Basis for a New Definition of Acid Dissociation Constant Part II

 Hello fam, we are making significant progress so far and that’s really impressive ! So far we have covered the basics of the concept of acids and we are moving on to look at the acid dissociation constant which is the dividing line between strong acids and weak acids.

Let me do a little bit of recap from yesterday’s lesson. We agreed that what makes an acid to “behave” like an acid is because it is able to interact in a special way with water via dative bonding. When this occurs, hydrogen ion from the acid can accept lone pairs from oxygen and then we would have H3O+ in which all three atoms of hydrogen and one atom of oxygen obey the octet rule respectively.
Let me harp a little more on the idea of ‘lone pair’ because I believe the term may not have been well understood in yesterday’s lesson. Let me take you back to the electron configuration of oxygen; 1s^2 2s2 2px^2 2py^1 2pz^1(written in the most basic form as 1s^2 2s^2 2p^4). Recall that the p sublevel is composed of three orbitals having exactly the same energy(degenerate); px, py, pz. The orbital is a region in space where there is a high probability of finding an electron in an atom. We would deal with this definition in detail when we discuss the historical development of the atomic theory.
The 1s^2 orbital is found in the innermost shell of the oxygen atom (n =1 level) that does not participate in bonding. The second shell in oxygen contains the electrons in the second energy level(n =2) which are; 2s^2 2px^2 2py^1 2pz^1(or 2s^2 2p^4 for short). Water is formed when electrons are shared between two atoms of hydrogen containing one electron each and the single electrons in the 2py^1 2pz^1orbitals of oxygen. This is the ordinary covalent bonding in water.
This implies that we have yet four electrons that occur in pairs in the 2s^2 and 2px^2 orbitals of oxygen. These are the two “lone pairs” that I referred to yesterday. The term ‘lone’ connotes the idea that they are not combining with any other atom. They are electron pairs that belongs exclusively to oxygen. The two electrons that occur singly in oxygen that eventually combine with two different hydrogen atoms to form water would then constitute the “shared pair” or the “bond pair” of the water molecule.
I hope I have cleared the air on that concept? Don’t hesitate to reach out to me if confusion still persists in any way.
Now, we agreed that we have the real scenario in the acid solution as a dynamic equilibrium between our hypothetical weak acid HA unionized molecules, hydronium ion(H3O+) and the acid anion(A-). This would give us the equation that we have as;
HA(aq) + H2O(l) H3O+(aq) + A-(aq) (As described yesterday)
Then we can write a new definition for the equilibrium expression for the dissociation of the acid as;
Ka = [H3O+] [A-]/[HA] [H2O]
Water is the solvent in this context hence it is present in large excess. We do not include reactants that are in large excess such as solvents in the equilibrium expression because their concentrations do not change appreciably in the reaction due to the fact that the amount of that particular substance (the solvent) in the system is large.
Let me give you an example that you can relate with in order to buttress my point. If I give you a bucket full of sand, then I remove a couple of grains of sand from the bucket, does the number of grains removed make any difference in the amount of sand in the bucket? Your guess is as good as mine. The answer is No! The bucket of sand would look so much the same after removing a few grains of sand showing that the system was not significantly impacted by the change. I hope we are together on that point? What if I add another couple of grains of sand to the bucket, would you notice the change? The answer of course is No. This is because the bucket contains the grains of sand in large excess. As such, removal or addition of a small amount or grains of sand to the bucket would make no observable difference when considering the quantity of sand in the bucket. Reach out to me if any confusion still persists.
Having justified the removal of water from the equilibrium expression, we can now write the new equilibrium expression for the dissociation or ionization of the weak acid HA as;
Ka = [H3O+] [A-]/[HA]
We now come to the conclusion that the value of the acid dissociation constant Ka depends on;
1) Concentration of the unionized weak acid [HA] at equilibrium
2) Concentration of the hydronium ion (H3O+) at equilibrium
3) Concentration of the anion in the weak acid (A-) at equilibrium.
Tomorrow we are going to get into how to calculate the acid dissociation constant so we are headed for a bit of mathematical stuff. Tighten your seatbelt guys, its going to be fun!
Salient Point;
The acid dissociation constant can also be expressed in the form;
Ka = [H3O+] [A-]/[HA]
Hope you enjoyed our lesson today?
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