Zadatak:Ravnoteza u homogenim sistemima

Submitted by morissey on Wed, 12/18/2013 - 19:58
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Molim vas za pomoc :) Izracunati pH vrednost dobijenog mesanjem 25cm3 rastvora HCl koncentracije 0,5mol/dm3; 10cm3 rastvora NaOH koncentracije 0,5mol/cm3 i 15cm3 vode? 

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milicaradenkovic - Wed, 12/18/2013 - 20:18

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

n(kiseline)=c*V=0.025*0.5=0.0125 mol

n(baze)=c*V=0.5*0.01=0.005 mol u visku je kiselina

n1(kiseline)=n(baze)=0.005 mol

n2(kiseline)=n(kiseline)-n1(kiseline)=0.0125-0.005=0.0075 mol

Vu=V1+V2+V3=10+25+15=50 ml

c(kiseline)=n2(kiseline)/Vu=0.0075/0.050=0.15 mol/l

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

{H+}=c(kiseline)=0.15 mol/l

pH= -log{H+}= -log{0.15}=0.824

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morissey - Sat, 12/21/2013 - 19:12

Hvala :-)