Zadatak Mesanje pri cemu dolazi do hemijske reakcije

Submitted by nobody on Sat, 06/14/2014 - 14:34
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  Pomešaju se 50,0 cm3 rastvora HCl masene koncentracije 182,3 mg/cm3, 25,0 cm3 25,1 % rastvora NaOH gustine 1,275 g/cm3 i 25,0 cm3 20,0 % rastvora NaCl, gustine 1,148 g/cm3. Kolika je koncentracija rastvora u mol/dm3 s obzirom na sve komponente?

(c(Cl−) = 3,48 mol/dm3; c(H3O+) = 0,50 mol/dm3; c(Na+) = 2,98 mol/dm3)

Treba mi pomoc, hvala unapred.

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milicaradenkovic - Sat, 06/14/2014 - 15:34

HCl+NaOH------------------->NaCl+H2O

n(Na+)=n(NaCl)+n(NaOH)

n(NaOH)=ms/Mr=w*mr/Mr=w*ro*V/Mr=0.251*0.025*1275/40=0.2 mol

n(NaCl)=w*ro*V/Mr=0.025*0.20*1148/58.5=0.098 mol

n(Na+)=0.098+0.2=0.298 mol

Vu=V1+V2+V3=50+25+25=100 cm3

c(Na+)=n(Na+)/Vu=0.298/0.1=2.98 mol/dm3

n(Cl-)=n(NaCl)+n(HCl)=0.25+0.098=0.348 mol

n(HCl)=(y/Mr)*V=(182.3/36.5)*0.05=0.25 mol

c(Cl-)=n(Cl-)/Vu=0.348/0.1=3.48 mol/dm3

c(H+)=n(H+)/Vu

n(H+)=n2(HCl) jer je u visku HCl

n(H+)=n(HCl)-n(NaOH)=0.25-0.2=0.05 mol

c(H+)=0.05/0.1=0.5 mol/dm3

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nobody - Sat, 06/14/2014 - 15:56

Hvala...