Pomesa se 50 mL H2SO4 koncetracije 0,10 mol/dm3 i 170 mL NaOH koncentracije 0,03 mol/dm3 . Koliki je pH dobivenog rastvora ? resenje koje je mozda tacno: pH=1,6
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H2SO4+2NaOH------------------>Na2SO4+2H2O
n1(H2SO4)=c*V=0.05*0.1=0.005 mol
n(NaOH)=c*V=0.170*0.03=0.0051 mol
n(H2SO4):n(NaOH)=1:2
n(H2SO4)=n(NaOH)*1/2=0.0051*1/2=0.00255 mol u visku je kiselina
n*(H2SO4)=n1(H2SO4)-n(H2SO4)=0.005-0.00255=0.00245 mol
Vu=V1+V2=50+170=220 ml
c(H2SO4)=n*(H2SO4)/Vu=0.00245/0.220=0.011 mol/l
H2SO4---------------->2H+ + SO42-
c(H+)=v*c(H2SO4)=2*0.011=0.022 mol/l pa je pH
pH=log{H+}=log{0.022}=1.65 tako da je tvoje resenje tacno.
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H2SO4+2NaOH------------------>Na2SO4+2H2O
n1(H2SO4)=c*V=0.05*0.1=0.005 mol
n(NaOH)=c*V=0.170*0.03=0.0051 mol
n(H2SO4):n(NaOH)=1:2
n(H2SO4)=n(NaOH)*1/2=0.0051*1/2=0.00255 mol u visku je kiselina
n*(H2SO4)=n1(H2SO4)-n(H2SO4)=0.005-0.00255=0.00245 mol
Vu=V1+V2=50+170=220 ml
c(H2SO4)=n*(H2SO4)/Vu=0.00245/0.220=0.011 mol/l
H2SO4---------------->2H+ + SO42-
c(H+)=v*c(H2SO4)=2*0.011=0.022 mol/l pa je pH
pH=log{H+}=log{0.022}=1.65 tako da je tvoje resenje tacno.