This question can be interpreted two ways. I will assume you want to know how much NH4I solution must be added to react with ALL of the Pb(NO3)2?
Pb^2+ + 2I^- ==> PbI2
millimols Pb^2+ = mL x M = ?
millimols I^- necessary to react with all of the Pb^2+ is twice that.
Then mL NH4I needed = millimols/M = ?
What volume of a 0.430 M NH4I solution is required to react with 289 mL of a 0.560 M Pb(NO3)2 solution?
1 answer