Proizvod rastvorljivosti

Submitted by amna0504 on Tue, 11/19/2013 - 20:37
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Za rastvaranje 1g PbJ2 treba 1470 ml vode. Koliki je proizvod rastvorljivosti?

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milicaradenkovic - Tue, 11/19/2013 - 20:56

n(PbI2)=m/Mr=1/461

n(PbI2)=0.00217 mol

0.00217 mol--------------------1.470 l

x mol-----------------------------1 l

x=0.00217*1/1.47=0.00148 mol/l

{Pb2+}={I-}=0.00148 mol/l

PbI2------------------->Pb2+ + 2I-

R={Pb2+}*{I-}*2

R={0.00148}*3=3.24*10(-9)

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milicaradenkovic - Tue, 11/19/2013 - 21:13

Izvini greska je u formuli

R={Pb2+}*{2*I-}*2

R={0.00148}*{2*0.00148}*2

R=1.3*10(-8)