Jonske ravnoteze

Marijana maya's picture

Izracunati pH  rastvora na 20 stepeni C dobijenog mesanjem istih zapremina rastvora HF ciji je pH-2,00 odnosno pH je 3,00.Ka(HF)je 6,7*10-4.

Marijana maya's picture

Resenje je 2,15

milicaradenkovic's picture

HF----------------->H+ + F-

Ka={H+}1*{F-}1/{HF}1

{H+}1={F-}1=0.01 mol/l jer je pH=2

[HF}1=0.149 mol/l

{H+}2={F-}2=0.001 mol/l jer je pH3

Ka+{H+}2*{F-}2/{HF}2

pa je {HF}2=0.00149 mol/l 

{HF}u.={HF}1+{HF}2=0.149+0.00149=0.15049 mol/l

{HF]u.={HF}u./2 jer imas 2 zapremine

{HF}u.=0.15049/2=0.0752 mol/l

Ka={H+}*{F-}/{HF}u.

{H+}={F-}=x

6.7*10(-4)=x*2/0.0752

x*2=5.038*10(-5)

x=(koren(5.038*10(-5))=0.0071

{H+}=0.0071 mol/l

pH= -log{H+}= -log{0.0071}=2.15