Pufer

Submitted by banejeca943 on Sat, 01/04/2014 - 19:32
banejeca943's picture
Forums: 
Opšta diskusija

Koliko cm3 rastvora NaOH masene koncentracije 40 g/dm3 treba dodati u 100cm3 rastvora CH3COOH koncentracije 2 mol/dm3 da bi se dobio pufer cije je pH=5? Kk(CH3COOH)=1.8*10^-5 mol/dm3

milicaradenkovic's picture

milicaradenkovic - Sat, 01/04/2014 - 20:04

CH3COOH+NaOH---------------->CH3COONa+H2O

CH3COOH--------------->CH3COO- + H+

Kk=n{CH3COO-}*{H+}/n{CH3COOH}

c(NaOH)=y/Mr=40/40=1 mol/l

1.8*10(-5)=1*10(-5)*n(CH3COO-)/0.2

n(CH3COOH)=V*c=0.1*2=0.2 mol

n(CH3COO-)=0.36 mol

CH3COONa------------>CH3COO- + Na+

n(CH3COONa)=n(CH3COO-)=0.36 mol

n(NaOH)=n(CH3COONa)=0.36 mol

V(NaOH)=n(NaOH)/c(NaOH)=0.36/1=0.36 l=360 ml .Ne znam da li je tacan rezultat pa ti proveri.

banejeca943's picture

banejeca943 - Sun, 01/05/2014 - 12:38

Nije tacan, ovako sam i ja dobila. Tacan je 128.6 ml. U svakom slucaju, hvala.

milicaradenkovic's picture

milicaradenkovic - Sun, 01/05/2014 - 17:06

Greska je u kolicini kisleine

n(CH3COOH)=0.2-x

n(CH3COO-)=x

1.8*10(-5)=1*10(-5)*x/(0.2-x)

3.6*10(-6)-1.8*10(-5)*x=1*10(-5)*x

3.6*10(-6)=2.8*10(-5)*x

x=3.6*10(-6)/2.8*10(-5)=0.1286 mol

n(CH3COO-}=0.1286 mol

n(CH3COONa)=n(CH3COO-)=0.1286 mol

n(NaOH)=n(CH3COONa)=0.1286 mol

V(NaOH)=n(NaOH)/c(NaOH)=0.1286/1=0.1286 l=128.6 ml. Izvini na gresci.