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Submitted by Vuk Biljuric on Sat, 05/03/2014 - 02:10
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Odredi molarnu masu neke dvokisele baze ako 0.514g cistog uzorka te baze neutralise 30cm3 rastvora neke jednobazne kiseline koncetracije 0.2mol/dm3.

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milicaradenkovic - Sat, 05/03/2014 - 09:01

M(OH)2+2HX------------------------->MX2+2H2O

n(HX)=c*V=0.03*0.2=0.006 mol

n(M(OH)2)=n(HX)/2=0.006/2=0.003 mol

Mr=m/n(M(OH)2)=0.514/0.003=171 g/mol

Mr=Ar(M)+2*Ar(O)+2*Ar(H)

171=M+2*16+2*1

M=171-32-2=137 g/mol i to je barijum i formula je Ba(OH)2