Analiticke reakcije

Submitted by Vlada92 on Sat, 09/23/2017 - 10:57
Vlada92's picture
Forums: 
Opšta diskusija

Izracunati pH vrednost rastvora dobijenog razblazivanjem 2,50 cm3 vodenog rastvora amonijaka, masenog udela 33% do 250cm3. 

Resenje zadatka pH= 11,08

milicaradenkovic's picture

milicaradenkovic - Sat, 09/23/2017 - 11:36

NH4OH------------->NH4 + +OH-

(OH-)+(koren(Kb*c))

c=n/V=ms/Mr*V=mr*w/Mr*V

mr=ro*V=0.89*2.5=2.225 g NH3

c=2.225*0.33/35*0.25=0.084 mol/l NH3

(OH-)=(koren(1.8*10(-5)*0.084))=0.001229 mol/l

pOH= -log(OH-)=2.92 a pH=14-pOH=14-2.92=11.08