rstvor

Tina Trezic's picture

u 100 cm3 rastvora NaOH(c=0,1 mol/dm3) dodato je 50 cm3 rastvora Hcl(c=0,5 mol/dm3) izracunati pH vrednost rastvora.resenje je 1,00

milicaradenkovic's picture

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

n(baze)=c*V=0.1*0.1=0.01 mol

n(kiseline)=c*V=0.5*0.05=0.025 mol u visku je kiselina

n2(kiseline)=n(kiseline)-n(baze)=0.025-0.01=0.015 mol

Vu=V1+V2=100+50=150 ml

c(kiseline)=n2(kiseline)/Vu=.015/0.15=0.1 mol/l

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

{H+}=c{kiseline}=0.1 mol/l

pH= -log{H+}=1