A student finds that 22.06 mL of 0.7270 M lead nitrate is needed to precipitate all of the iodide ion in a 50.00-mL sample of an unknown. What is the molarity of the iodide ion in the student's unknown?

Answer 1

balance the equation I assume Lead(II) nitrate.

Pb(NO3)2 + 2I->>PbI2 + 2NO3-

so you need one mole of lead nitrate for each two moles of iodide ions.

molaratyleadnitrate*volume=2*molarity*50ml

molarity=.7270*22.06/100