Quick note: In this article, [H+] and [H3O+] are used interchangeably. It is a better habit to write [H3O+] as it is more accurate than simply giving [H+]. lg refers to log10.
Introduction
Just this afternoon, a friend of mine sent me the following question (adapted). It’s more of an ‘interdisciplinary’ type of question, and it requires some skills from organic chemistry to solve it, as well. See if you are able to solve it (answers in the next article):
“A 20mM histidine monohydrochloride solution (Mr = 209.63) is prepared. 20ml of it is used in a titration with 0.05M NaOH. It has an initial pH of 4.00. In a previous experiment, 2 pKa values of histidine were determined: 6.00 and 9.35. Calculate the pH of the solution after addition of 12ml of 0.05M NaOH.”
For reference, the structure of histidine is provided below (Fig. 1).
Fig. 1: Structure of histidine.
If you are able to solve this, well, you can skip this article. Or maybe skim through it just for a quick revision. If you can’t, by the end of this article, you should be able to. So without further ado, let us begin.
We begin with a clarification of the terms we will use throughout this article. For those who are already familiar with the terms pH, pKa and Kw, you can skip this part.
Obviously the simplest term is the pH. We may have heard that it refers to ‘potential hydrogen’. Simply stated, the pH refers to the negative log (logarithm) of the concentration of hydrogen ions (measured in mol dm-3). The log is base 10.
From the above, and considering that pH values range from 0 to 14, we can conclude that the concentration of hydrogen ions ranges from 100 to 10-14 (considering the negative sign). This means that the strongest base would still have hydrogen ions, but very little (10-14).
The pKa may seem similar to the pH but they are not directly related. We will learn more about it later. As for Kw, it refers to the ionic product of water. Due to the amphiprotic properties of water, it may undergo the following reaction (Fig. 2):
2H2O ⇆ H3O+ + OH-
Fig. 2: Autoionization of water.
We may have noticed that this process is known as the autoionization of water. The principle for Kw comes from this. The ionic product of water is given by the concentration of H+ (or H3O+) ions multiplied by the concentration of OH- ions (Fig. 3).
Kw = [H3O+][OH-]
Fig. 3: The ionic product of water.
Kw is always a constant at a given temperature, and is 1.0 x 10^-14 mol2 dm-6 at 25C (this is provided in the data booklet). Finally, do take note that the square brackets ‘[]’ refer to the concentration of the involved ions.
Question 1: Calculate the pH and concentration of OH- ions of a 0.05 mol dm-3 HCl solution at 298K.
Upon further inspection this question consists of two parts which may be separately answered: the calculation of the pH and the calculation of the concentration of OH-. We will thus treat each part separately.
For the calculation of the pH, we will apply the following formula, which has to be memorized (do not worry, a whole list will be provided at the end):
pH = -lg[H3O+]
Fig. 4: Formula for calculation of pH.
We know that HCl is a strong acid and that it will always completely dissociate to form one H+ per HCl molecule. As such, [HCl] = [H+] or [H3O+] = 0.05 mol dm-3. And given this, simply taking the negative log of 0.05 will give us the answer of 1.3 (which is decently accurate).
If we were given a strong base, we could apply the same method to calculate [OH-], and then [H3O+] from there.
If you have been reading through the entire article so far, you should be able to solve this second part. Since we know the concentration of [H3O+] ions (0.05 mol dm-3), are you then able to solve the second part?
For this part, we have to apply the formula that we have previously learnt, that is, the formula for the ionic product (Kw) of water. As we recall, it is Kw = [H3O+][OH-]. Since we have the [H3O+] and Kw (which is 1.0 x 10^-14, since 298K is 25oC), we can go ahead and calculate [OH-] to be 2.0 x 10^-13, which is the correct answer.
Question 2: Calculate the pH of a 0.20 mol dm-3 ethanoic acid solution at 298K (given that Ka of ethanoic acid is 1.8 x 10^-5 mol dm-3 at 298K).
