Question

Conductivity of 0.001M acetic acid at a certain temperature is 5.07×10^-5 S/cm.if limiting molar conductivity of acetic acid at the same temperature is 390 S cm^2/mol.the dissociation constant of acetic acid at that temperature is?

Answer 1

We're at a warmer temperature than `25^@ "C"`.

---

We know that the limiting molar conductivity, `Lambda_m^0`, is simply the molar conductivity as if the electrolyte ionized completely. `Lambda_m` is the value measured normally...

Thus, it is weighted by the fraction of dissociation `alpha` (Eq. 7) to give the actual molar conductivity `Lambda_m` at the given concentration:

`Lambda_m = alphaLambda_m^0`

The conductivity you gave is not in the proper units yet.

`5.07 xx 10^(-5) "S/""cm" xx cancel"1 L"/("0.001 mols") xx (1000 cancel"mL")/cancel"1 L" xx ("1 cm"^3)/cancel"1 mL"`

`= "50.7 S"cdot"cm"^2"/mol"`

And now this can be compared to `Lambda_m^0` to find the fraction of dissociation:

`alpha = Lambda_m/Lambda_m^0`

`= ("50.7 S"cdot"cm"^2"/mol")/("390 S"cdot"cm"^2"/mol")`

`= 0.13`

By definition, the dissociation of acetic acid is given by

`"CH"_3"COOH"(aq) rightleftharpoons "CH"_3"COO"^(-)(aq) + "H"^(+)(aq)`

with mass action expression

`K_a = (["CH"_3"COO"^(-)]["H"^(+)])/(["CH"_3"COOH"])`

Some fraction `alpha` of the starting concentration becomes the acetate and proton products, and that much is lost RELATIVE to `100%` of the starting acid, so...

`K_a = ((alphacdot["CH"_3"COOH"]_0)(alphacdot["CH"_3"COOH"]_0))/((1 - alpha)cdot["CH"_3"COOH"]_0)`

`= (alpha^2["CH"_3"COOH"]_0)/(1 - alpha)` (cf. Eq. 21)

`= (0.13^2 cdot "0.001 M")/(1 - 0.13)`

And we obtain an acid dissociation constant of:

`color(blue)(K_a = 1.94 xx 10^(-5))`

at whatever temperature this is.

The actual value is around `1.76 xx 10^(-5)` at `25^@ "C"`, and dissociation is endothermic, so a higher `K_a` means this temperature is warmer than `25^@ "C"`.