Question

1. If you started with 1.75g of lithium chloride and added that to 250. mL of a 0.150M solution of lead(II) acetate

   1. Write the reaction

   2. What would be the grams of precipitate produced

   3. Calculate the final molarity of the chloride ion

   4. Calculate the final molarity of the lead ion

Answer 1

a) 2Liclcag) + Pb(oAc)aca) Pbcla(5) + 2 LioAC (ag) b> moles of O.1SO O.2 SO mol e As peo equation, 2 rooles of Licl seacts coith i mole of pbfofAc)2. Given IS gLic 42 39 Lic mole Of of Lic mole 42.39 mole of Licl O. O428 o.S Dole pb(0Ac) mole of LicI musi eaci coith eacts coith (o.04128 xo S) mole mole Lic : o.0428 Pb(OAc) = o.0206A oole Pb(OAc) Gut heae o.0375 roole pbCoAC)a is used and Lic is liroiting deagen :. P6COAC)2 is eccess Lic gives i oole pbCla O.O4128x) . o o48 mole Lic) gives Dole of Pb cla = o.02064 mo1e pbcl2 :. o.02064 mole Pbcla - 278. i X o.o2064 5.439g = 5.74 8 Grsam of ppt

Moles of PbGla = CODC. Of PbGl x vol of Soln in Lit' O. O2064 mole Conc. of Pbcla O. 250 L O.o825 ool /L PbclaJ Pb2+ aci PbGla C Pb2 0.0826 rool/L CJ [P = O.165 mol/L