Woah! This question seems like a big jump from the previous one. Ethanoic acid is a weak acid, which means it does not completely dissociate, thus nullifying the methods we have used previously. Furthermore, what does Ka even mean?
We will explain the terms slowly. Ka refers to the acid dissociation constant, as calculated by the formula below (Fig. 5). It applies mostly for weak acids. In the formula, the concentrations of the dissociated conjugate base and acid have to be applied while the concentration of the acid is at the bottom.
The stronger the acid, the higher the Ka. This is because more dissociation will occur and thus there will be a higher concentration of A- and H+ rather than HA.
Ka = [A-][H+]/[HA]
Fig. 5: Formula for calculation of acid dissociation constant.
Because of the relationship discussed above, Ka is a measure of acid strength (the lower the Ka, the stronger the acid. However, it is inconvenient to use this because of the small values that result (as can be seen for ethanoic acid in the question). Instead, we use pKa, which is derived from Ka by the formula below (Fig. 6).
pKa = -lg Ka
Fig. 6: Formula for calculation of pKa from Ka.
Now, let us move on to solve the question. For this type of question, we draw an I.C.E table (Fig. 7, Initial, Change, Equilibrium), and make the assumption that contribution of protons by autoionization of water is negligible. We also let the proton concentration at equilibrium be x. The numbers indicated refer to the concentrations of the various compounds.
Note that the initial concentrations of CH3COO- and H+ are zero, because the reaction has not started yet!
Fig. 7: I.C.E table.
Now that we have the values for [A-], [H+] and [HA], as well as the value for Ka, we can proceed to solve for x by substituting these values: 1.8 x 10^-5 = x^2/(0.20 - x). We can solve this, although it may seem a little inconvenient.
In chemistry, however, we further simplify things by equating 0.20 - x to approximately 0.20. We can do this because the dissociation expected from a weak acid is very low, making x (the dissociation) negligible. Thus 1.8 x 10^-5 = x^2/0.20. Then we simply solve to give x = 1.90 x 10^-3.
However, that is not the end! The question asks us for the pH. Previously, we discussed the formula pH = -lg[H3O+], which we can use here again by subbing x into the equation. The pH would then be the negative logarithm of x. Doing the relevant calculations gives us 2.72, which makes sense for a weak acid.
In general, it is possible for two other variations of this question to appear. In each of these questions, two out of three of [HA], pH and Ka will be given, while the remaining one will have to be solved. This gives rise to three different types of questions.
We have already given how to solve pH given [HA] and Ka. When pH and either [HA] and Ka are given, we simply find [H+] by the following formula, derived from pH = -lg[H+] (Fig. 8). Then we sub this [H+] and the other given value into the acid dissociation constant equation (Fig. 5).
[H+] = 10-pH
Fig. 8: Derived formula from Fig. 4.
Question 3: Calculate the pH of a 0.05 mol dm-3 sodium ethanoate solution at 298K (given that the Kb of the ethanoate ion is 5.56 x 10-10 mol dm-3 at 298K).
Hmm… we know we can no longer use Ka, and furthermore a different constant, Kb is given. This is known as the base dissociation constant. The formula is now a little bit different, as shown in Fig. 9. Similarly to Ka, the higher Kb, the stronger the base.
Kb = [HA][OH-]/[A-]
Fig. 9: Formula for calculation of base dissociation constant.
Now, let us proceed to calculate out the answer, as we can still compute and substitute the values in. First, we let the hydroxide concentration at equilibrium be x. And for this, we can also draw out an ice table as demonstrated in Fig. 10.
Fig. 10: I.C.E table.
By subbing in the relevant values for Kb and the concentrations, we again obtain something similar to the previous question, 5.56 x 10^-10 = x^2/(0.05 - x). We also assume 0.05 - x is approximately 0.05, by the same principles. Further calculations yield the value of x to be 5.27 x 10-6.
