MCAT General Chemistry Review
Part I Review
Chapter 10: Acids and Bases
There are many ways that drugs can enter the human body. The route of administration of a drug is the path by which that drug comes into contact with the body. According to the U.S. Food and Drug Administration’s Data Standards Manual, there are no fewer than 110 distinct ways in which a drug can come in contact with and/or enter the human body in a local or systemic manner, including the catch-all route of “other.” Some drugs can be applied as drops, salves, or creams to mucus membranes. Others are injected. Some employ a transdermal patch, while others are eaten, drunk, or inhaled. You will be challenged in medical school to learn and recall the routes of administration for many commonly prescribed medications and treatments. This is one of the more daunting memorization tasks that will be demanded of you during medical school and residency.
The route of administration of a drug compound is related to both the location of its target tissue (local or systemic), as well as the chemical and physical properties of the compound. For example, compounds that are water-soluble can be administered intravenously (an aqueous solution dripped directly into the bloodstream), while those that are lipid-soluble can be administered transcutaneously (from, say, a patch or a cream) or orally (in a pill or liquid suspension). The polarity, size, and charge of the drug compound will determine its solubility in polar or nonpolar environments and will be major contributing factors to the most effective and efficient route of administration.
Whether a drug compound has an ionic charge is usually a function of the acidic or basic nature of the compound. For example, a basic organic compound that is water-insoluble when neutral can be reacted with an acid to form a salt that, because it is ionic, will be water-soluble. Correspondingly, an acidic organic compound that is water-insoluble when neutral can be reacted with a base to form a water-soluble salt. On the other hand, the protonated (acidic) form of an organic compound can be reacted with a base to neutralize the compound and release it from its salt, changing (and usually reversing) its solubility in water.
Medical professionals aren’t the only ones concerned about drug solubilities and routes of administration. There’s a science to illegal drugs, too. People in the general population who use illegal drugs (and those who produce them) are knowledgeable about their available forms, as well as the most effective and efficient modes of delivery. One of the clearest examples of this is the difference in the ways that people use the two forms of cocaine. C17H21NO4 (cocaine) is a large alkaloid compound derived from the coca plant. It is a central nervous stimulant that has been used medicinally, ceremonially, and recreationally since at least the pre-Columbian era. About 125 years ago, Pope Leo XIII purportedly carried around a hipflask filled with cocaine-laced wine called Vin Mariani. The 1886 original recipe for what is now the world’s most famous cola included coca leaves (from which this famous cola derives its name). Cocaine was once used to treat heroin addiction. Sigmund Freud wrote rhapsodically about its ability to cause “exhilaration and lasting euphoria.” By 1903, however, the American Journal of Pharmacy was warning that most cocaine abusers were “bohemians, gamblers, high- and low-class prostitutes, night porters, bell boys, burglars, racketeers, pimps, and casual laborers.”
Today, cocaine is used primarily in two different forms. Most commonly, the alkaloid compound is reacted with hydrochloric acid (which protonates its tertiary amine functional group), extracted with water, and dried to an aqueous soluble powder (cocaine hydrochloride); this powder either is snorted (insufflated) into the nasal cavity, where it is absorbed into the capillary beds, or is injected directly into the venous circulation. The salt form, however, because it has a very high boiling point close to the temperature at which it burns, cannot be smoked. To produce a form of cocaine that can be vaporized and inhaled from a pipe, the cocaine hydrochloride must be reacted with a base, typically either ammonia (to produce pure “freebase cocaine”) or sodium bicarbonate (to produce the less-pure “crack” cocaine). The base reacts with the protonated tertiary amine, removing the hydrogen ion to re-form the neutral alkaloid compound. The freebase cocaine is water-insoluble and usually extracted with ether, or it is left in the aqueous solution, which is heated and evaporated. The freebase or crack form of cocaine has a much lower boiling point; consequently, it can be smoked without risk of burning (combusting).
What a difference a little hydrogen ion can make! The complexities of drug delivery can in part be related to the presence or absence of the hydrogen ion—a mere proton! In this chapter, our focus will be those two classes of compounds—acids and bases—that are involved in so many important reactions. Acid-base reactions are an important focus for the MCAT; in fact, the neutralization reaction is one of the most commonly tested reaction types on Test Day. We will begin with a review of the different definitions of acids and bases and their properties, including the characterization of acids and bases as either strong or weak. Focusing on weak acids and bases, we will discuss the significance of the equilibrium constants, Ka and Kb, for acids and bases, respectively. Finally, we will review acid-base titrations and buffer systems.
Over the last century, chemists have used different definitions to identify compounds as acids or bases. Three definitions have been proposed, and each is progressively more inclusive: Every Arrhenius acid (or base) can also be classified as a Brønsted-Lowry acid (or base), and every Brønsted-Lowry acid (or base) can also be classified as a Lewis acid (or base).
The first definitions of acids and bases were formulated by Svante Arrhenius toward the end of the 19th century. Arrhenius defined an acid as a species that dissociates in water to produce a hydrogen ion, H+, and a base as a species that dissociates in water to produce a hydroxide ion, OH-. These definitions, though useful for many reactions, fail to describe acidic and basic behavior in nonaqueous media.
This is the most specific definition of acids and bases and is the least useful on the MCAT.
A more general definition of acids and bases was proposed independently by Johannes Brønsted and Thomas Lowry in 1923. A Brønsted-Lowry acid is a species that donates hydrogen ions, while a Brønsted-Lowry base is a species that accepts hydrogen ions. The advantage of this definition over Arrhenius’s is that it is not limited to aqueous solutions. For example, OH-, NH3, and F- are all Brønsted-Lowry bases because each has the ability to accept hydrogen protons. However, neither NH3 nor F- can be classified as Arrhenius bases because they do not dissociate to produce OH-ions in aqueous solutions. You’ll notice, however, being the perceptive student that you are, that according to both of these definitions, there’s only one way for a species to be an acid, and that is to produce a hydrogen ion. The only difference between the two definitions for acidic compounds is the requirement (or lack thereof) of an aqueous medium in the Arrhenius definition. Most acid-base chemistry on the Physical Sciences section of the MCAT will involve the transfer of hydrogen ions in accordance with the Brønsted-Lowry definitions.
The Brønsted-Lowry definition is a more general description and much more useful and common on the MCAT than the other two. It is all about the proton (H+).
Brønsted-Lowry acids and bases always occur in pairs because the definitions require the transfer of a proton from the acid to the base. These are conjugate acidbase pairs (see below). For example, H3O+ is the conjugate acid of the base H2O, and NO2- is the conjugate base of HNO2.
H3O+ (aq) H2O (aq) + H+ (aq)
HNO2 (aq) NO 2- (aq) + H+ (aq)
At approximately the same time as Brønsted and Lowry, Gilbert Lewis also proposed definitions for acids and bases. Lewis defined an acid as an electron-pair acceptor (acid = acceptor) and a base as an electron-pair donor. Lewis’s are the most inclusive definitions: Every Arrhenius acid is also a Brønsted-Lowry acid, and every Brønsted-Lowry acid is also a Lewis acid (and likewise for the bases). However, the converse is not true: The Lewis definition encompasses some species not included within the Brønsted-Lowry definition. For example, BCl3 and AlCl3 are species that can each accept an electron pair, which qualifies them as Lewis acids, but they will not donate a hydrogen ion, which disqualifies them as Brønsted-Lowry acids (or Arrhenius acids, for that matter). On the MCAT, you may encounter Lewis acids more often in the Biological Sciences section, specifically in the organic chemistry reactions for which Lewis acids act as catalysts, such as in the anti-addition of diatomic halogens to alkenes.
