Mathematical Problems Involving Acid Dissociation Constant

 Hello fam, we are here again! Today we would delve further into the concept of the acid dissociation constant. It is imperative that we note that the acid dissociation constant is an example of an equilibrium constant. It reflects the concentrations of the species involved when equilibrium has been attained.


Remember that equilibrium is attained in the system when the rate of forward reaction is equal to the rate of reverse reaction. This implies that the reactants are being converted to products just as fast as the products are being changed back to reactants. The implication of this is that the concentration of each of the species in the reaction system remains fairly the same unless something disturbs this equilibrium.


Before we proceed further in this discussion, it is important to note that the value of Ka does not depend on the initial concentration of the acid. This is because, we can only define the Ka at a point when the concentration of the system is no more changing drastically. The initial concentration of the acid is the concentration of the acid before it was added to water.


When you add the acid HA of known concentration to water, a reversible reaction is set up as we discussed yesterday as follows;

HA(aq) + H2O(l) > H3O+(aq) + A-(aq)

Ka = [H3O+] [A-]/[HA]

We can only define the Ka when HA and H2O is converting to H3O+ and A- at the same rate at which H3O+ and A- are combining to form HA and H2O again (Refer to yesterday’s lesson to know why H2O is excluded from the equilibrium expression). That is, THERE IS NO NET CHANGE IN THE CONCENTRATION OF SPECIES IN THE SYSTEM. This last statement is very important hence it was written in uppercase characters for emphasis.


Let us assume that I have 2 moles/l of weak acid HA which is dissolved in water to give hydronium ions and A- anions. This is the initial concentration of the acid. If I am now told that the concentration of the ions H3O+ and A- at equilibrium is 0.008 moles/l each then I would have to set up what is generally called an ICE table. 

I stands for initial concentration

C stands for change in concentration

E stands for equilibrium concentration


When we add the acid to water, there was initially zero concentration of H3O+ and A- before the dissociation began. All we had was the initial concentration of the acid given as 2 moles/l or 2M. As dissociation progresses, the concentration of the acid HA would decrease (hence we write -x in the change row) by an amount x as it dissociates while the concentration of the H3O+ and A- would increase(hence we write +x in the change row) by an amount x as they are formed in solution. The equilibrium amount of the acid is usually obtained by subtracting the equilibrium concentration of the ions (x) from the initial concentration of the unionized acid.


On the other hand, the equilibrium concentration of the A- and H3O+ ions is obtained by adding the equilibrium concentration of the ions (x) to zero (the initial concentration of the ions before dissociation began).


We then have the ICE table as;

                HA(aq)  + H2O(l) - H3O+ (aq) + A-(aq)

I               2                                      0                    0

C             -x                                     +x                  +x

E          2 – x                                   x                    x


Ka = (x) (x)/(2 – x)

Ka = x^2/2 - x

Where x = concentration of ions H3O+ and A- at equilibrium = 0.08 M

Ka = (0.08) (0.08)/(2 – 0.08)

Ka = (0.08)^2/ 2 – 0.08

Ka = 0.0033 

Hence, the acid dissociation constant for the acid HA that we have considered here is 0.0033.


We can now make a generalization which is called the Ostwald’s dilution law for any weak acid HA. The law states that, if we have x moles of weak acid dissolved in V liters of water, Ka can be obtained as;

Ka = x^2/(1 – x)/V

This law works best in cases where the equilibrium concentration of the weak acid is less than 1.


This is the simplest kind of problems that we can find about Ka, more complicated problems exist that involve a term called the pH and we shall consider these as we progress in the series. Tomorrow, we shall look at the kind of problems where Ka is provided and we need to obtain the equilibrium concentration based on the details provided in the question. Next tomorrow, we shall look at percentage dissociation of a weak acid. Hope you are enjoying the ride? See you then!


SALIENT POINT:

The acid dissociation constant can only be calculated using the equilibrium concentrations of HA, H3O+ and A-.


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


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