Lecture 5 - Acids and Bases (Part 2)

Lecture Recording:

 

Lecture Summary/Notes:

Finally got these up! Enjoy! 

Ka, Kb, pH, pOH

If we  have a general equilibrium reaction:

Where Ka is the acid dissociation constant, we can use that value to solve for [H+], which in turn can be used to find pH. **p.710 in the textbook has a table of Ka, Kb, pKa, and pKb values of some weak acids and bases**


NOTE: Kw/Ka = Kb. Remember Kw = 1.0 x 10^-14. Here is a derivation of that equation:

Textbook Example - P.E.A. and P.E.B. (page 712)

P.E.A. - We are given the initial value of HOCl and the pH. We find the final conditions, saying that we use an x amount of HOCl and form x amount of H+ and OCl-. We write out the equilibrium equation in terms of x. We then find the x value (pink cloud) by solving for [H+] ( x = [H+] ). We then substitute the x value back into the equilibrium equation to solve for Ka.

P.E.B. - In this case, we are solving for the equilibrium constant of a base (Kb). We find the final conditions of the equilibrium reaction, and write out the equilibrium equation in terms of x. We then solve for x (in this case, [OH-] = x). We are given the pH, but need the pOH, so we subtract the pH from 14 to get pOH. We then isolate [OH-] to get x, and then substitute it back into the equilibrium equation to solve for Kb.

Textbook Example - P.E.B. (page 713)

P.E.B. - We are given the mass (1.0 g), we find the molar mass (180.2 g/mol) of Acetylsalicylic acid so that we can find the # of moles (5.55 x 10^-3). We then divide it by the given volume (0.325 L) to find the concentration. We then use this value as the initial concentration of the acid, and then find our final conditions. We then, again, form the equilibrium equation. In this case we are given the Ka, so we set the equation equal to the value, and then solve for x using the quadratic formula. We end up with 2 x values, and eliminate the negative value (cannot have negative concentration), leaving us with the x value, which is = to [H+]. We then use that value to find pH. (Thank you, anonymous!)

Simplifying Assumption

This is super useful for when we are solving for x, given the Ka/Kb value. The assumption is that if the   equilibrium constant is very very small compared to the initial concentration of the acid/base, then the reaction will go to such a small extent that it will hardly change the initial concentration of the acid/base. See the following example:

The red says: This reaction will go to such a small extent, that the x value is going to be a number significantly less than 0.1. We can therefore make a simplifying assumption. Thus, we can just ignore the x in the denominator. Think about it this way: if x is 0.001, and the initial concentration is 0.1, 0.1-0.001=0.099, which we will make 0.1 anyways (rounded for proper sig figs). So we can -assume- that it will hardly affect it, and ignore it in the denominator, so that we can solve for x in the numerator more easily. Get it? :) 

Condition for using the simplifying assumption:
The initial concentration is at least 100 times greater than the equilibrium constant.
(So [HA]0 >100Ka, or [Hb]0>100Kb)

Percent Ionization

Percent ionization is the percentage of the acid that actually dissociates.

Textbook Example - P.E.A. and P.E.B. (page 717)

P.E.A. - If we write out the dissociation reaction of HF, and then find the final conditions of the equilibrium reaction (in terms of x). We then write out the equilibrium equation, setting it equal to the given Ka. In this case, we are able to use the simplifying assumption since the initial concentration (0.20M) is 100X greater than the Ka. So we then solve for x very easily, and set it equal to both the H+ and F- concentrations. F- is the conjugate base of HF. We then found the percent ionization of HF by setting the [F-]/[HF]0 and multiplying it by 100%. We get a final answer of 5.6%. This means (green text) that only 5.6% of the HF molecules will dissociate and make F- molecules. The rest stay as HF molecules. This is a very WEAK acid! For 0.020M, we would go through the whole process again, solving for [H+] where in the denominator, instead of 0.2, we would have 0.02. Solving for x, we would get 0.0036M, and then we would have percent ionization = (0.0033/0.020) x 100% to give us 18.2% (thank you, anonymous!).

P.E.B. - For this question we are given the percent ionization and the initial concentration of the acid. We first solved for the concentration of the conjugate base. (on the left is the percent ionization equation). We know that x amount of conjugate base is formed, so the found concentration of the conjugate base is = x. We then substitute x into the equilibrium equation [Ka = (x^2)/([HA]-x)] to find the Ka value.

Polyprotic Acids

Polyprotic acids are acids with more than one proton (H). Two examples of polyprotic acids are H2SO4 (sulphuric acid) and H3PO4 (Phosphoric acid). With these acids, the number of dissociations possible is equal to the number of H's in the acid. See below:

There are two possible dissociations for sulphuric acid

There are three possible dissociations for phosphoric acid. For each dissociation, there is a separate dissociation constant. (If you can't read it) Ka1 = 7.1 x 10^-3, Ka2 = 6.3 x 10^-8, Ka3 = 4.2 x 10^-13. After every dissociation, the Ka value drops considerably. WHY? It's because if you deprotonate the acid over and over, you have to lose an ion, which drops the acidity by lots!

**p. 718 in the textbook has a table of ionization constants of some polyprotic acids**
**If you dissolve a polyprotic acid in water, you only need to consider the FIRST dissociation. If you want to know the concentration of the second dissociation anion, it = Ka2.**