Question

A chemist combines 53.7 mL of 0.260 M sodium chloride with 21.8 mL of 0.510 M lead nitrate. (a) How many grams of lead chloride will precipitate? (b) What is the final molarity of the Na+ ion? (c) What is the final molarity of the lead or chloride ion, whichever one is in excess?

Answer 1

Pb(NO3)2 (aq) + 2NaCl(aq) -----------------> PbCl2(s) + 2NaNO3(aq)

no of moles of NaCl = molarity * volume in L

                                 = 0.26*0.0537   = 0.014moles

no of moles of Pb(NO3)2 = molarity * volume in L

                                          = 0.51*0.0218 = 0.01112 moles

2 moles of NaCl react with 1 mole of Pb(NO3)2

0.014 moles of NaCl react with = 0.014*1/2   = 0.007 moles of Pb(NO3)2 is required.

Pb(NO3)2 is excess reactant

a.2 moles of NaCl react with Pb(NO3)2 to gives 1 moles of PbCl2

0.014 moles of NaCl react with Pb(NO3)2 to gives = 0.014*1/2     = 0.007 moles of PbCl2

mass of PbCl2 = no of moles * gram molar mass

                           = 0.007*279   = 1.953g

b. Total volume of solution = 53.7 + 21.8   = 75.5ml = 0.0755

   molarity of NaCl   = no of moles of NaCl/ total volume in L      = 0.014/0.0755   = 0.185 M

NaCl ----------> Na^+ (aq) + Cl^- (aq)

0.185M                0.185M       0.185M

molarity of Na^+ = 0.185M

c. molarity of PbCl2 = no of moles of PbCl2/total volume in L

                                = 0.007/0.0755    = 0.0927M

pbCl2 --------------> Pb^2+ (aq) + 2Cl^-

0.0927M                0.0927M       2*0.0927M

molarity of Pb^2+   = 0.0927M

molarity of Cl^-     = 2*0.0927M   = 0.1854M