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Autor: Jovana1996, Uto, 01/10/2013 - 23:00.

Odrediti pH u rastvoru dobijenom mesanjem 100 ml rastvora HCl (w=1%,ro=0,98g/ml) i 200 ml rastvora HNO3 (w=2% ro=1,2 g/ml)

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Autor: micaloknica, Sre, 02/10/2013 - 08:48.

Ja mislim da se zadatak radi na sledeci nacin mada nisam sigurna

HNO3-------------->H+ + NO3-

mr=ro*V=1.2*200=240 g

ms=mr*w=240*0.02=4.8 g a n(HNO3)=ms/Mr=4.8/63=0.0762 mol

HCl----------------->H+ + Cl-

mr=ro*V=0.98*100=98 g

ms=w*mr=98*0.01=0.98 g a n(HCl)=ms/Mr=0.98/36.5=0.0268 mol

n(H)=n(HCl)+n(HNO3)=0.0762+0.0268=0.103 mol

Vu=V1+V2=100+200=300 ml

c(H+)=n(H+)/Vu=0.103/0.3=0.343 mol/l pa je pH

pH=- log(H+)=-log(0.343)=0.46