Lead ions can be precipitated from aqueous solutions by the addition of aqueous iodide: Pb2+(aq) +...
Lead ions can be precipitated from aqueous solutions by the
addition of aqueous iodide:
Pb2+(aq) + 2 I-(aq) →
PbI2(s)
Lead iodide is virtually insoluble in water so that the reaction
appears to go to completion. How many milliliters of 3.550 M
HI(aq) must be added to a solution containing 0.100 mol of
Pb(NO3)2 to completely precipitate the
lead?
Homework Answers
Number of moles of Pb(NO3)2 is 0.100 moles
Number of moles of Pb2+=Number of moles of Pb(NO3)2
Number of moles of Pb2+=0.100 moles
Number of moles of I- required =2 x Number of moles of Pb2+
Number of moles of I- required = 2x 0.100= 0.200 moles
molarity = number of moles * volume in L
Volume of 3.550 M HI = Number of moles of I- / molarity
Volume of 3.550 M HI= 0.200/3.550 = 0.05634 L = 56.34 mL
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Lead ions can be precipitated from aqueous solutions by the
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