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0:00
Welcome to Unit 8 review. In this unit we will be discussing acids and bases. We have already defined this terminology, but now it’s time to look at this concept in a lot more depth. Specifically we will refine our understanding of these terms, then discuss pH and pOH for strong acids and bases, weak acids and bases, buffers, titrations, molecular structure, pKa, more on buffers, the Henderson-Hasselbalch equation, and buffer capacity.
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So let’s get started by recalling what we already know about acids and bases, and what these substances will do. Recall that the Bronsted-Lowry model focuses on proton transfer as the defining characteristic of an acid-base reaction. Protons are simply hydrogen nuclei, and in the Bronsted-Lowry model, an acid is something that can donate a proton, while a base is something that can accept a proton. This means that an
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acid-base reaction is one in which one compound transfers a proton to another compound. Acids and bases can be neutral compounds, or they can be charged species, either cations or anions. When a molecule acts as an acid, it will lose a proton, and in doing so it will become its conjugate base. This is because if the resulting molecule acts as a base and gains a proton, it will become the original acid molecule again, which is why there are equilibrium
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arrows here. Here is a list of acids and their conjugate bases. Note that as each acid loses an H plus, the resulting base must have a charge that is one less than the acid, so a neutral acid becomes negatively charged, or a positively charged acid becomes neutral. If a molecule acts as a base, it will gain a proton, and in doing so it will become its conjugate acid. This is because if the resulting molecule acts as an acid and loses a proton, it will become
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the original molecule of base again. Looking at this list of bases and their conjugate acids, if a molecule gains an H plus, the resulting acid must have a charge that is one more than the base, so a neutral base becomes positively charged, and a negatively charged base becomes neutral. Now that we understand these definitions, we should be able to look at an acid-base reaction and identify the acid, base, conjugate acid, and conjugate base. See if you can label this
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one. If we track the protons, clearly water has acted as an acid, giving up a proton, and ammonia was the base, accepting the proton. This makes hydroxide the conjugate base, and ammonium is the conjugate acid. Now let’s get a little more technical. When a strong acid dissociates in aqueous solution, every acid molecule will transfer a proton onto a water molecule, forming a hydronium ion, or H3O+. Here is an acid-base reaction involving
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hydrochloric acid, which is a strong acid. Because it is a strong acid, every molecule of acid transfers a proton to a water molecule. Here are some common strong acids. There are weak acids as well, and these also transfer protons to water molecules, but they do not react completely, they will establish some kind of equilibrium between the unionized acid and the conjugate base. Similarly, a strong base will steal a proton from a water molecule to produce a hydroxide ion.
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Some strong bases are comprised of hydroxide ions themselves. Weak bases, just like weak acids, only partially react. We should recall that water can be referred to as amphoteric. This means it can act as either an acid or a base. We saw how water acts as a base in the presence of an acid, becoming hydronium, and in the presence of a base, water can act as an acid, becoming hydroxide. Earlier we learned about chemical equilibria, so it’s time to learn about the intersection
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of this concept with acids and bases by looking at acid-base equilibria. Again, strong acids deprotonate entirely, but weak acids will only deprotonate to an extent, which means that an equilibrium is established between the ionized and unionized form of the acid. Looking at this equilibrium, we also know how to generate the equilibrium expression, only this time instead of Kc, we can use Ka, which is essentially the same thing, we just
