraction system are obviously not viable, though I can't imagine testing for lead is that complex considering you can buy test strips from the hardware store.

Edit: Mauricehas confirmed my suspicion that the included procedure will not work for the above reasons. I am still interested if anyone knows of a procedure that works in the ppm range.


1 Answer 1


The mentioned reference is valid for titrating lead solutions which don't contain other ions. If the $\ce{Pb^{2+}}$ solution contains also some other ions, like $\ce{Ca^{2+}}$ or $\ce{Mg^{2+}}$, the procedure will titrate the total amount of these ions. Furthermore, it is valid for concentrations of the order of $50$ to $1000$ mg/L $\ce{Pb}$ if the reference volume to be measured is $20$ mL. If the solutions you want to titrate contains less $\ce{Pb^{2+}}$, this amount is too small to be determined by the given method.

  • 1
    $\begingroup$ Thank you, I suspected as much. I will edit the question to ask directly for a new procedure. $\endgroup$ Feb 12, 2023 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.