This is the most general and inclusive description. Lewis acids tend to show up as catalysts in organic chemistry reactions.
NOMENCLATURE OF ARRHENIUS ACIDS
The name of an Arrhenius acid is related to the name of the parent anion (the anion that combines with H+ to form the acid). Acids formed from anions whose names end in –ide have the prefix hydro- and the ending –ic.
Acids formed from oxyanions are called oxyacids. If the anion ends in -ite (less oxygen), then the acid will end with –ous acid. If the anion ends in –ate (more oxygen), then the acid will end with –ic acid. Prefixes in the names of the anions are retained. Some examples include the following:
There are always some exceptions to the rules. For instance, MnO4- is called permanganate even though there are no “manganate” or “manganite” ions, but this will not be important to memorize for Test Day.
Properties of Acids and Bases
Acids and bases are usually characterized according to their relative tendencies either to donate or to accept hydrogen ions. Furthermore, aqueous acid and base solutions can be characterized according to their concentrations of hydrogen and hydroxide ions.
AUTO-IONIZATION OF WATER AND HYDROGEN ION EQUILIBRIA
Since many acid-base reactions take place in water—and on the MCAT this is almost exclusively the case—it is very important that you understand the behavior of acidic and basic compounds vis-à-vis the acid-base behavior of water. Only then can you fully appreciate the meaning and significance of such terms as strong acid or weak base or measurements of pH or pOH.
The Acid-Base Behavior of Water
Water is a member of a unique class of compounds in the world of acids and bases. The H2O molecule can act as either an acid or a base, depending on the acid-base nature of the species with which it is reacting. Water acts as an acid by donating one of its hydrogen ions, and it acts as a base by accepting a hydrogen ion. This leads us to the definition of an amphoteric species: one that in the presence of a base reacts like an acid and, in the presence of an acid, reacts like a base. As an amphoteric compound, water can react with itself, in a process called auto-ionization, in the following manner:
One water molecule donates a hydrogen ion to another water molecule to produce the hydronium ion (H3O+) and the hydroxide ion (OH+). By the way, some of you may be used to seeing the hydrogen ion represented simply as H+, rather than as H3O+. This is fine, but it’s important to remember that the proton is never just “free floating” in the solution; it’s always attached to water or some other species that has the ability to accept it. Auto-ionization of water is a reversible reaction; therefore, the above equation is an equilibrium expression for this reversible reaction. For pure water at 298 K, the water dissociation constant, Kw, has been experimentally determined and is
Kw = [H3O+][OH-] = 10-14 at 25°C (298 K)
Because each mole of water that auto-ionizes produces one mole each of hydrogen (or hydronium) ions and hydroxide ions, the concentrations of the hydrogen ions and hydroxide ions are always equal in pure water at equilibrium. Thus, the concentration of each of the ions in pure water at equilibrium at 298 K is 10-7 mol/L.
The concentrations of the two ions will not always be equal. In fact, they will only be equal when the solution is neutral. Nevertheless, the product of their respective concentrations must always equal 10-14 when the temperature of the solution is 298 K. For example, if a species is added to pure water and that species donates hydrogen ions to the water (i.e., the species is an acid), then the hydrogen ion concentration will increase, causing the water system to shift to reverse the auto-ionization process. The result is a decrease in the hydroxide ion concentration and a return to the equilibrium state. This is nothing other than Le Châtelier’s principle in action as we’ve seen time and time again: The addition of product to a system at equilibrium (in this case, the addition of H+ to the water system at equilibrium) causes the system to shift in direction away from the products, toward the reactants. The shift away from the product side necessarily leads to a decrease in the concentration of the hydroxide ion such that the product of the concentrations of the dissolved ions equals the Kw. The addition of a species that accepts hydrogen ions (i.e., a base), resulting in a decrease in the hydrogen ion concentration, will cause the water system to shift forward to replace the hydrogen ions. The increase in auto-ionization will necessarily lead to an increase in the hydroxide ion concentration and a return to the equilibrium state.
Before we introduce the scales used to measure the concentrations of hydrogen ions and hydroxide ions in different acid-base solutions, we want to emphasize the important thermodynamic principle, often unnoticed by students, contained in the water dissociation constant (Kw) expression. The Kw is an equilibrium constant; unless the temperature of the water is changed, the value for Kw cannot be changed. Thus, the product of the concentrations of the hydrogen ions and the hydroxide ions in the aqueous solution at 298 K must always equal 10-14. At different temperatures, however, the value for Kw changes. At temperatures above 298 K, the value for Kw will increase, a direct result of the endothermic nature of the auto-ionization reaction.
pH and pOH Scales
The concentrations of hydrogen ions and hydroxide ions in aqueous solutions can vary significantly, and the vastness of the range makes measurements on a linear scale unmanageable. The scales of concentrations for acidic and basic solutions are condensed into something more manageable by being expressed in logarithmic terms, just like the decibel scale for sound intensity. These logarithmic scales are the pH and the pOH scales for the concentrations of the hydrogen and hydroxide ions, respectively.
We find that in many cases, the reactivity of a reaction involving an acid is not a function of hydrogen ion concentration but instead the logarithm of the hydrogen ion concentration (just as loudness of sound is a function of the logarithm of sound intensity). As a result, we often use the logarithmic pH and pOH scales to express the concentrations of the hydrogen and hydroxide ions, respectively.
pH and pOH are specific calculations of the more generic “p-scale.” A p-scale is defined as the negative logarithm of the number of items. There’s no reason why this logarithmic calculation cannot be applied to the population values of all the world’s countries [pPop = -log(population)] or to the number of hairs on the head of every living human [pHair = -log(hairs)]. The log scale system could be applied to anything, but let’s be real: pPop and pHair aren’t going to earn us any Test Day points. It’s much more valuable for us to understand the significance of the p-scale expression for the concentrations of hydrogen ions and hydroxide ions in acid and base aqueous solutions.
The pH of a solution is given by
pH = -log[H+] = log(1/[H+])
Likewise, the pOH of a solution is given by
pOH = -log[OH-] = log(1/[OH-])
For pure water at equilibrium at 298 K, the concentration of the hydrogen ion equals the concentration of the hydroxide ion and is 10-7 mol/L. Therefore, pure water at 298 K has a pH of 7 and a pOH of 7. (Note: -log 10-7 = 7.)
From the water dissociation constant expression (Kw = [H3O+][OH-] = 10-14), we find that
pH + pOH = 14 (for aqueous solutions at 298 K)
This equation demonstrates a fundamental property of logarithms: The log of a product is equal to the sum of the logs; that is, log(xy) = log x + log y.
For any aqueous solution at 298 K, a pH less than 7 (or pOH greater than 7) indicates a relative excess of hydrogen ions, and the solution is acidic; a pH greater than 7 (or pOH less than 7) indicates a relative excess of hydroxide ions, and the solution is basic. A pH (or pOH) equal to 7 indicates equal concentrations of hydrogen and hydroxide ions, resulting in a neutral solution.