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call it the acid dissociation constant, because we are describing the degree to which the acid will ionize. The rest will look exactly the same as what we know. If something were to act as a base, we could do the exact same thing, using Kb, the base dissociation constant. So now let’s go ahead and see how we can determine either Ka or Kb for a weak acid or weak base, given the equilibrium concentrations for a given solution. The main component in vinegar is acetic acid,
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which is what gives it its sour taste. At equilibrium, a vinegar solution contains the following concentrations of acetic acid, hydronium, and the acetate ion, so what will be the Ka value for acetic acid? Well let’s first write the acid-base equilibrium for acetic acid. That looks like this, with acetic acid donating a proton to a water molecule to form hydronium and the acetate anion. And the Ka value relates to the equilibrium concentrations in this way, just as it
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does for any other acid. Luckily we are given equilibrium concentrations, so all we have to do is plug in these values, and do the arithmetic. We will get 1.77 x 10-5 as the Ka value for acetic acid. Now let’s try doing a calculation for a weak base. Caffeine, or C8H10N4O2, is a weak base. If a solution of caffeine at equilibrium displays the following concentrations for caffeine, the conjugate acid of caffeine, and the hydroxide ion, what is the Kb for caffeine? Once again,
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let’s understand that this is the acid-base equilibrium that is happening in this solution, and this is the corresponding Kb expression. All we have to do is plug in our equilibrium concentrations, and again do the arithmetic. We should get 2.5 x 10-4 as the Kb for caffeine. So we now know that if we have an acid-base equilibrium featuring a weak acid like acetic acid, we will be able to calculate the Ka for that acid by plugging in all of the equilibrium
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concentrations into the Ka expression, and solving for Ka. But the thing is, we usually won’t have this data. Often the most practical way to gain information about an acidic or basic solution is to simply to measure the pH, which equals the negative log of the hydronium concentration. And we want to be able to determine the Ka or Kb for a weak acid or base by using pH data. Let’s say we have a solution of nitrous acid with a concentration of 0.0516 molar, and we measure the
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pH of this solution to be 2.34. What is therefore the Ka for nitrous acid? Well let’s write out the complete chemical equilibrium, whereby nitrous acid donates a proton to water to form hydronium and the nitrite ion. We are going to have to set up an icebox, so let’s construct that now. So what can we fill in first? Well let’s see what we have. We have an initial concentration for the acid,
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that’s 0.0516 molar, so that can go here. We also know that at first, before any of the acid reacts, there will be essentially no hydronium, and there will be no conjugate base yet, so we have zeroes in these boxes. Everything is in a one to one ratio stoichiometrically, so filling in the other two rows will be as simple as possible. Now we also have the pH, which is a way to express hydronium concentration. So let’s use that to determine the hydronium concentration
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at equilibrium. Since pH is the negative log of hydronium concentration, the way we get hydronium concentration is 10 to the negative pH, and that gives us 0.0046 molar as the equilibrium concentration for hydronium, so let’s put that in this box. We also know that for every molecule of acid that reacts, we will get precisely one hydronium and one molecule of the conjugate base, so that means the nitrite ion must also have this concentration, since it must be present
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in the same amount as hydronium. That also means this same number was the change in concentration for these species, which means the concentration of nitrous acid must have decreased by precisely the same amount. That will put the equilibrium concentration of nitrous acid at 0.047 molar. Now as we know, to calculate Ka we simply plug these equilibrium values into the Ka expression, do the arithmetic, and we end up with a value of 4.5 x 10-4 as the Ka value for nitrous acid.
