find the volume of water added to 300 ml 0.2 m ch3cooh solution so that degree of ionization become double

Hello,

Ethanoic acid is a weak acid and dissociates:

CH3COOHCH3COO−+H+

For which Ka=[CH3COO−][H+][CH3COOH]=1.8×10−5

These are equilibrium concentrations.

The degree of dissociation α is the fraction of the amount of electrolyte which dissociates. If C is the initial concentration of the acid then we can write an ICE table based on mol/l

∴ Ka=C2α2(C−Cα)=C2α2C(1−α)=Cα2(1−α)

Because the dissociation is small i.e 10−9<Ka<10−4 then we can assume that (1−α)⇒1 .

∴ Ka=Cα2

α=√KaC

α=√1.8×10−50.2=9.486×10−3

Now we double this value to get α⋅ :

α⋅=2×9.486×10−3=18.97×10−3

To get the new concentration we get:

18.97×10−3=√1.8×10−5C⋅

360×10−6=1.8×10−5C⋅

C⋅=1.8×10−5360×10−6=0.05xmol/l

This is the new concentration we need. To find how much water we need to add to achieve this dilution we can say:

C1V1=C2V2

∴ 300×0.2=0.05×V2

V2=300×0.20.05=1200xml

This means we need to add a further 1200 - 300 = 900 ml.

Hope this helps you out.

Good luck!