Estimating p-Scale Values
An essential skill that you must hone for Test Day, applicable to many problems involving acids and bases, is the ability to convert pH, pOH, pKa, and pKb values quickly into nonlogarithmic form and vice versa.
When the original value is a power of 10, the operation is relatively straightforward: Changing the sign on the exponent gives the corresponding p-scale value directly. For example,
If [H+] = 0.001 or 10-3, then the pH = 3 and pOH = 11. If Kb = 1.0 × 10-7, then pKb = 7.
Other important properties of logarithms include these: Log x n = n Log x and Log 10 x = x. From these two properties, one can derive the particularly useful relationship that will be seen on Test Day (and we can see in the example):-Log 10-x=x.
More difficulty arises (in the absence of a calculator or a superhuman ability to calculate logarithms in your head) when the original value is not an exact power of 10. The MCAT is not a math test, and it is not primarily interested in determining your ability to perform mathematical calculations. Exact calculation of the logarithmic value of a number that is not an integer power of 10 will be excessively onerous, if not outright impossible. The test writers are interested, however, in testing your ability to apply mathematical concepts appropriately in solving certain problems. Fortunately, there is a simple method of approximation that will be foolproof for Test Day.
If the nonlogarithmic value is written in proper scientific notation, it will look like n × 10-m, where n is a number between 1 and 10. Using the basic log rule that the log (xy) = log x + log y, we can express the negative log of this product, -log (n × 10-m), as -log (n) - log (10-m). Since log refers to the common log with base of 10, we can simplify -log (n) - log (10-m) to -log (n) - (-m), or m - log (n). Now, since n is a number between 1 and 10, its logarithm will be a fraction between 0 and 1 (note: log 1 = 0 and log 10 = 1). Thus, m - log (n) will be between (m - 1) and (m - 0). Furthermore, the larger n is (that is, the closer to 10), the larger the fraction, log (n) will be; consequently, the closer to (m - 1) our answer will be. If this is too much to remember—and given the amount of information you need to remember for Test Day, it might be—all you need to remember is this: When the nonlogarithmic value is n × 10-m, the logarithmic value will be between (m - 1) and m. It’s that simple!
Example: If Ka = 1.8 × 10-5, then pKa = 5 - log 1.8. Because 1.8 is small, its log will be small, and the answer will be closer to 5 than to 4. (The actual answer is 4.74.)
Learning how to estimate when using logarithms is an important skill that can save us a lot of time on Test Day.
STRONG ACIDS AND BASES
Strong acids and bases are those that completely dissociate (or nearly so) into their component ions in aqueous solution. For example, when sodium hydroxide, NaOH, is added to water, the ionic compound dissociates, for all intents and purposes, completely according to the net ionic equation:
NaOH (s) = Na+ (aq) + OH- (aq)
Hence, in a 1 M solution of NaOH, complete dissociation gives 1 M OH-. The pH and pOH for this solution can be calculated as follows:
pH = 14 - pOH = 14 - (-log[OH-]) = 14 + log (1) = 14 + 0 = 14
Virtually no undissociated strong acid or base, such as NaOH, remains in solution, and we can consider the dissociation of strong acids and bases essentially as going to completion. In the NaOH example given above, you should note that in calculating the pH, we assumed that the concentration of OH- associated with the auto-ionization of water is negligible compared to the concentration of OH- due to the addition of the strong base. The contribution of OH- and H+ ions to an aqueous solution from the auto-ionization of water can be neglected only if the concentration of the acid or base is significantly greater than 10-7 M. Keeping this in mind as you solve acid-base problems on Test Day will help you avoid “silly” mistakes. For example, with your brain on “autopilot,” you might calculate the pH of a 1 × 10-8 M solution of HCl (a strong acid) as 8, because -log (10-8) = 8. But a pH of 8 can’t possibly describe an acidic solution (at least not at 298 K), because the presence of the acid will cause the hydrogen ion concentration to increase above 10-7 and the pH must be below 7.
Always be sure to think about the answer and whether it makes sense on Test Day. As seen in this paragraph, a basic pH for an acidic solution should make you think about what might be wrong with your answer.
So what went wrong in this case? The error was in not recognizing that the acid compound concentration is actually ten times less than the equilibrium concentration of hydrogen ions in pure water generated by water’s autodissociation. Consequently, the hydrogen ion concentration from the water itself is significant and can’t be ignored. Now, the addition of the acid results in the common ion effect (Le Châtelier’s principle in action in ionic solutions) and causes the water system to shift away from the side of the ions, thereby reducing the concentration of hydrogen ions and hydroxide ions. The reversal of auto-ionization is thermodynamically favored to return the water system to equilibrium, and we can express this mathematically as
Kw = [H3O+][OH-] = [x + (1.0 × 10-8)](x) = 10-14
where x = [H3O+] = [OH-] from the auto-ionization of water. Solving for x (using a calculator!) gives x = 9.5 × 10-8 M. The total concentration of hydrogen ions is [H+]total = 9.5 × 10-8 + 1.0 × 10-8 = 1.05 × 10-7 M. Notice that this indeed is slightly less than what the value would be if the common ion effect were not acting here [(1.0 × 10-7)+ (1.0 × 10-8) = 1.1 × 10-7]. The pH of this acid solution can now be calculated as pH = -log (1.05 × 10-7) = 6.98. This value is slightly less than 7, as it should be expected for a very dilute acidic solution. The point of all of this is: Don’t put your brain on autopilot on Test Day. Be alert and keep thinking critically, no matter how familiar the problem setups might seem to you!
The Kw (like all equilibrium constants) will change if the temperature changes and, in turn, will change the pH scale. So be careful on the MCAT because our pH scale of 1-14 is only valid at 25°C.
Strong acids commonly encountered in the laboratory and on the MCAT include HCl (hydrochloric acid), HBr (hydrobromic acid), HI (hydroiodic acid), H2SO4 (sulfuric acid), HNO3 (nitric acid), and HClO4 (perchloric acid). Strong bases commonly encountered include NaOH (sodium hydroxide), KOH (potassium hydroxide), and other soluble hydroxides of group IA and IIA metals. Calculation of the pH and pOH of strong acids and bases assumes complete dissociation of the acid or base in solution: [H+] = normality of strong acid and [OH-] = normality of strong base.
WEAK ACIDS AND BASES
Before we go any further in our discussion of acids and bases as “strong” or “weak,” we want to ensure that you are making the mental distinction between the chemical behavior of an acid or base with respect to its tendency to dissociate (e.g., strong bases completely dissociate in aqueous solutions) and the concentration of acid and base solutions. Although we may casually describe a solution’s concentration as “strong” or “weak,” it is preferable to use the terms concentrated and dilute, respectively, because they are unambiguously associated with concentrations, not chemical behavior.
Continuing our focus on the chemical behavior of acids and bases, we now must consider those acids and bases that only partially dissociate in aqueous solutions. These are called weak acids and bases. For example, a weak monoprotic acid, HA, will dissociate partially in water to achieve an equilibrium state:
HA (aq) + H2O (l) H3O+ (aq) + A- (aq)
Since the system exists in an equilibrium state, we can write the dissociation equation to determine the acid dissociation constant Ka as
The smaller the Ka, the weaker the acid, and consequently, the less it will dissociate. Note that the concentration of water, while not seemingly included in the dissociation constant expression, is actually incorporated into the value of Ka (Keq [H2O] = Ka).
