The Ml of 0.10 Pb(NO3)2 that are required to react with 75 Ml of 0.20M Nai is 75 ml
calculation
Pb(NO3)2 +2Nai → Pbi2 + 2NaNo3
find the number of moles of Nai used
moles= molarity x volume in liters
volume in liters = 75/1000=0.075 moles
mole =0.20 x0.075 = 0.015 moles
by use of mole ratio between Pb(NO3)2 to Nai which is 1 :2 the moles of Pb(NO3)2 = 0.015 x1/2 =7.5 x10^-3 moles
volume (ml) )= moles/molarity x 1000
=volume = (7.5 x10^-3) / 0.10 = 75 Ml