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We now know what an acid-base reaction is, and can perform relevant calculations, but how do we know if an acid or base is strong of weak? Let’s cover some basic guidelines. Very generally speaking, the strength of an acid or base is proportional to the stability of its conjugate. If an acid loses a proton and becomes something very stable, like a chloride or iodide ion, the acid will be very strong, as deprotonation is very favorable. If an acid loses
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a proton and becomes something unstable, like a hydroxide ion, the acid must be very weak, as deprotonation is unfavorable and thus unlikely to occur, like it would be for water. So strong acids and bases have weak conjugates. But to expand from this general guideline, there are few more details to consider. First, we can note the location on the periodic table of the central atom in the acid or base. If we are looking at the binary acids in group 17,
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we can note that acidity increases going down the table. This has to do with the size of the anion that will be the conjugate base after deprotonation. The chloride ion is much larger than the fluoride ion, and this affects its ability to accommodate the negative charge. Bromide is larger, and iodide is larger still, which is why hydrobromic acid is stronger than hydrochloric acid, and hydroiodic acid is one of the strongest acids known. We can also
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say that acid strength increases moving to the right. This has to do with electronegativity. The more electronegative the element, the better it can accommodate a negative charge. So if methane were to deprotonate, carbon would be very bad at accommodating the negative charge. Whereas nitrogen is slightly better, oxygen is better still, and fluorine is quite good at accommodating this charge. So the acidity of a central atom increases going down and to the right on the
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periodic table. By the same token, basicity must therefore increase in the opposite directions, because more electropositive elements are more capable of accommodating positive charges. Next we can look at molecules that contain hydroxyl groups. These could be acidic, basic, or amphoteric depending on the nature of the central atom. In order to decide the acidity or basicity of the compound, we have to know whether one bond will break, resulting in hydroxide ions,
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or another bond will break, resulting in the transfer of a proton to a water molecule, forming a hydronium ion. This depends on the strength of these bonds, which is determined by the electronegativity of the central atom. If it has a low electronegativity, it will not hang on to the electrons in this bond very well. That bond will break, resulting in hydroxide ions, which means the compound was a base. If it has a high electronegativity, it will hold
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onto the electrons in that bond very tightly, meaning this other bond is more likely to break, generating hydronium, meaning the compound was an acid. In the latter case, with electronegative central atoms, we are necessarily describing oxyacids. Apart from the strength of an oxyacid increasing along with increasing electronegativity for the central atom, acidity will also increase as the oxidation number of the central atom increases. Nitric acid is a stronger acid
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than nitrous acid because the nitrogen is plus 5 rather than plus 3. And sulfuric acid is stronger than sulfurous acid for the same reason. So when we are looking at acids and bases, there are factors in terms of molecular structure that determine their strength. We must look at the location of the central atom on the periodic table, because this will determine characteristics like ionic radius and electronegativity. We must consider the oxidation state of the central atom.
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And there are other factors like the extent of resonance stabilization that may apply, although that is a bit beyond our scope here. Now it’s time to return to the concept of pH and introduce the related parameter, pOH, and the kinds of calculations we can do with these two values. To set up the appropriate context, we must understand that pure water will autoionize to a small degree.
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This means that it will react with itself, and one water molecule will transfer a proton to another water molecule to generate one hydronium and one hydroxide. This is an acid-base equilibrium, so it must have an equilibrium constant, and for the autoionization of water it has a special symbol, Kw, which will be 1 x 10-14 at room temperature. This value will increase as temperature increases, because at higher temperatures, molecules are moving with more kinetic energy,
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and are thus more likely to transfer a proton upon colliding. We should be able to use information about Kw at various temperatures to calculate the hydronium and hydroxide concentrations in a sample of water. Let’s say we are looking at a sample of pure water at room temperature. We already mentioned that at room temperature the Kw for the autoionization of water will be 1 x 10-14. What will be the concentration of each ion? Well we know that the product of their concentrations
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must equal Kw, and we also know that the hydronium and hydroxide concentrations must be the same, because when a proton is transferred from one molecule to another, we will generate precisely one hydronium ion and one hydroxide ion, so these substances must be present in identical amounts. That means that Kw can be said to be equal to hydronium concentration squared. So we take the square root of both sides, and we can see that hydronium concentration must be equal to 10-7,