A weak monovalent base, BOH, undergoes dissociation to yield B+ and OH- in solution:
BOH (aq) B+ (aq) + OH- (aq)
And the base dissociation constant Kb can be calculated according to
Weak acids and bases are often seen on the MCAT in the Biological Sciences section.
The smaller the Kb, the weaker the base, and consequently, the less it will dissociate. As with the acid dissociation expression, the base dissociation expression incorporates the concentration of the water in the value of the Kb itself.
Generally speaking, we can characterize a species as a weak acid if its Ka is less than 1.0 and as a weak base if its Kb is less than 1.0. On the MCAT, molecular (that is, nonionic) weak bases are almost exclusively amines.
CONJUGATE ACID/BASE PAIRS
Since Brønsted and Lowry define an acid-base reaction as one in which a hydrogen ion (proton) is transferred from the acid to the base, the acid and base always occur in pairs called conjugates. A conjugate acid is the acid formed when a base gains a proton, and a conjugate base is the base that is formed when an acid loses a proton. For example,
HCO3- (aq) + H2O CO32- (aq) + H3O+ (aq)
Be aware of the relationship between conjugate acids and bases because you will need to recognize these on the MCAT. Taking a proton from a molecule will give you the conjugate base, and putting a proton on will give you the conjugate acid!
The CO32- is the conjugate base of the HCO3- acid, and the H3O+ is the conjugate acid of the H2O base. To find the Ka, we consider the equilibrium concentrations of the dissolved species:
The reaction between bicarbonate (HCO3-) and water is reversible, so we can write it as follows:
CO32- (aq) + H2O (aq) HCO3- (aq) + OH- (aq)
And write the Kb for CO32- as follows:
In a conjugate acid-base pair formed from a weak acid, the conjugate base is generally stronger than the conjugate acid. Note that this does not necessarily mean that a weak acid will produce a strong conjugate base or that a weak base will produce a strong conjugate acid, although it is always the case that a strong acid will produce a weak conjugate base (e.g., HCl/Cl-) and a strong base will produce a weak conjugate acid (e.g., NaOH/H2O). As it turns out, for HCO3- and CO32-, the reaction of CO32- with water to produce HCO3- and OH- occurs to a greater extent (that is, is thermodynamically more favorable) than the reaction of HCO3- and water to produce CO 32- and H3O+.
When you add the previous two reversible reactions, you see that the net is simply the dissociation of water:
H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)
Since the net reaction is the auto-ionization of water, the equilibrium constant for the reaction is Kw = [H3O+][OH-] = 10-14, which is the product of Ka and Kb. Remember: The product of the concentrations of the hydrogen ion and the hydroxide ion must always equal 10-14 for acid or base aqueous solutions. Because water itself is an amphoteric species (both a weak acid and a weak base), all acid-base reactivity in water ultimately reduces to the acid-base behavior of water, and all acid or base aqueous solutions are governed by the dissociation constant for water. Thus, if the dissociation constant for either the acid or its conjugate base is known, then the dissociation constant for the other can be determined using this equation:
Ka (conjugate acid) × Kb(conjugate base) = Kw = 10-14
As you can now see, Ka and Kb are inversely related. In other words, if Ka is large, then Kb is small, and vice versa.
APPLICATIONS OF KaAND Kb
The most common use of the acid and base dissociation constants is for the determination of the concentration of one of the species in the solution at equilibrium. On Test Day, you may be challenged to calculate the concentration of the hydrogen ion (or pH), the concentration of the hydroxide ion (or pOH), or the concentration of either the original acid or base. One such example is provided below to show you the important steps in solving quickly and correctly these types of problems.
To calculate the concentration of H+ in a 2.0 M aqueous solution of acetic acid, CH3COOH (Ka = 1.8 × 10-5), first write the equilibrium reaction:
CH3COOH (aq) H+ (aq) + CH 3COO- (aq)
Next, write the expression for the acid dissociation constant:
Once again, we are dealing with an equilibrium process here and can apply those principles to the problems dealing with weak acids and bases on the exam.
Acetic acid is a weak acid, so the concentration of CH3COOH at equilibrium is equal to its initial concentration, 2.0 M, less the amount dissociated, x. Likewise [H+] = [CH3COO-] = x, because each molecule of CH3COOH dissociates into one H+ ion and one CH3COO- ion. Thus, the equation can be rewritten as follows:
This estimation makes life easy and will get you the correct answer on Test Day.
We can approximate that 2.0 - x 2.0 because acetic acid is a weak acid and only slightly dissociates in water. This simplifies the calculation of x:
The fact that [x] is so much less than the initial concentration of acetic acid (2.0 M) validates the approximation; otherwise, it would have been necessary to solve for x using the quadratic formula. That sounds rather unpleasant, doesn’t it? Fortunately for you, the MCAT test writers select examples of weak acids and bases that allow you to make this approximation. (A rule of thumb is that the approximation is valid as long as x is less than 5 percent of the initial concentration.)
Acids and bases may react with each other, forming a salt (and often, but not always, water), in what is termed a neutralization reaction (see Chapter 4, Compounds and Stoichiometry). For example,
HA + BOH BA + H2O
Remember the reaction types discussed in Chapter 4? Well, here is our neutralization reaction.
The salt may precipitate out or remain ionized in solution, depending on its solubility and the amount produced. Neutralization reactions generally go to completion. The reverse reaction, in which the salt ions react with water to give back the acid or base, is known as hydrolysis.
Four combinations of strong and weak acids and bases are possible:
1. Strong acid + strong base: e.g., HCl + NaOH NaCl + H2O
2. Strong acid + weak base: e.g., HCl + NH3 NH4Cl
3. Weak acid + strong base: e.g., HClO + NaOH NaClO + H2O
4. Weak acid + weak base: e.g., HClO + NH3 NH4ClO
The products of a reaction between equal concentrations of a strong acid and a strong base are a salt and water. The acid and base neutralize each other, so the resulting solution is neutral (pH = 7), and the ions formed in the reaction do not react with water.
The product of a reaction between a strong acid and a weak base is also a salt, but often no water is formed because weak bases are usually not hydroxides. However, in this case, the cation of the salt will react with the water solvent, re-forming the weak base in hydrolysis. For example,
HCl (aq) + NH3 (aq) NH4+ (aq) + Cl- (aq) Reaction I
NH4+ (aq) + H2O (l) NH3 (aq) + H3O+ (aq) Reaction II
NH4+ is the conjugate acid of a weak base (NH3), which is stronger than the conjugate base (Cl-) of the strong acid HCl. NH4+ will then transfer a proton to H2O to form the hydronium ion. The increase in the concentration of the hydronium ion will cause the water system to shift away from auto-ionization, thereby reducing the concentration of hydroxide ion. Consequently, the concentration of the hydrogen ion will be greater than that of the hydroxide ion at equilibrium, and as a result, the pH of the solution will fall below 7.
