Analiticka hemija

Submitted by Sofija333 on Wed, 08/30/2017 - 09:02
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100 cm3 zasicenog rastvora barijum-fluorida upareno je do suva. Masa zaostalog taloga iznosi 0,1138g. Izracunati proizvod rastvorljivosti barijum-fluorida.

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milicaradenkovic - Mon, 09/04/2017 - 13:49

m(taolga)=0.1138 g

BaF2-------------->Ba2+ + 2F-

(Ba2+)=(F-)=R

Ksp=(Ba2+)*(2*F-)2=4*R3

n=m(taloga)/Mr=0.1138/175=0.00065 mol

c=n/V=0.00065/0.1=0.0065 mol/l

Ksp=4*(0.0065)3=1.1*10(-6) mol3/dm9 BaF2

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Sofija333 - Tue, 09/05/2017 - 19:00

Hvala Milice :))