Percent Dissociation of a Weak Acid

 Hello fam! I trust you woke up strong and ready for the day today. I didn’t see your comments in answer to the question of yesterday so I would post the detailed answer at the end of today’s lesson.


Like I told you, we would deal with the idea of percent ionization of a weak acid today. Remember that we have established long before that a weak acid would only dissociate to a very small extent when dissolved in water. The extent to which the weak acid can ionize or dissociate in solution can be expressed as a percentage. This percentage is what we call the percentage dissociation of the weak acid.


In the simplest terms, the percentage dissociation is the percentage of the ions A- and H3O+ that are formed when the acid is dissolved in water. Here I have written a simplified formula for the percentage dissociation of a weak acid. I have avoided all the intricate processes of deriving the final equation and I have simply given the result of the derivation for simplicity.


The formula for the percentage dissociation of a weak acid is;

 α = √Ka/C * 100

α = Percentage dissociation

Ka = Acid dissociation constant

C = Initial concentration of the acid.


Example 1;

Dissociation constant of acetic acid is 1.8 * 10^-5. Calculate the percent dissociation of acetic acid in 0.01M solution.


Solution;

Using the formula given;

α = √Ka/C * 100

α = Unknown

Ka= 1.8 * 10^-5

C = 0.01 M

α =√1.8 * 10^-5/0.01 * 100

α = 4.2%


Now try this one;

A weak acid with a dissociation constant 1.54 * 10^-5 M is 1.26% dissociated. What is the concentration of the hydrogen ions in the solution?


Hope you are following the lessons? Feel free to ask your questions in the comments section or via messenger, whatsapp, calls or text messages.


Solution to yesterday’s problem;

Setting up the ICE table;

             HC2H3O2(aq) + H2O(l)   <-> C2H3O2^-(aq) + H3O^+(aq)

I            0.25                                               0                         0

C          -x                                                    +x                      +x

E        0.25 – x                                            x                        x

Ka = [C2H3O2^-] [H3O^+]/[ HC2H3O2]

[C2H3O2^-] = [H3O^+] = x

1.8 * 10^-5 = x^2/0.25 – x

1.8 * 10^-5 (0.25 – x) = x^2

4.5 * 10^-6 - 1.8 * 10^-5x = x^2

x^2 + 1.8 * 10^-5x - 4.5 * 10^-6 = 0

x = 0.0021 

[C2H3O2^-] = [H3O^+] =  0.0021M 

The concentration of the hydronium ion is  0.0021M


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