On the other hand, when a weak acid reacts with a strong base, the pH of the solution at equilibrium is in the basic range because the salt hydrolyzes to re-form the acid, with the concurrent formation of hydroxide ion from the hydrolyzed water molecules. The increase in the concentration of the hydroxide ion will cause the water system to shift away from auto-ionization, thereby reducing the concentration of the hydrogen ion. Consequently, the concentration of the hydroxide ion will be greater than that of the hydrogen ion at equilibrium, and as a result, the pH of the solution will rise above 7. Consider the reaction of acetic acid CH3COOH (weak acid) with sodium hydroxide NaOH (strong base):
CH3COOH (aq) + NaOH (aq) Na+ (aq) + CH3COO- (aq) + H2O (l) Reaction I
CH 3COO- (aq) + H2O (l) CH3COOH (aq) + OH- (aq) Reaction II
The pH of a solution containing a weak acid and a weak base depends on the relative strengths of the reactants. For example, the acid HClO has a Ka = 3.2 × 10-8, and the base NH3 has a Kb = 1.8 × 10-5. Thus, an aqueous solution of HClO and NH3 is basic because the Ka for HClO is less than the Kb for NH3 (that is to say, HClO is weaker as an acid than NH3 is as a base, and consequently, at equilibrium, the concentration of hydroxide ions will be greater than the concentration of hydrogen ions in the aqueous solution).
Polyvalence and Normality
The relative acidity or basicity of an aqueous solution is determined by the relative concentrations of acid and base equivalents. An acid equivalent is equal to one mole of H+ (or H3O+) ions; a base equivalent is equal to one mole of OH- ions. Some acids and bases are polyvalent; that is, each mole of the acid or base liberates more than one acid or base equivalent. For example, the divalent acid H2SO4 undergoes the following dissociation in water:
H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)
HSO4- (aq) + H2O (l) H3O+ (aq) + SO42- (aq)
One mole of H2SO4 can produce two acid equivalents (2 moles of H3O+). You’ll notice, if you look closely at the dissociation reaction for sulfuric acid, the first dissociation goes to completion but the second dissociation goes to an equilibrium state. The acidity or basicity of a solution depends upon the concentration of acidic or basic equivalents that can be liberated. The quantity of acidic or basic capacity is directly indicated by the solution’s normality (see Chapter 9, Solutions). For example, since each mole of H3PO4 can yield three moles (equivalents) of H3O+, a 2 M H3PO4 solution would be 6 N (6 normal).
Another measurement useful for acid-base chemistry is equivalent weight. Chapter 4 defined and discussed this term extensively, and Chapter 9 reviewed it briefly. The gram equivalent weight is the mass of a compound that produces one equivalent (one mole of charge). For example, H2SO4(molar mass: 98 g/mol) is a divalent acid, so each mole of the acid compound yields two acid equivalents. The gram equivalent weight is 98/2 = 49 grams. That is, the complete dissociation of 49 grams of H2SO4 will yield one acid equivalent (one mole of H3O+). Common polyvalent acids include H2SO4, H3PO4, and H2CO3. Common polyvalent bases include CaCO3, Ca(OH)2, and Mg(OH)2. Magnesium hydroxide is the active ingredient in the thick, “milky” over-the-counter preparation that is used as an antidote for occasional acid indigestion and diarrhea. Calcium carbonate (CaCO3) is the active ingredient in the over-the-counter tablet preparation also used to treat an upset tummy.
An amphoteric, or amphiprotic, species is one that reacts like an acid in a basic environment and like a base in an acidic environment. In the Brønsted-Lowry sense, an amphoteric species can either gain or lose a proton. Water is the most common example. When water reacts with a base, it behaves as an acid:
H2O + B- HB + OH-
When water reacts with an acid, it behaves as a base:
HA + H2O H3O+ + A-
The partially dissociated conjugate base of a polyvalent acid is usually amphoteric (e.g., HSO4- can either gain an H+ to form H2SO4 or lose an H+ to form SO 42-). The hydroxides of certain metals (e.g., Al, Zn, Pb, and Cr) are also amphoteric. Furthermore, species that can act as either oxidizing or reducing agents (see Chapter 11, Redox Reactions and Electrochemistry) are considered to be amphoteric as well, because by accepting or donating electron pairs, they act as Lewis acids or bases, respectively. A simple way of remembering the meaning of the term is to think of amphibian animals, such as frogs. Amphibians live both on land and in water. Water is the most common amphoteric species, being able to act as both an acid and a base, and amphibians (both land and water living) = amphoteric compounds (both acid and base). As you might suspect, this is no mere linguistic coincidence; amphi comes from the Greek prefix meaning “on both sides” or “of both kinds.”
Recall that we spoke about gram equivalent weights in Chapter 4 and about normality and all units of concentration in Chapter 9.
Titration and Buffers
Titration is a procedure used to determine the molarity of a known reactant in a solution. There are different types of titrations, including redox, acid-base, and complexometric (metal ion). The MCAT will likely test your understanding of titration procedures and calculations in the context of acid-base solutions. (Metal ion titration may appear on Test Day, but it is less commonly tested and will likely have an informative passage attached.) Titration is a common laboratory procedure, and almost certainly you have performed at least one titration (probably acid-base) in your academic career.
Titrations are accomplished by reacting a known volume of a solution of unknown concentration (called the titrand) with a known volume of a solution of known concentration (called the titrant). In acid-base titration, the equivalence point is reached when the number of acid equivalents present in the original solution equals the number of base equivalents added, or vice versa. It is important to emphasize that, while a strong acid/strong base titration will have an equivalence point at pH 7, the equivalence point need not always occur at pH 7. Also, when titrating polyprotic acids or bases (discussed later in this chapter), there are several equivalence points, as each different acidic or basic conjugate species is titrated separately. In problems involving titration or neutralization of acids and bases, the equation to remember is
NaVa = NbVb
where Na and Nb are the acid and base normalities, respectively, and Va and Vb are the volumes of acid and base solutions, respectively. (Note that as long as both volumes use the same units, the units used do not have to be liters.)
This formula should remind you of Chapter 9 ... If not, turn back and take a look!
The equivalence point in an acid-base titration is determined in two common ways: either evaluated by using a graphical method, plotting the pH of the titrand solution as a function of added titrant by using a pH meter (see Figure 10.1), or estimated by watching for a color change of an addedindicator. Indicators are weak organic acids or bases that have different colors in their protonated and deprotonated states. Because they are highly colored, indicators can be used in low concentrations and therefore do not significantly alter the equivalence point. The indicator must always be a weaker acid or base than the acid or base being titrated; otherwise, the indicator would be titrated first! The point at which the indicator actually changes color is not the equivalence point but rather the end point. If the indicator is chosen correctly and the titration is performed well, the volume difference (and therefore the error) between the end point and the equivalence point is usually small and may be corrected for or simply ignored.
A useful set of compounds (indicators) will change color as it goes between its conjugate acid and base forms:
This allows us to use it to follow a titration, and we can see that since it is an equilibrium process that we can apply Le Châtelier’s principle. Adding H+ shifts equilibrium to the left. Adding OH- removes H+ and therefore shifts equilibrium to the right.
