# Question #90a90

##### 1 Answer

#### Answer:

#### Explanation:

First thing first, I think that you have some incorrect data in your question.

For example, if you use the information provided by the question, the *molar conductivity* of the chloroacetic solution will have different value than what you list here,

I assume that this value was supposed to be the *molar conductivity at infinite dilution*,

More specifically, this value should either be

#Lamda_0 = 362 * 10^(-4)"S m"^2"mol"^(-1)#

or

#Lamda_0 = "362 S cm"^2"mol"^(-1)#

With this being said, the equation that establishes a relationship between *conductivity*, *molar conductivity*,

#color(blue)(Lamda = k/c)" "# , where

This is where things usually get a little tricky. You need to make sure that you use the right units. For example, molar concentration must be used in *moles per cubic meter*, which means that you must convert the given *moles per liter*

#0.0625"moles"/color(red)(cancel(color(black)("liter"))) * (10^3color(red)(cancel(color(black)("liters"))))/"1 m"^3 = 0.0625 * 10^(3)"mol m"^(-3)#

Likewise, notice that the conductivity uses

#3.319 * 10^(-3)"S"/color(red)(cancel(color(black)("cm"))) * (10^2color(red)(cancel(color(black)("cm"))))/"1 m" = 3.319 * 10^(-1)"S m"^(-1)#

The molar conductivity of the solution will thus be

#Lamda = (3.319 * 10^(-1)"S m"^(-1))/(0.0625 * 10^3"m"^(-3)) = 5.3104 * 10^(-3) "S m"^(-2)"mol"^(-1)#

Now, the equation that connects the *degree of ionization* of a weak electrolyte,

#alpha = Lamda/Lamda_0#

In your case, the degree of dissociation for chloroacetic acid will be - keep in mind that **unitless**!

#alpha = (5.3104 * 10^(-3)color(red)(cancel(color(black)("S m"^(-2)"mol"^(-1)))))/(362 * 10^(-4)color(red)(cancel(color(black)("S m"^(-2)"mol"^(-1))))) = color(green)(0.147)#

The *acid dissociation constant*, **Ostwald's dilution law**, which establishes a relationship between the dissociation constant and the degree of ionization of a weak electrolyte

#color(blue)(K_a = alpha^2/(1 - alpha) * c)#

In your case, the value of *moles per liter*

#K_a = (0.147^2)/(1-0.147)^2 * "0.0625 M" = color(green)(1.86 * 10^(-3)"M")#

The listed value for the acid dissociation constant of chloroacetic acid is

You can check this result by using the molar conductivity

#color(blue)(K_a = Lamda^2/((Lamda_0 - Lamda) * Lamda_0) * c)#

This calculation will give you

#K_a = 1.58 * 10^(-3)"M"#