Hidroliza

oxygenius2's picture

Izračunati masu natrijum-formijata, HCOONa, koja je rastvorena u 4,60 dm3 vode, ako se zna da Ph vrednost dobijenog rastvora iznosi 8,00. Ka(HCOOH)=1.7*10-4 mol/dm3

milicaradenkovic's picture

HCOOH--------------------->HCOO- + H+

Ka={HCOO-}*{H+}/{HCOOH}

pH=8 pa je {H+}=1*10(-8) mol/l

Kb=Kw/Ka=1*10(-14)/1.7*10(-4)=5.88*10(-11)

HCOO-+H2O---------------->HCOOH+ + OH-

Kb={HCOOH}*{OH-}/{HCOO-}

pOH=6 pa je {OH-]=1*10(-6) mol/l

{HCOOH}={OH-}=1*10(-6) mol/l

5.88*10(11)=(1*10(-6))2/(HCOO-)

(HCOO-)=1*10(-12)/5.88*10(-11)=0.017 mol/l

n(HCOO-)=(HCOO-)*V=0.017*4.60=0.0782 mol

n(HCOOONa)=n(HCOO-)=0.0782 mol

m(HCOONa)=n(HCOONa)*Mr=0.0782*58=5.318 g