Acid-base titrations can be performed for different combinations of strong and weak acids and bases. The most useful combinations are strong acid/strong base, weak acid/strong base, and weak base/strong acid. Weak acid/weak base titrations can be done but are not usually accurate (and therefore almost never performed), because the pH curve for the titration of a weak acid and weak base lacks the sharp change that normally indicates the equivalence point. Furthermore, indicators are less useful because the pH change is more gradual.
STRONG ACID AND STRONG BASE
Let’s consider the titration of 10 mL of a 0.1 N solution of HCl with a 0.1 N solution of NaOH. Plotting the pH of the reaction solution versus the quantity of NaOH added gives the curve shown in Figure 10.1.
Our tug-of-war analogy for a bond between two atoms can be recycled in a different way with titrations. Whichever is stronger (our acid or our base) will determine the equivalence point for the titration. Here they are equal, so the equivalence point is at a neutral pH.
Because HCl is a strong acid and NaOH is a strong base, the equivalence point of the titration will be at pH 7, and the solution will be neutral. Note that the end point shown is close to, but not exactly equal to, the equivalence point; selection of a better indicator, one that changes colors at, say, pH 8, would have given a better approximation. Still, the amount of error introduced by the use of an indicator that changes color around pH 11 rather than, say, pH 8 is not especially significant: a mere fraction of a milliliter of excess NaOH solution.
In the early part of the curve (when little base has been added), the acidic species predominates, so the addition of small amounts of base will not appreciably change either the [OH-] or the pH. Similarly, in the last part of the titration curve (when an excess of base has been added), the addition of small amounts of base will not change the [OH-] significantly, and the pH remains relatively constant. The addition of base most alters the concentrations of H+ and OH- near the equivalence point, and thus the pH changes most drastically in that region. The equivalence point for strong acid/strong base titration is always at pH 7 (for monovalent species).
If you are using a pH meter so that you can chart the change in pH as a function of volume of titrant added, you can make a good approximation of the equivalence point by locating the midpoint of the region of the curve with the steepest slope.
WEAK ACID AND STRONG BASE
Titration of a weak acid, HA (e.g., CH3COOH), with a strong base, such as NaOH, produces the titration curve shown in Figure 10.2.
Note that any monoprotic weak acid titrated with a strong base will give a similar curve. The exact pH of the equivalence point depends on the identity of the weak acid, but the “tug-of-war” rule tells us it will be above 7.
Let’s compare Figure 10.2 with Figure 10.1. The first difference we should notice is that the initial pH of the weak acid solution is greater than the initial pH of the strong acid solution. This makes sense because we know that weak acids don’t dissociate to the same degree that strong acids do; therefore, the concentration of H3O+ will generally be lower (and pH will be higher) in a solution of weak acid. The second difference we should catch is the shapes of the curves. The pH curve for the weak acid/strong base titration shows a faster rise in pH for given additions of base. The pH changes most significantly early on in the titration, and the equivalence point is in the basic range. The third difference we should notice is the position of the equivalence point. While the equivalence point for strong acid/strong base titration is pH 7, the equivalence point for weak acid/strong base is above 7 (in the basic range). This is because the reaction between the weak acid (HA) and strong base (OH-) produces a stronger conjugate base (A-) and a weaker conjugate acid (H2O). This produces a greater concentration of hydroxide ions than that of hydrogen ions at equilibrium (due to the common ion effect on the auto-ionization of water). The equivalence point for weak acid/strong base titration is always in the basic range of the pH scale.
WEAK BASE AND STRONG ACID
The appearance of the titration curve for a weak base titrand and strong acid titrant will look like the “inversion” of the curve for a weak acid titrand and strong base titrant. The initial pH will be in the basic range (typical range: pH 10-12) and will demonstrate a fairly quick drop in pH for additions of the strong acid. The equivalence point will be in the acidic pH range, because the reaction between the weak base and strong acid will produce a stronger conjugate acid and weaker conjugate base. The stronger conjugate acid will result in an equilibrium state with a concentration of hydrogen ions greater than that of the hydroxide ions. The equivalence point for weak base/strong acid titration is always in the acidic range of the pH scale.
POLYVALENT ACIDS AND BASES
The titration curve for a polyvalent acid or base looks different from that for a monovalent acid or base. Figure 10.3 shows the titration of Na2CO3 with HCl in which the divalent (the term diprotic is equivalent) acid H2CO3 is the ultimate product.
In region I, little acid has been added, and the predominant species is CO32-. In region II, more acid has been added, and the predominant species are CO32- and HCO3-, in relatively equal concentrations. The flat part of the curve is the first buffer region (see next section), corresponding to the pKaof HCO3- (Ka = 5.6 × 10-11 implies pKa = 10.25).
Region III contains the equivalence point, at which all of the CO32- is finally titrated to HCO3-. As the curve illustrates, a rapid change in pH occurs at the equivalence point; in the latter part of region III, the predominant species is HCO3-.
In region IV, the acid has neutralized approximately half of the HCO3-, and now H2CO3 and HCO3- are in roughly equal concentrations. This flat region is the second buffer region of the titration curve, corresponding to the pKa of H2CO3 (Ka = 4.3 × 10-7 implies pKa = 6.37). In region V, the equivalence point for the entire titration is reached, as all of the HCO3- is finally converted to H2CO3. Again, a rapid change in pH is observed near the equivalence point as acid is added.
The titrations of the acidic and basic amino acids (which have acidic or basic side chains, respectively) will show curves similar to the one shown in Figure 10.3. But rather than two equivalence points, there will in fact be three: one corresponding to the titration of the carboxylic acid and a second corresponding to the titration of the amino acid, both of which are attached to the central carbon, and a third corresponding either to the acidic or basic side chain.
A buffer solution consists of a mixture of a weak acid and its salt (which consists of its conjugate base and a cation) or a mixture of a weak base and its salt (which consists of its conjugate acid and an anion). Two examples of buffers that are common in the laboratory and commonly tested on the MCAT are a solution of acetic acid (CH3COOH) and its salt, sodium acetate (CH3COO- Na+), and a solution of ammonia (NH3) and its salt, ammonium chloride (NH4+Cl-). The acetic acid/ sodium acetate solution is an acid buffer, and the ammonium chloride/ammonia solution is a base buffer. Buffer solutions have the useful property of resisting changes in pH when small amounts of strong acid or base are added. Consider a buffer solution of acetic acid and sodium acetate (the sodium ion has not been included because it is not involved in the acid-base reaction):
CH3COOH (aq) + H2O (l) H3O+ (aq) + CH3COO- (aq)
When a small amount of strong base, such as NaOH, is added to the buffer, the OH- ions from the NaOH react with the H+ ions present in the solution; subsequently, more acetic acid dissociates (the system shifts to the right), restoring the [H+]. The weak acid component of the buffer acts to neutralize the strong base that has been added. The resulting increase in the concentration of the acetate ion (the conjugate base) does not yield as large an increase in hydroxide ions as the unbuffered strong base would. Thus, the addition of the strong base does not result in a significant increase in [OH-] and does not appreciably change pH. Likewise, when a small amount of HCl is added to the buffer, H+ ions from the HCl react with the acetate ions to form acetic acid. Acetic acid is weaker than the added hydrochloric acid (which has been neutralized by the acetate ions), so the increased concentration of acetic acid does not significantly contribute to the hydrogen ion concentration in the solution. Because the buffer maintains [H+] at relatively constant values, the pH of the solution is relatively unchanged.
