Question

Calculate the number of milliliters of 0.639 M Ba(OH)2 required to precipitate all of the Pb2+ ions in 148 mL of 0.798 M Pb(NO3)2 solution as Pb(OH)2. The equation for the reaction is: Pb(NO3)2(aq) + Ba(OH)2(aq) Pb(OH)2(s) + Ba(NO3)2(aq) mL Ba(OH)2

Answer 1

184,8 mL