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


Hope you enjoyed our lesson today?


Follow  our page  and share!


Do you have assignments, term papers, projects and difficult homework questions? Contact us for help now! We also offer home and remote tutoring services to students at all levels in science subjects


Our articles and other important details are available at our blog:

www.tolearnchem.blogspot.com

Comments

Post a Comment

Popular posts from this blog

Definition of acids 2 - Introduction to Lewis Acids

Image
 Hello fam! Welcome to today’s lesson. We started to look at the concept of acid definition in yesterday’s lesson. We saw yesterday that acids have been defined by Arrhenius and BrØnsted – Lowry. The Arrhenius definition is limited in the sense that it does not account for behavior of acids in nonaqueous media. Did you try the assignment yesterday? The answers are attached to this post. I believe that the concept of proton donation is now clear to you after looking at the images attached. Reach out to me if you have any challenge. Now let us talk about the Lewis definition of acids. This definition of acids was put out by G. N Lewis in 1923. There is a foundation that we need in order to have an understanding of the idea that was put forward by Lewis.  There is something called an orbital in an atom. It is a region in space where there is a high probability of finding the electron. Now let us assume that your office is at No.38 Bronx street. I may not find you everywhere in the compoun
Read more

Calculations on pKa and pH

Image
 Hello! I heartily welcome you all to today’s lesson. We are going to look at the last two calculations that pertain to pH, acids and bases today and tomorrow we would commence our discussion on salts to pave way for a discussion on buffers. I hope all the topics so far are making sense? Your feedback would be highly appreciated. Example 8; What is the pH of 0.2 M solution of sodium acetate. The pKa of acetic acid is 1.8 * 10^-5. Solution; We have that the hydrolysis of sodium acetate occurs as follows; CH3CH2COONa(aq) <-> CH3CH2COO^-(aq) + Na^+(aq)    CH3CH2COO^-(aq) + H2O(l) <-> CH3CH2COOH + OH^-(aq) We can see that that the acetate ion acted as a base hence it accepted a proton. We would have to obtain the Kb Kb = 1 * 10^-14/Ka Kb = 1 * 10^-14/1.8 * 10^-5 Kb = 5.6 * 10^-10 Setting up the ICE TABLE, we have that;          CH3CH2COO^-(aq) + H2O(l) <-> CH3CH2COOH + OH^-(aq) I            0.2                                                      0                       
Read more

Bonding in Complex Salts

Image
 Hello! yesterday, we started looking at the complex salt which we called tetramine copper II chloride with the formula; [Cu(NH3)4]Cl2. We did say that this is a complex salt because of the presence to the complex cation [Cu(NH3)4]^2+. I want to state that if we had a complex anion, the salt would also be a complex salt. A typical example of this is; K4 [Fe(CN)6]. The complex ion is  [Fe(CN)6]4- which is anionic and the counter ions in this case are potassium cations. I told you that we will look at the bonding in the complex salt [Cu(NH3)4]Cl2 and why it does not produce simple ions in water. I will try to explain the basics of the concept today as we would still look at coordination chemistry in more detail sometime later.  Now, the complex salt [Cu(NH3)4]Cl2 has both ionic and covalent bonds. The four ammonia molecules are attached to the central copper II ion by a coordinate covalent bond(see the page timeline for an explanation of what a coordinate covalent bond is). We showed the
Read more