In the human body, one of the most important buffers is the H2CO3/HCO3- conjugate pair in the plasma component of the blood. CO2 (g), one of the waste products of cellular respiration, has low solubility in aqueous solutions. The majority of the CO2 transported from the peripheral tissue to the lungs (where it will be exhaled out) is dissolved in the plasma in a “disguised” form. CO2 (g) and water react in the following manner:
CO2 (g) + H2O (l) H2CO3 (aq) H+ (aq) + HCO3- (aq)
Carbonic acid (H2CO3) and its conjugate base, bicarbonate (HCO3-), form a weak acid buffer for maintaining the pH of the blood within a fairly narrow physiological range. The most important point to notice about this system for pH homeostasis is its direct connection to the respiratory system. In conditions of metabolic acidosis (excess of plasma H+), for example, the respiratory rate (breathing rate) will increase in order to “blow off” a greater amount of carbon dioxide gas; this causes the system to shift from the right to the left, thereby restoring the normal physiological pCO2 and in doing so, reducing the [H+] and buffering against dramatic and dangerous changes to the blood pH. One of the more interesting topics to ponder about the blood buffer system is why a weak acid buffer system was selected (evolutionarily speaking) as a primary mechanism for human blood pH homeostasis at around pH 7.4 (weakly basic). Buffers have a definite and narrow range of optimal buffering capability (pKa ±1), and pH 7.4 is actually slightly above the outer limit of buffering capability for the carbonic acid/bicarbonate system (pKa = 6.1). It’s an interesting question—and one to which there is an answer—but we don’t want to ruin all the surprises of medical school, so we will leave it unanswered for the time being.
You can think of buffer systems as kitchen sponges that soak up strong acids and strong bases to minimize changes in pH for a solution.
The Henderson-Hasselbalch equation is used to estimate the pH or pOH of a solution in the buffer region where the concentrations of the species and its conjugate are present in approximately equal concentrations. For a weak acid buffer solution,
The Henderson-Hasselbalch equation is also useful in the creation of buffer solutions other than those formed during the course of a titration. By careful selection of the weak acid (or base) and its salt, a buffer at almost any pH can be produced.
Note that when [conjugate base] = [weak acid] (in a titration, halfway to the equivalent point), the pH = pKa because log 1 = 0. Buffering capacity is optimal at this pH.
Likewise, for a weak base buffer solution,
and pOH = pKb when [conjugate acid] = [weak base]. Buffering capacity is optimal at this pOH.
One of the subtleties of buffer systems and Henderson-Hasselbalch calculations that usually goes unnoticed or misunderstood by students is the effect of changing the concentrations of the conjugate pair but not changing the ratio of their concentrations. Clearly, changing the concentrations of the buffer components in such a way that results in a change in their ratio will lead to a change in the pH of the buffer solution. But what about changing the concentrations while maintaining the ratio of the buffer components? For example, what is the effect on the system of doubling the concentrations of the acid and the base (thereby maintaining a constant ratio of the two)? Because we are taking the log of the ratio of the components, the logarithmic value will not change as long as the ratio doesn’t change. If the ratio of the buffer components doesn’t change, the pH of the buffer solution doesn’t change. Nevertheless, something has changed, to be sure. The buffering capacity—the size of that kitchen sponge—has changed. Doubling the concentrations of the buffer components produces a buffer solution with twice the buffering capacity. The kitchen sponge is twice as big and can soak up twice as much acid or base.
Blood pH is maintained in a relatively small range (slightly above 7) by a bicarbonate buffer system. Too great a change in either direction would lead to acidosis or alkalosis.
In this chapter, we have reviewed the important principles of acid-base chemistry. We clarified the differences among the three definitions of acids and bases, including the nomenclature of some common Arrhenius acids. We investigated important properties of acids and bases, including the important acid-base behavior of water (auto-ionization) and hydrogen ion equilibria. We explained the mathematics of the pH and pOH logarithmic scales and demonstrated a useful Test Day shortcut for approximating the logarithmic value of hydrogen ion or hydroxide ion concentrations. Strong acids and bases are defined as compounds that completely dissociate in aqueous solutions, and weak acids and bases are compounds that dissociate only partially (to an equilibrium state). We discussed neutralization and salt formation upon reaction of acids and bases, and finally, we applied our fundamental understanding of acid-base reactivity to titrations, useful for determining the concentration of a known acid or base solution, and to weak acid and weak base buffers, useful for minimizing changes in pH upon addition of strong acid or base.
Wow! This was a long chapter, packed with a lot of concepts and information. You’ve certainly earned a break after this. You’ve just accomplished a major task in the overall effort to earn points on Test Day. You may not understand everything that you’ve just read, and you probably don’t remember everything—and that’s okay. Now that you’ve read through these concepts, take some time to work through some of your MCAT practice passages and questions related to these topics so that the concepts settle and solidify. If you find that you need to review for a second or third time some of the discussion points related to acid-base chemistry, we’re always here for you in these pages.
If you’ve been paying attention (and we’re sure you have been), you’ve probably noticed that you’re now one chapter away from completing this review of general chemistry. While we don’t want to offer our congratulations prematurely, we want to acknowledge all the hard work you’ve invested in this process. Keep it up: Success on Test Day is within your grasp!
CONCEPTS TO REMEMBER
There are three definitions of acids and bases: Arrhenius—acids produce hydrogen ions, and bases produce hydroxide ions in aqueous solutions; Brønsted-Lowry—acids donate hydrogen ions, and bases accept hydrogen ions; and Lewis—acids accept electron pairs, and bases donate electron pairs.
Water is an amphoteric species (both acid and base) and auto-ionizes to produce equilibrium concentrations of hydrogen ions and hydroxide ions equal to 10-7 M at 298 K. The Kw for water at 298 K is 10-14. All aqueous acid or base solutions are defined by the equilibrium constant for water.
pH and pOH scales are logarithmic and express the negative log of the molar concentration of the hydrogen ions or hydroxide ions, respectively. For aqueous solutions at 298 K, a pH less than 7 is acidic, and a pH greater than 7 is basic; a pH 7 is neutral.
Strong acids and bases dissociate completely in aqueous solutions. Examples of strong acids include HCl, H2SO4, and HNO3. Examples of strong bases include NaOH and KOH.
Weak acids and bases dissociate incompletely in aqueous solutions. They have Ka’s or Kb’s less than 1. Examples of weak acids include CH3COOH, H2CO3, and H2O. Examples of weak bases include NH3 and H2O.
Brønsted-Lowry acid-base reactions always involve chemical pairs called conjugates: Strong acid produces weak conjugate base; strong base produces weak conjugate acid; a weaker acid produces a stronger conjugate base; a weaker base produces a stronger conjugate acid.
An equivalent is equal to one mole of charge (hydrogen ions or hydroxide ions). Equivalent weight is the mass in grams of an acid or base compound that yields one acid or base equivalent (i.e., one mole of charge). Normality is the number of acid or base equivalents per liter of solution.
Acid-base titration is used to determine the molar concentration of a known acid or base solution. The equivalence point is the point in the titration at which the equivalents of acid equal the equivalents of base.
