Question
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?
Answers
bobpursley
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
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