But now we have [OH-]. To get the pH, we must have [H+]. So how do we obtain the [H+] value? By using our ionic product equation, of course! Kw = [H+][OH-]. We know Kw (since it’s at 298K) and [OH-], thus we can calculate [H+] to be 1.897 x 10^-9. And from there we can calculate the pH, which is 8.72.
Question 4: Calculate the pH of a buffer solution that contains 0.05 mol dm-3 of the weak acid HA and 0.15 mol dm-3 of the sodium salt NaA (given Ka of HA = 7.5 x 10^-4 mol dm-3).
Before we move on to answer this question, let us first unpack the meaning of a ‘buffer solution’. A buffer solution simply refers to a solution which resists changes in pH. There are two types of buffer solutions: an acidic buffer solution, and an alkaline one (not a basic one).
The acidic buffer solution is (obviously) acidic, having a pH of below 7. It consists of a mixture of a weak acid and its conjugate base. Crucially, however, the Ka of the acid must be greater than the Kb of the conjugate base. This is because only when Ka > Kb can the mixture be acidic.
For the alkaline buffer solution, the pH has to instead be above 7. The mixture this time is between a weak base and its conjugate acid. The Ka of the acid here must be lower than that of the Kb of the base, so as to allow the buffer solution to become basic.
So, how does the buffer solution work? The buffer solution has both an acid and a base. For example, the base and the conjugate acid could be NH3 and NH4+. As such, if we attempt to introduce a change in pH, i.e. by adding an acid, it would react with the base and the pH would not change. It would work similarly for adding a base.
Let’s move on to solve the question. How do we begin? We can apply the knowledge of the previous question and construct an I.C.E table for HA ⇆ H+ + A-; however, there is a better and faster way, by applying what is known as the Henderson-Hasselbalch equation (Fig. 11).
pH = pKa + lg([A-]/[HA})
Fig. 11: Henderson-Hasselbalch equation for acid buffers.
In the question, we are provided with the values of [A-] and [HA]. Furthermore, we can obtain the pKa from the Ka, which is given. As such, we begin by calculating the pKa (Fig. 6). Then, we can simply calculate the pH by doing the relevant calculations. For those who want the answer, it is -lg(7.5 x 10^-4) + lg(0.15/0.05) = 3.60 (3sf).
The same principle is followed for basic buffer solutions, but in that case the Henderson-Hasselbach equation would be different, as follows (Fig. 12). In that case we will be given different values to be substituted into the equation.
pH = pKa + lg(conjugate base / conjugate acid)
Fig. 12: Henderson-Hasselbalch equation for basic buffers.
Question 5: Calculate the pH of the resultant solution when the respective volumes of 0.50 mol dm-3 NaOH (aq) are added to 25.0 cm3 of 0.50 mol dm-3 CH3COOH (aq): (i) 0.00 cm3, (ii) 15.00 cm3, 25.00 cm3 and 35.00 cm3 (given Ka of CH3COOH is 1.8 x 10-5 mol dm-3 at 298K).
This question is not an easy one to answer, but when we figure out the answer to this one, we should be able to go ahead and answer the question provided at the beginning of this article. (i) is the easy one, given that there is no NaOH. Thus the hydrogen ions contributing to the pH should solely be from the dissociation of CH3COOH.
As such, we draw out an I.C.E table (Fig. 13) for the dissociation of CH3COOH, where x is the concentration of protons at equilibrium:
Fig. 13: I.C.E table.
Also, recall that we are looking at concentrations here. So we do not need to take into account the volume, because the concentration of protons in CH3COOH will remain constant for all its values (thus returning only a single pH value).
The values we get are then substituted into Ka and solved similarly to Question 2, by approximating 0.50 - x to be 0.50. We can then calculate the pH to be 2.52. The workings will not be shown because they have already been covered in Question 2.
For (ii), it gets more interesting. First, we know that NaOH and CH3COOH undergo an acid-base reaction to produce a salt which does not have a pH. As such, we can determine the pH using Fig. 11 (Henderson-Hasselbalch equation), where the salt is the conjugate base and the ethanoic acid is the conjugate acid.
