Question

Question #7f749

Answer 1

`3.260 * 10^(-3)"S m"^2"mol"^(-1)`

Explanation:

The idea here is that you need to use Kohlrausch's law of independent migration of ions to find the limiting molar conductivity, or LMC, for the butyrate ions.

You know that for dilute solutions, the molar conductivity can be determined by taking into account individual contributios of ions.

The thing to notice here is that you're dealing with a weak acid, a weak electrolyte which by definition does not dissociate completely in aqueous solution.

The formula that allows you to find the limiting molar conductivity for a weak electrolyte solution made up of `i` ions looks like this

`Lamda_"m"^@ = sum_i(lamda_i xx nu_i)" "`, where

`nu_i` - the number of ions you get per formula unit (or molecule);
`lamda_i` - the molar conductivity of an ion `i`.

As you can see, the molar conductivity of a weak electrolyte depends on its concentration.

Now, butyric acid, `"C"_4"H"_8"O"_2`, will dissociate to give protons, `"H"^(+)`, and butyrate anions, `"C"_4"H"_7"O"_2^(-)`.

`"C"_4"H"_8"O"_2 rightleftharpoons "H"^(+) + "C"_4"H"_7"O"_2^(-)`

So, how many ions do you get per molecule? Well, you get one proton and one butyrate anion.

This means that you have

`Lamda_"butyric" = 1 xx lamda_"protons" + 1 xx lamda_"butyrate"`

Therefore, you can say that

`lamda_"butyrate" = Lamda_"butyric" - lamda_"protons"`

`lamda_"butyrate" = 3.823 * 10^(-2)"S m"^2"mol"^(-1) - 3.497 * 10^(-2)"S m"^2"mol"^(-1)`

`lamda_"butyrate" = color(green)(3.260 * 10^(-3)"S m"^2"mol"^(-1))`

To get the ion transfer numbers, also called the transference numbers, you need to use the limiting molar conductivities of the two ions.

Now, looking at the LMC values for the two ions, you can predict that most of the conductivity will be carried out by the protons.

So, the ion transfer number for the protons is

`t_"+" = (lamda_"protons")/(lamda_"protons" + lamda_"butyrate")`

`t_"+" = (3.497 * 10^(-2)color(red)(cancel(color(black)("S m"^2"mol"^(-1)))))/(3.497 * 10^(-2)color(red)(cancel(color(black)("S m"^2"mol"^(-1)))) + 3.260 * 10^(-3)color(red)(cancel(color(black)("S m"^2"mol"^(-1)))))`

`t_"+" = 0.9147`

This means that about `91.47%` of all conductance is carried out by the protons. The remaining `8.53%` will be carried out by the butyrate anions.

`t_(-) = 0.0853`