pH rastvora

Submitted by MARIJA 84 on Mon, 01/27/2014 - 10:14
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Izracunati pH rastvora dobijenog mesanjem 50ml sumporne kiseline koncentracije 0,2mol/L i 50ml natrijum-hidroksida konc.0,3mol/L.

H2SO4+2NaOH=Na2SO4+2H20

N H2SO4=0.05*0.2=0.01mol

nNaOH=0.05*0.3=0.015mol  u vusku je baza

sledi da reaguje 0,0075mol H2SO4

Vuk=0,05+0.05=0.1l

C kiseline=0.0075/0.1=0.075 mol/L

pH=-log0.075=1.13

jel ovo tacno?

milicaradenkovic's picture

milicaradenkovic - Mon, 01/27/2014 - 10:29

Nije.

H2SO4+2NaOH---------------->Na2SO4+2H2O

n(baze)=c*V=0.05*0.3=0.015 mol

n(kiseline)=c*V=0.05*0.2=0.01 mol

u visku je kiselina

n1(kiseline)=n(baze)/2=0.015/2=0.0075 mol

n2(kiseline)=n(kiseline)-n1(kiseline)=0.01-0.0075=0.0025 mol

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

c(kiseline)=n2(kiseline)/Vu=0.0025/0.1=0.025 mol/l

H2SO4------------------->2H+ + SO42-

{H+}=2*c(kiseline)=2*0.025=0.05 mol/l

pH= -log{H+}=1.3

Simalux's picture

Simalux - Sun, 09/12/2021 - 19:31

Kako može biti kiselina u višku kada baza ima više molova?