Question #90a90

Question

Question #90a90

1 Answer

Impact of this question

5978 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-25T00:06:13

$α = 0.147$ $alpha = 0.147$
$K_{a} = 1.86 \cdot 10^{- 3}$ $K_a = 1.86 * 10^(-3)$

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, ${362 S m}^{2} {mol}^{- 1}$ $"362 S m"^2"mol"^(-1)$ .

I assume that this value was supposed to be the molar conductivity at infinite dilution, $Λ_{0}$ $Lamda_0$ . Moreover, I think that the value you provided is incorrect.

More specifically, this value should either be

$Λ_{0} = 362 \cdot 10^{- 4} {S m}^{2} {mol}^{- 1}$ $Lamda_0 = 362 * 10^(-4)"S m"^2"mol"^(-1)$

or

$Λ_{0} = {362 S cm}^{2} {mol}^{- 1}$ $Lamda_0 = "362 S cm"^2"mol"^(-1)$

With this being said, the equation that establishes a relationship between conductivity, $k$ $k$ , and molar conductivity, $Λ$ $Lamda$ , looks like this

$Λ = \frac{k}{c}$ $color(blue)(Lamda = k/c)" "$ , where

$c$ $c$ - the molarity of the solution.

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

$0.0625 \frac{moles}{liter} \cdot \frac{10^{3} liters}{{1 m}^{3}} = 0.0625 \cdot 10^{3} {mol m}^{- 3}$ $0.0625"moles"/color(red)(cancel(color(black)("liter"))) * (10^3color(red)(cancel(color(black)("liters"))))/"1 m"^3 = 0.0625 * 10^(3)"mol m"^(-3)$

Likewise, notice that the conductivity uses ${cm}^{- 1}$ $"cm"^(-1)$ , so convert it to

$3.319 \cdot 10^{- 3} \frac{S}{cm} \cdot \frac{10^{2} cm}{1 m} = 3.319 \cdot 10^{- 1} {S m}^{- 1}$ $3.319 * 10^(-3)"S"/color(red)(cancel(color(black)("cm"))) * (10^2color(red)(cancel(color(black)("cm"))))/"1 m" = 3.319 * 10^(-1)"S m"^(-1)$

The molar conductivity of the solution will thus be

$Λ = \frac{3.319 \cdot 10^{- 1} {S m}^{- 1}}{0.0625 \cdot 10^{3} m^{- 3}} = 5.3104 \cdot 10^{- 3} {S m}^{- 2} {mol}^{- 1}$ $Lamda = (3.319 * 10^(-1)"S m"^(-1))/(0.0625 * 10^3"m"^(-3)) = 5.3104 * 10^(-3) "S m"^(-2)"mol"^(-1)$

Now, the equation that connects the degree of ionization of a weak electrolyte, $α$ $alpha$ , and its molar conductivity looks like this

$α = \frac{Λ}{Λ_{0}}$ $alpha = Lamda/Lamda_0$

In your case, the degree of dissociation for chloroacetic acid will be - keep in mind that $α$ $alpha$ must come out to be unitless!

$α = \frac{5.3104 \cdot 10^{- 3} {S m}^{- 2} {mol}^{- 1}}{362 \cdot 10^{- 4} {S m}^{- 2} {mol}^{- 1}} = 0.147$ $alpha = (5.3104 * 10^(-3)color(red)(cancel(color(black)("S m"^(-2)"mol"^(-1)))))/(362 * 10^(-4)color(red)(cancel(color(black)("S m"^(-2)"mol"^(-1))))) = color(green)(0.147)$

The acid dissociation constant, $K_{a}$ $K_a$ , can be calculated using Ostwald's dilution law, which establishes a relationship between the dissociation constant and the degree of ionization of a weak electrolyte

$K_{a} = \frac{α^{2}}{1 - α} \cdot c$ $color(blue)(K_a = alpha^2/(1 - alpha) * c)$

In your case, the value of $K_{a}$ $K_a$ will be - use the molarity in moles per liter

$K_{a} = \frac{{0.147}^{2}}{{(1 - 0.147)}^{2}} \cdot 0.0625 M = 1.86 \cdot 10^{- 3} M$ $K_a = (0.147^2)/(1-0.147)^2 * "0.0625 M" = color(green)(1.86 * 10^(-3)"M")$

The listed value for the acid dissociation constant of chloroacetic acid is $1.4 \cdot 10^{- 3} M$ $1.4 * 10^(-3)"M"$ , so the result is good enough.

You can check this result by using the molar conductivity

$K_{a} = \frac{Λ^{2}}{(Λ_{0} - Λ) \cdot Λ_{0}} \cdot c$ $color(blue)(K_a = Lamda^2/((Lamda_0 - Lamda) * Lamda_0) * c)$

This calculation will give you

$K_{a} = 1.58 \cdot 10^{- 3} M$ $K_a = 1.58 * 10^(-3)"M"$

Science

Math

Humanities

... and beyond

Question #90a90

1 Answer

Explanation:

Related questions

Impact of this question