[Danial]

do i treat this question normally, but multiply the ionic strength of acetic acid by 0.048?

[Borek]

What do you mean by "normally"? What is ionic strength definition?

[Danial]

Would I treat it like

0.5 (0.048 x (0.0078 x  (acitic acid z^2s) + (0.5 x (dioxanes z^2)

for acetic acid C2H4O2  C (0.0078 x 2 x 1^2) +  H (0.0078 x 4 x 1^2) + O (0.0078 x 2 x 2^2)
dioxane C4H8O2    C (0.5x4x1) + H ( 0.5 x 8 x 1) + O (0.5 x 2 x 4)

so the ionic strength = 0.5 ( 0.048 x 0.1014 + 10) = 5
that seems pretty high though
is it because of the way im doing the z calculations?
like cause its C2, (2 x concentration x 1^2)?
but im not sure what to do with acetic acid as it has no charge, so i broke it up into C H and O
would z=0 for acetic acid?
What about dioxanes charge?

[Borek]

If something has no charge, c_iz_i² is zero, so you just ignore it.

When you calculate ionic strength you carry summation for all IONS present. Use information provided to calculate concentration of ions.

[Danial]

so the answer to that question is juts 0?
dioxane is neutral aswell ?

[Borek]

so the answer to that question is juts 0?

How is dissociation fraction defined? What does it mean "4.8% of the acid dissociated"?

dioxane is neutral aswell ?

Yes.

[Danial]

does it mean 4.8% of the acetic acid dissociated in the dioxane? ohh would create a charge on 4.8% of both the compounds?


0.5 ( 0.048 x 0.0078 x 1^2)

[AWK]

0.5 ?

[Borek]

does it mean 4.8% of the acetic acid dissociated in the dioxane?

No, in water. But you are on the right track.

However, you need to sum effect of both ions, H⁺ and CH₃COO^-.

[Danial]

thanks heaps for your help Borek

so it'll end up 0.5 ( 0.048 x 0.0078 x 1^2 x 2)

[Borek]

Looks OK now.

[AWK]

Note - over 100 years ago G.N. Lewis remarked that for electrolyte 1:1 (your case) its ionic strength is equal to its concentration (0.048 x 0.0078). This comes from his formula (0.5x2=1)