### Acids and Bases on a Snowy Valentine's Day

Schools can make up for snow days. Children in kindergarten and elementary schools, for instance, may be celebrating Valentine's Day in their classrooms on Monday (President's Day has been designated as a snow makeup day). Making up, however, does not address the need for distributed practice in learning. Students need to spread their effort over time as opposed to cramming. School closings are disruptive. We may easily cover topics at a later date, but making up can not address the loss in pace. This is certainly one avenue through which technology can help. Of course, unlike being inside the classroom, students are not required to be on their seats at a given time. The instructor likewise would not know if you have silenced your cellphone, closed *Facebook*, and turned off email notifications. But since you are here, hopefully, I have your attention.

Practice problems as well as an online homework are still up. In addition, we do have a textbook that you could open and read. What we may be missing from not spending time together for 50 minutes inside a classroom is my perspective (and perhaps, my boring jokes). This post is then my attempt to provide a direction, a starting point for you to have before reading the text and solving problems. This requires additional initiative or motivation on your part. It may or may not help, but in the past, I have heard that it somehow helps. Below is an email I received a year ago:

It has worked at least once, so perhaps it could help us well. At least, it could put chemistry into our minds during this seven day break from a face-to-face lecture inside Reiss 112.

We are currently covering in class acid-base equilibria. As I have mentioned several times, this chapter and the next one are extensions of the chapter on chemical equilibrium. The concepts are the same, we are simply delving deeper into the specific case of reactions that involve proton transfer (acid-base reactions). The math behind the quantitative treatment of equilibrium should be familiar.

Hopefully, at this time, we already know by heart the Bronsted-Lowry definition of acids and bases: An acid is a proton-donor and a base is a proton acceptor. In aqueous solutions of either an acid or a base, the following equilibria are therefore relevant.

With pure water, we only had to concern ourselves with one reaction and its corresponding equilibrium expression: (taken from the previous lecture)

Thus, introducing an acid HA or a base B adds a reaction that needs to be considered, and of course, this comes with its own equilibrium expression. For instance, with an aqueous solution of an acid, we need to worry about

Thus, one needs to ask if HA produces much more than this or not (In the case of a base, B, one should look at the amount of OH-). How much [H+] is provided by HA depends on two factors: the magnitude of its dissociation constant

The following is the problem:

And here is my detailed answer:

The way I answered this problem (specifically part (b)) does not fit what this post described for weak acids or bases. The problem above in fact requires that we identify first what equilibrium expression we need to use. I should also note that the answer you would find in the textbook's solutions manual (posted in Blackboard) is unfortunately incorrect. Therefore, if you are able to follow what I did, I think that is a good and clear indication that you do understand what this chapter is all about.

Practice problems as well as an online homework are still up. In addition, we do have a textbook that you could open and read. What we may be missing from not spending time together for 50 minutes inside a classroom is my perspective (and perhaps, my boring jokes). This post is then my attempt to provide a direction, a starting point for you to have before reading the text and solving problems. This requires additional initiative or motivation on your part. It may or may not help, but in the past, I have heard that it somehow helps. Below is an email I received a year ago:

Hello Professor de Dios, You don't know me and I am not a student of yours, but in a way I am becoming one.... I have been struggling through my own Chem2 class this semester and found your lessons online. I wanted to thank you so much for posting exactly what I have been looking for: common sense explanations to what I am supposed to be learning. I am a 40-something mother who is going back to school for Physical Therapy after many years in the corporate world. I am taking classes at a local community college working on my re-req's and here, I am dealing with an ineffective, unprofessional, and frequently scatter brained teacher. As students we are often exchanging glances wondering what is going on or what exactly he means. The dean has apologized explaining he's all they have.... You however, have changed all that!! I thank you from the bottom of my heart for taking the time to post your lessons and letting an interloper experience how a real chemistry class should sound. Best regards,XXXXXX

It has worked at least once, so perhaps it could help us well. At least, it could put chemistry into our minds during this seven day break from a face-to-face lecture inside Reiss 112.

We are currently covering in class acid-base equilibria. As I have mentioned several times, this chapter and the next one are extensions of the chapter on chemical equilibrium. The concepts are the same, we are simply delving deeper into the specific case of reactions that involve proton transfer (acid-base reactions). The math behind the quantitative treatment of equilibrium should be familiar.

Hopefully, at this time, we already know by heart the Bronsted-Lowry definition of acids and bases: An acid is a proton-donor and a base is a proton acceptor. In aqueous solutions of either an acid or a base, the following equilibria are therefore relevant.