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which will be the concentration of hydroxide as well. We could do the same thing for another temperature and Kw. Let’s say at 80 degrees Celsius, the Kw is 2.4 x 10-13. What will be the concentrations here? Again, that means that the Kw is equal to the square of either concentration, so we must simply take the square root and see that each concentration will be equal to 4.9 x 10-7. So now we know that a pure sample of water will contain a tiny bit of hydronium ions
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and hydroxide ions, due to the autoionization of water. We will use hydronium concentration and hydroxide concentration to figure out the acidity or basicity of any solution, as compared to these values for water. We consider water to be neutral, as it has an equal number of hydronium and hydroxide ions, since every time an ionization occurs, it generates one of each. This means that if there is more hydronium than hydroxide in a solution, we will consider the
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solution to be acidic, and if there is more hydroxide than hydronium, we will consider the solution to be basic. In order to assess the acidity or basicity of a solution, we will therefore have to be able to discuss hydronium and hydroxide concentration. As we have seen, these concentrations usually have values in scientific notation, which isn’t very palatable, so we have developed a way to translate these values into numbers that are easier to communicate. We will
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use a logarithmic scale, since concentrations can often span many orders of magnitude. This will be done using what is called the p function, where pX will represent the negative log of X. We can apply this in a number of different ways. The pH of a solution is the negative log of the hydronium concentration. So pH will measure the acidity of the solution. Since a hydronium concentration
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above 10-7 indicates an acidic solution, it will be a pH below 7 that indicates an acidic solution, because of how logarithms work. A pH above 7 will indicate a basic solution. Likewise, we can calculate the pOH of a solution by taking the negative log of the hydroxide concentration. This will tell us the basicity of a solution. Here it’s the reverse of pH. Low pOH is basic, high pOH is acidic. We can also arrive at hydronium or hydroxide concentration given the pH or pOH,
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we just rearrange these equations. When no base is listed for a logarithm, it is implied to be 10, so that means that if we first divide both sides by negative one, we can then show that 10 to the negative pH will give us the hydronium concentration. Likewise, for pOH, dividing first by negative one means that 10 to the negative pOH will give us the hydroxide concentration.
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Lastly, because of these relationships, we can derive one last equation, and that is as follows. pH plus pOH will equal 14, since Kw equals 10-14. With all of these relationships we should be able to get from any one of these values to any other in a given solution. Let’s do a quick example. Say we have 0.0235 molar HCl solution. What is its pH, and what is its pOH? Well as we said, pH is equal to the negative log of the hydronium
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concentration. HCl is monoprotic and a strong acid, which means it will completely deprotonate, and this will therefore also be the concentration of hydronium. We just take the log base ten, flip the sign, and we should get 1.63 as our answer. Now we also said we wanted the pOH, luckily this is also very easy to get from the pH. If pH plus pOH equals 14, then let’s just subtract the pH from 14 to get 12.37, and that will be the pOH.
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Let’s now return to the concept of titration and get a more quantitative understanding of this technique. As we recall, titration is a form of stoichiometry, where a neutralization reaction can be monitored with a colored indicator to determine the concentration of an acid or base. The indicator will turn the solution a certain color once a precise stoichiometric amount of acid or base has been added so as to precisely
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neutralize the other substance. But now that we have a better understanding of acid-base calculations, we can take a closer look at the pH as it changes during a titration. First let’s examine the simplest situation, the titration of a strong acid with a strong base, by looking at this titration curve. Let’s say we have 25 milliliters of 0.1 molar hydrochloric acid, and we are titrating with 0.1 molar sodium hydroxide. At first there is only hydrochloric acid in solution,
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so the pH starts off quite acidic, around 1. As base is added, some of the acid is neutralized, slowly raising the pH. Once 25 milliliters are added, this is enough to reach the equivalence point. This is where there is precisely enough base present to neutralize all the acid, which means there will be no acid or base, only water and the sodium chloride salt that results from the acid-base reaction, so the pH will be completely neutral at this point. Then as more base is added,
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the pH begins to climb into basic territory, as there is no acid left to neutralize it. When titrating using any strong acid and strong base, the equivalence point will always be 7, but things get trickier when one of the species is weak. Let’s examine a situation where we keep the previous volumes and concentrations the same, but this time a weak acid like acetic acid is titrated with strong base. Because acetic acid is weak, not all of it will ionize, which means there
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is less hydronium in solution than for a solution with hydrochloric acid of the same concentration. So we begin at a pH of 3 this time. We can see that as sodium hydroxide is added, the pH will climb more rapidly than the previous example. The equivalence point will also climb higher than the previous example, because even when enough base has been added to neutralize all the acetic acid,
