Question

A student prepares a aqueous solution of acetic acid . Calculate the fraction of acetic acid that is in the dissociated form in his solution. Express your answer as a percentage. You will probably find some useful data in the ALEKS Data resource.

Answer 1

Answer:

10.71%

Explanation:

The dissociation of acetic acid can be well expressed as follow:

CH₃COOH ⇄   CH₃COO⁻  + H⁺

Let assume that the prepared amount of the aqueous solution is 14mM since it is not given:

Then:

The I.C.E Table is expressed as follows:

                     CH₃COOH       ⇄   CH₃COO⁻        +           H⁺  

Initial              0.0014                       0                                0

Change            - x                           +x                               +x

Equilibrium   (0.0014 - x)                 x                                 x

Recall that:

Ka for acetic acid CH₃COOH  = 1.8×10⁻⁵

∴

$K_a = \dfrac{[x][x]]}{[0.0014-x]}$

$1.8*10^{-5} = \dfrac{[x][x]]}{[0.0014-x]}$

$1.8*10^{-5} = \dfrac{[x]^2}{[0.0014-x]}$

$1.8*10^{-5}(0.0014-x) = x^2$

$2.52*10^{-8} -1.8*10^{-5}x = x^2$

$2.52*10^{-8} -1.8*10^{-5}x - x^2 =0$

By rearrangement:

$- x^2 -1.8*10^{-5}x +2.52*10^{-8}= 0$

Multiplying through  by (-) and solving the quadratic equation:

$x^2 +1.8*10^{-5}x-2.52*10^{-8}= 0$

$(-0.00015 + x) (0.000168 + x) =0$

x = 0.00015 or x = -0.000168

We will only consider the positive value;

so x=[CH₃COO⁻] = [H⁺] = 0.00015

CH₃COOH = (0.0014 - 0.00015) = 0.00125

However, the percentage fraction of the dissociated acetic acid is:

$= \dfrac{ 0.00015}{0.0014}\times 100$

= 10.71%