Maseni odnos gubitaka

minion007's picture

Izracunati odnos masenih gubitaka Ag2CO3 pri ispiranju taloga sa 100cm3 destilovane vode i sa 100cm3 rastvora Na2CO3 c(1/2 Na2CO3)=0,1 mol/dm3

minion007's picture

?

milicaradenkovic's picture

Ne znam  sta znaci 1/2 Na2CO3 u zadatku?

minion007's picture

to znaci da je c(Na2CO3) =0.1x2

milicaradenkovic's picture

Ksp(Ag2CO3)=8.2*10(-12)

Ag2CO3----------------->2Ag+ + CO32-

Ksp=(Ag+)2*(CO32-)

(Ag+)=2*R

(CO32-)=R

Ksp=4*R3

R=(3koren(Ksp/4))=1.4*10(-4) mol/l

(CO32-)=1.4*10(-4) mol/l

(Ag2CO3)=(CO32-)=1.4*10(-4) mol/l

n(Ag2CO3)=(Ag2CO3)*V=1.4*10(-4)*0.1=1.4*10(-5) mol

m(Ag2CO3)=n(A2CO3)*Mr=276*1.4*10(-5)=0.00386 g

(Ag+)=Ksp/(CO32-)

(CO32-)=(Na2CO3)=0.1*2=0.2 mol/l

(Ag+)=8.2*10(-12)/0.2=4.1*10(-11)

(Ag2CO3)1=n(Ag+)/2=4.1*10(-11)/2=2.05*10(-11) mol/l

n(Ag2CO3)1=(Ag2CO3)1*V=2.05*10(-11)*0.1=2.05*10(-12) mol

m(Ag2CO3)1=n(Ag2CO3)1*Mr=276*2.05*10(-12)=5.66*10(-10) g

m(Ag2CO3)/m(Ag2CO301=0.00386/5.66*10(-10)=6.82*10(6)