We begin by finding how much of CH3COOH has to react in order to completely remove NaOH. Since they react in a 1:1 ratio, we can calculate the amount of NaOH to determine this. We convert 15.00 cm3 as given in (ii) to 0.015 dm3, which we multiply by 0.5 mol dm-3: 0.015 x 0.5 = 7.5 x 10-3. This is the amount of CH3COOH reacted.
The amount of CH3COOH unreacted (and amount of conjugate base formed) is thus the initial amount subtracted by 7.5 x 10^-3. The initial amount is also calculated by dividing 25.0 cm3 (of the CH3COOH) by 1000, before multiplying by 0.5. After subtraction, we calculate the amount of unreacted CH3COOH to be 5.0 x 10^-3.
Concentrations can be calculated by dividing the numbers we have by the volume of the solution. The volume would be 25 + 15 = 40, for the total volume of NaOH and CH3COOH.
Also take note that we have to divide by 1000 to convert the units to dm3. After dividing the amount of conjugate base and acid by the correct values (we need them for the Henderson-Hasselbalch equation), we get 0.188 and 0.125, respectively.
We can also calculate the pKa since we are given the Ka. Then, we substitute the values into the equation, giving the answer, 4.92.
We move on to look at (iii). Here we have 25.00 cm3 of NaOH. Using the method as previously discussed, we divide by this volume by 1000 to convert to dm3. Then, we multiply by 0.5. This gives 1.25 x 10^-2. The amount of initial CH3COOH is also 1.25 x 10^-2! This means that all of the NaOH and CH3COOH are reacted.
This leads to a solution of the salt, CH3COONa. However, we cannot jump to conclusions and state that there is no pH, because water remains in the mixture! The water can react with this salt by the following reaction (Fig. 14).
CH3COONa + H2O ⇆ CH3COOH + OH-
Fig. 14: Reaction between sodium ethanoate and water.
Since ethanoic acid is a weak acid while the hydroxide ion is a strong base, that leads to a weakly basic solution. We again rely on our old friend the I.C.E table to calculate the concentration of OH- ions (initial concentration of the salt is calculated by dividing 1.25 x 10^-2 by 50, and then multiplying by 1000 to convert to dm units).
Fig. 15: I.C.E table.
Note that H2O is regarded to be in excess and thus not included in the table. We calculate Kb using Fig. 9 and then substituting the relevant values, with 0.25 - x approximately equating to 0.25. We can obtain the value of Kb by using a special equation (Fig. 16).
Kw = Ka x Kb
Fig. 16: Formula for calculation of ionic product of water.
Thus Kb = Kw / Ka. We sub in the relevant values, which we obtain from the question (the ionic product is known). Kb is 5.56 x 10-10. Equating Kb to y^2/0.25 then gives us y = 1.18 x 10^-5. Thus the concentration of OH- at equilibrium is 1.18 x 10^-5.
We use a second special formula, pOH = -lg[OH-], giving us pOH = 4.93. Finally, we use a final third special formula, pOH + pH = 14. Thus pH = 9.07.
pOH = -lg[OH-]
Fig. 17: Formula for calculation of pOH.
pOH + pH = 14
Fig. 18: Formula for pH and pOH.
Finally, what about (iv)? The larger amount of NaOH means that there will be excess NaOH. Since it is a strong base, we ignore all other factors and directly calculate [OH-].
First we calculate unreacted NaOH by subtracting the relevant values (35/1000 x 0.5) - 1.25 x 10^-2. This value is then divided by total volume (25 + 35 = 60) to find concentration of hydroxide ions.
Then we use Fig. 17 and Fig. 18 to find the correct pH, which is 12.92. This means we have finished answering the question!
This article is getting a bit long, so I will say goodbye here. The answer for the question posed at the start will be explained in detail in the next article.