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there will then be acetate in solution, which is strong enough to react with water in solution to generate some hydroxide, meaning the solution will be basic at the equivalence point. After the equivalence point, the pH will continue to climb as more base is added. We can see that the behavior for each curve is different leading up to the equivalence point, due to the initial discrepancy in acidity, but once the acid is completely neutralized, the behavior
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after the equivalence point is identical, as this simply represents the addition of more hydroxide. Now that we know how pH varies during titration, it is time to learn how to do quantitative calculations regarding the pH of a solution at various points during a titration. These will essentially be stoichiometric calculations, because we will simply have to know how much acid or base we started with and how much acid or base was neutralized upon the addition of
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a certain volume of titrant, this allows us to calculate the amount of hydronium or hydroxide remaining in solution and therefore the pH. Let’s return to our previous example of 25 milliliters of 0.1 molar hydrochloric acid being titrated with 0.1 molar sodium hydroxide. We can calculate the pH of the solution after any amount of base has been added, so let’s do this in increments of 12.5
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milliliters, from zero to 37.5. This will be easy since both the acid and base are strong, so we can assume complete dissociation of both species. Before any base has been added, we simply have 0.1 molar hydrochloric acid, which will be equal to hydronium concentration, so we can just take the negative log to get 1 for the pH. After we start adding base, we need to figure out how many
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moles of acid are neutralized. 25 milliliters, or 0.025 liters, times 0.1 moles per liter, will be equal to 0.0025 moles of acid. If 12.5 milliliters of base are added, that will be 0.00125 moles of base. That is enough to neutralize half the acid, so we will get 0.00125 moles of acid left over, divided by the new volume of 0.0375 liters, and that gives us a hydronium concentration of 0.033,
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and a pH of 1.48. After 25 milliliters, we have precisely the same amount of acid and base, so we will get a completely neutral solution of pH 7. After 37.5 milliliters of base, now we will have excess hydroxide. 25 milliliters or 0.0025 moles were neutralized by the acid, so we have an additional 0.00125 moles of hydroxide remaining. Dividing by the total volume of acid plus base, which is 0.0625 liters gives us a hydroxide concentration of 0.02 molar, for a pOH of 1.7,
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and subtracting from 14 gives a pH of 12.3. Another important concept is buffers and buffer capacity, so let’s make sure to define these terms as well. A buffer is a mixture of a weak acid and its conjugate base, or a weak base and its conjugate acid, that bestows the solution with an ability to resist changes in pH when small amounts of acid or base are added to it. An example of a buffer solution would be one containing acetic acid and
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sodium acetate, which provides the acetate ion. This comprises a conjugate acid-base pair for acetic acid, which is a weak acid. Another such example involving a weak base and its conjugate acid would be a buffer solution made from ammonia and ammonium chloride, which provides the ammonium ion. Just one proton transfer separates these substances. Buffer solutions function as follows. The Ka of acetic acid is larger than the Kb of acetate, so the forward reaction of the
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top equilibrium will be more prevalent, and there will be hydronium ions in solution. Therefore if we add a little bit of strong base, it will react with the few hydronium ions present, and since the ionization of acetic acid is an equilibrium, removing hydronium will push the equilibrium to the right and more acetic acid will ionize to restore the hydronium concentration back to its original value, as we would expect from our knowledge of Le Chatelier’s principle.
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This means the pH will change very little, or not at all. Likewise, if we add a strong acid, it will simply react with the acetate ions instead of water molecules, which will only generate more acetic acid, so this too will have very little effect on the pH, since no hydronium or hydroxide is formed in that ionization. This diagram shows the ways that the equilibrium between acetic acid and acetate will shift if acid or base are added, and the direction it
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must shift after the fact in order to equilibrate. We can start with equal amounts of the conjugate acid-base pair. If we add more acid, we can see that the equilibrium shifts to the left, and it will move to achieve equilibrium. If we add base, the opposite will occur. Whichever way we go, the system will balance and resist change in pH. Let’s do one calculation. What would be the pH of a buffer solution that is 0.1 molar in acetic acid and 0.1 molar in sodium acetate,
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given the following Ka for acetic acid? We already know that the Ka for acetic acid is greater than the Kb for acetate, so we can conclude that the equilibrium will shift right. Let’s go ahead and put these initial concentrations into an icebox. That’ll be 0.1 for the acid and conjugate base, and there’s no hydronium yet, or it is at least negligible. Then, we will put negative X for the
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change in acetic acid, and X for the change of the products, since we are moving forward. Hence these will be the equilibrium concentrations. We plug those into the equilibrium expression and solve for X, which will be the hydronium concentration, and from there we just take the negative log to arrive at the pH, which will be 4.74. So by applying our knowledge of acid-base equilibria, we can understand how a buffer solution is able to maintain a constant pH value.
