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Submitted by Milica i hemija on Fri, 01/03/2014 - 16:33
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Koliko cm3 rastvora KOH koncentracije 0.2 mol/l treba dodati u 200cm3 rastvora CH3COOH masene koncentracije 60 g/l da bi se dobio pufer cije je PH= 4? Kk = 1,8* 10-5

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milicaradenkovic - Fri, 01/03/2014 - 17:49

c(CH3COOH)=y/Mr=60/60=1 mol/l

CH3COOH+KOH-------------->CH3COOK+H2O

CH3COOK------------->CH3COO- + K+

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

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

n(CH3COOH)=c*V=0.2*1=0.2 mol

1.8*10(-5)=1*10(-4)*n(CH3COO-)/0.2

pH=4 pa je {H+}=1*10(-4) mol/l

n(CH3COO-)=0.036 mol

n(CH3COOK)=n(CH3COO-)

n(KOH)=n(CH3COOK)=0.036 mol

V(KOH)=n(KOH)/{KOH}=0.036/0.2=0.18 l=180 ml. Nisam sigurna za resenje pa ti proveri.