With pure water, we only had to concern ourselves with one reaction and its corresponding equilibrium expression: (taken from the previous lecture)

Thus, introducing an acid HA or a base B adds a reaction that needs to be considered, and of course, this comes with its own equilibrium expression. For instance, with an aqueous solution of an acid, we need to worry about

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

in addition to

Kw = [H+][OH-]

The [H+] terms in the above two equations are identical since we are working with the same aqueous solution. Thus, aqueous equilibria really entail solving simultaneous equations. Chemistry, however, allows us to treat this differently. Such treatment requires that we can gauge beforehand which equation matters more. In this particular case, H+ can come from two different sources, the acid HA and the autoionization of water. Which equation matters therefore depends on which source is dominant. And at 25 degrees Celsius, we know that water produces:

[H+] = [OH-] = 1.0 x 10-7

*Ka*, and its initial concentration. The value of the dissociation constant is a measure of the strength of an acid. Thus, it is helpful to keep in mind the common strong acids we may encounter in aqueous environment:
The following table (using HCl as a specific example) illustrates the approach we will take when dealing with a strong acid.

The above shows that one simply assumes that a strong acid is fully dissociated and equate [H+] at equilibrium to the initial concentration of the strong acid. This makes the assumption that the [H+] from the strong acid is far greater than the protons coming from water alone. However, the last row in the above table demonstrates a situation where most of [H+] will now come from the autoionization of water since the initial concentration of the strong acid is too low. Within the region between 0.000001 and 0.000000001 M HCl, solving simultaneous equations may be required for accuracy. But if the initial concentration of HCl is at least 0.00001 M, we can neglect autoionization of water and when the initial concentration of HCl is less than 0.000000001 M, we can neglect what is is coming from HCl. This is the first example of how we may approach equilibrium problems and it starts with identifying the dominant expression.

The next part concerns weak acids. These are acids that do not dissociate completely. The following demonstrates the general approach.

The quadratic equation can be solved by using the above formula. Chemistry, however, allows us to take a short-cut, without sacrificing accuracy:

The above takes less time and if one keeps in mind that

*pH*is a power of 10 quantity, one can begin to think of the dissociation constant and*c*in orders of magnitude and realize that taking the square root is equivalent to dividing by two when dealing with logarithmic values. This allows for quick and reliable estimates of pH without using a calculator.
With a spread sheet, one can explore a wide range of initial concentrations of a weak acid to see the big picture of when this approximation is accurate enough:

A popular question a student may ask at this point is "When should we use the quadratic formula and when is the approximation sufficient?" Formulating a guiding rule at this point is not really necessary since the approximation can be quickly assessed with a small amount of effort - just take the square root of the product of the dissociation constant and the initial concentration. (

*Simply go ahead, ask for forgiveness later - you may not need to.*) If this square root is less than 5 percent of the initial concentration then the resulting answer is accurate to two significant figures, as illustrated in the above table. At 0.01 M (eight row in the above table), the answers using either method lead to the same value for [H+], 0.00042 (two significant figures). This row corresponds to 4.15% dissociation (This is essentially*x/c*, a measure of how big*x*is compared to*c*, the heart of our approximation). The next row which shows a percent dissociation of 5.82% illustrates a difference that is starting to show up in [H+] in the least significant digit (0.00013*versus*0.00012).
Lastly, the percent ionization,

describing the fraction of the acid that actually dissociated, depends on both dissociation constant and initial concentration. Remember our approximation:

so percent ionization is roughly described by the following:

Thus, "Weak acids become strong when they are dilute". Of course, this is likewise explained by Le Chatelier's principle:

*Dilution involves concentration, it is not related to the strength of an acid. The strength of an acid is measured purely by how much (or what fraction) of it ionizes. Only a small fraction of a weak acid ionizes in solution. However, as the initial concentration of the weak acid is decreased, the fraction that dissociates becomes increasingly closer to 100%. The reason why this occurs is kinetics. As the solution becomes more dilute, it becomes less likely for H*

^{+}to see and find an A^{-}, which is required to reform the acid HA. Thus, it is simply a manifestation of Le Chatelier's principle: Bigger volume will favor the side of the reaction that has more particles.
For bases, one simply switches from [H+] to [OH-]:

It is good if I still have your attention up to this point. To end this post, I am going to use a specific problem from the text. It is question 113 at the end of chapter 16. It is perhaps the most challenging question at the end of the chapter. It involves an amino acid, glycine, and hinges on the fact that glycine, like other amino acids, can act either as an acid or base. Having both functions allows glycine to react with itself:

The following is the problem:

And here is my detailed answer:

The way I answered this problem (specifically part (b)) does not fit what this post described for weak acids or bases. The problem above in fact requires that we identify first what equilibrium expression we need to use. I should also note that the answer you would find in the textbook's solutions manual (posted in Blackboard) is unfortunately incorrect. Therefore, if you are able to follow what I did, I think that is a good and clear indication that you do understand what this chapter is all about.

*Happy Valentine's Day!*