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As we can see, buffer solutions are able to resist a change in pH when small amounts of strong acid or base are added to them due the equilibrium that exists between the weak acid or base and its conjugate. But as we have said, this is for small amounts of acid or base. Every buffer solution has a limited capacity to resist a change in pH, and once a certain amount of acid or base has been
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added, it will no longer be able to function. As we add more and more acid or base, the buffering action of the solution will diminish, and once all of the weak acid or weak base has been exhausted, the solution will no longer have any buffer capacity, and there will be no resistance to change in pH. As we can see here, in the beaker on the left we have a buffered solution at pH 8, and the addition of a small amount of acid will not have a substantial effect on the pH,
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which is why the beaker in the middle experiences only a small change in pH despite the addition of acid. But once we add too much acid, as we can see from the dramatic increase in volume in the beaker on the right, the buffer capacity of the solution is exhausted and the pH drops all the way to 1.53. We will define the buffer capacity of a solution as being the amount of acid or base that can be added to a certain volume of the buffer solution before the pH changes significantly,
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usually by at least one pH unit. Buffer capacity will certainly depend on the amount of acid and base that are in solution, which is why a buffer solution that is one molar in acetic acid and one molar in sodium acetate will have a much greater buffer capacity than a solution that is 0.1 molar in each substance. These solutions will have the same pH because they involve precisely the same equilibrium. But the more acetic acid molecules and acetate ions there are,
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the more molecules of strong acid or base that can be neutralized, and hence a greater resistance to change in pH as acid or base is continually added. In discussing buffer solutions, there is an equation that must be mentioned, and it is the Henderson-Hasselbalch equation. Before we discuss the implications of this equation, let’s first see how it is derived. If we start from the Ka expression for a generalized acid, we might want to solve for hydronium concentration in order
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to discuss the pH of a solution. To do this, we multiply both sides by the concentration of HA, and divide both sides by the concentration of A minus. That will give us this. Now we know to get pH we simply take the negative log, so let’s take the negative log of both sides. The log of the product on the right side can become the sum of two different logs because of the rules of logarithms, so this is what we end up with. From this we can simply express
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a few things differently. We know that the negative log of hydronium concentration is pH. The negative log of Ka is called the pKa of an acid. And because of the rules of logarithms, we can change this negative log to a positive log, for mathematical simplicity, if we flip this fraction upside down. That gives us the Henderson-Hasselbalch equation, which is useful for preparing or analyzing buffer solutions. There are two contexts in which this can be used. First,
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when looking at an existing buffer solution, if the pKa of the acid involved is known, and we have a way to measure the concentration of both the acid and the conjugate base, then we can plug these values in to solve for the pH of the solution. But in addition to this, when preparing buffer solutions, we know that we will want to use the acid and conjugate base in equal amounts. If we do so, then this numerator and denominator will be equal, meaning this
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fraction will be equal to one, and the log of 1 is always zero, so this term disappears completely, leaving us with pH = pKa. That means that when preparing a buffer solution, if the acid and conjugate base are used in equal concentrations, the pH of the solution will be equal to the pKa of the acid. Since there are so many acids with pKa’s tabulated that we can refer to, this will allow us to generate a buffer solution of virtually any pH by selecting an acid with a pKa that is
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equal to the desired pH of the buffer solution. And that concludes our Unit 8 review. I’ll see you in Unit 9, where we will discuss some applications of thermodynamics.
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