Strong acid/strong base titration has an equivalence point at pH 7. Strong acid/weak base titration has an equivalence point at pH less than 7. Strong base/weak acid titration has an equivalence point at pH greater than 7. Indicators approximate the equivalence point by a steady color change at end point.
Buffers are weak acid/conjugate base or weak base/conjugate acid systems that act to absorb strong acids or bases from solutions, thereby minimizing pH changes within the buffered region. Buffers are most effective within ±1 of pKa or pKb.
EQUATIONS TO REMEMBER
1. Which of the following is not a Brønsted-Lowry base?
2. What is the pH of a solution containing 5 mM H2SO4?
3. Which of the following is chloric acid?
4. Which of the following bases is the weakest?
5. The function of a buffer is to
A. speed up reactions between acids and bases.
B. resist changes in pH when small amounts of acid or base are added.
C. slow down reactions between acids and bases.
D. maintain a neutral pH.
6. What is the pH of the following solution shown below?
pKb = 3.45
[NH4+] = 70 mM
[NH3] = 712 mM
Answer questions 7-9 based on the titration curve of acid X shown below:
7. What is the approximate value of the first pKa?
8. Where is the second equivalence point?
A. pH = 3.0
B. pH = 4.1
C. pH = 5.9
D. pH = 7.2
9. What is the approximate value of the second pKa?
10. What is the approximate gram equivalent weight of phosphoric acid?
A. 25 g
B. 33 g
C. 49 g
D. 98 g
11. What is the [H+] of a 2 M aqueous solution of a weak acid “HXO2” with Ka = 3.2 × 10-5?
A. 8 .0 × 10-3 M
B. 6.4 × 10-5 M
C. 1.3 × 10-4 M
D. 4 .0 × 10-3 M
12. A solution is prepared with an unknown concentration of a theoretical compound whose Ka is exactly 1.0. What is the pH of this solution?
A. Higher than 7
B. Exactly 7
C. Lower than 7
D. Impossible to determine
13. Which of the following is NOT a characteristic of an amphoteric species?
A. Can act as a base or an acid, depending on its environment.
B. Can act as an oxidizing or reducing agent, depending on its environment.
C. Is always protic.
D. Is always a nonpolar species.
14. What is the approximate pH of a 1.2 × 10-5 M aqueous solution of NaOH?
Small Group Questions
1. The Henderson-Hasselbach equation cannot be applied to all acids and bases. Explain why not.
2. Phosphoric acid is a polyprotic acid with Ka’s of 7.1 × 10-3, 6.3 × 10-8, and 4.5 × 10-13. Estimate its pKb values.
Explanations to Practice Questions
A Brønsted-Lowry base is defined as a proton acceptor. (A), (B), and (C) can each accept a proton. (D), HNO2, cannot.
First, we’ll convert the concentration to 5 × 10-3 M. Next, since sulfuric acid is a strong acid, we can assume that both protons will dissociate. So the concentration of hydrogen ions is really 2 × 5 × 10-3, which simplifies to 1 × 10-2. The equation for pH is pH = -log[H+]. If [H+] = 1 × 10-2 M, then pH = 2.
Answering this question is simply a matter of knowing nomenclature. Acids ending in –ic are derivatives of anions ending in –ate, while acids ending in –ous are derivatives of anions ending in –ite. ClO3- is named chlorate because it has more oxygen than the other occuring ion, ClO2-, which is named chlorite. Therefore, HClO3 is chloric acid.
Members of the IA and IIA columns on the periodic table combined with OH- are always strong bases. This means (A) and (D) can be eliminated. (B) and (C) are both weak bases, but (B) is weaker than (C) because a positive charge would sit solely on ammonia’s nitrogen. Methylamine, on the other hand, has a neighboring carbon (remember, alkyl groups are electron donating), which will help reduce the positive charge’s stress on nitrogen.
The purpose of a buffer is to resist changes in the pH of a reaction. Buffers will not affect the kinetics of a reaction, so (A) and (C) are wrong. (D) is correct only in specific circumstances that require a pH of 7. Many natural buffer systems maintain pHs in the acidic or basic ranges.
The question is asking for pH, but because of the information given, we must first find the pOH and then subtract it from 14 to get the pH. The equation for pOH is:
pOH = pKb + log. When the given values are substituted into this equation ([conjugate acid] = 70mM; [weak base] = 712mM), we find that pOH = 2.44, so the pH = 14 - 2.44 = 11.56
The first pKa in this curve can be estimated by eye. It is located between the starting point (when no base had been added yet) and the first equivalence point. This point is approximately at 7-8 mL added, which corresponds to a pH of approximately 1.9. Notice that this region experiences very little change in pH. This is the defining characteristic of a buffer zone!
The second equivalence point is the midpoint of the second quick increase in slope. This corresponds approximately to pH = 5.9.
The value of the second pKa is found at the midpoint between the first and second equivalence points. In this curve, that corresponds to pH = 4.1. Just like the first pKa, it is in the center of a flat buffering region.
Gram equivalent weight is the weight (in grams) that would release 1 acid equivalent. Because H3PO4 contains 3 acid equivalents, we find the gram equivalent weight by dividing the mass of one mole of the species by 3. Therefore, (B) is the correct answer.
This question requires the application of the acid dissociation formula:
Weak acids do not dissociate completely; therefore, all three species that appear in the formula will be present in solution. Hydrogen ions and conjugate base anions dissociate in equal amounts, so [H+] = [X-]. We don’t know exactly how much they’ll dissociate though, so we’ll just stick in an x for both of those species. How much HX do we have? The original concentration minus x.
Each of these x terms is quite small. The numerator becomes more significant because we’ll end up with x2, but we can simply ignore the x in the denominator. Let’s also plug in our Ka value:
Now all we have to do is isolate x, which leads us to find that the [H+] is 8 .0 × 10-3 M.
A higher Ka implies a stronger acid. Consider the following theoretical reaction, which defines the Ka of acid HA, HA H+ + A-. In such a reaction, Ka = [H+][A-] / [HA]. A Ka near 1 therefore implies that there are enough hydrogen ions present to affect the pH significantly. Weak acids usually have a Ka that is several orders of magnitude below 1. Yet a detailed understanding of Ka is not necessary to answer this question. According to the pH scale, which sets the Ka of water at 10-7, a compound with a Ka above 10-7 is acidic; even if the acid is very weak, it will still cause the pH to drop below 7.
An amphoteric species is one that can act either as an acid or a base, depending on its environment. Proton transfers are classic redox reactions, so (A) and (B) are true. (C) must be true because, by definition, an amphoteric molecule needs to have a proton to give up in order to act like an acid. (D) is false, and thus the correct answer, because amphoteric species can be either polar or nonpolar in nature. Some examples: HSO4-, NH3, H2O.
NaOH Na+ + OH-
The balanced equation shows the same coefficients in front of each of the three species. So if the initial concentration of NaOH is 1.2 × 10-5, then the concentration of Na+ and OH- must be also, because NaOH will completely dissociate.
Next, we’ll find the pOH from [OH-]. The pOH lies between 4 and 5. How did we find that? Well, 1.2 × 10-5 is between 10-5 and 10-4. So its logarithm is between 4 and 5. Let’s estimate 4.8.
We’re ready to calculate pH from pOH.
pH = 14 - pOH = 14 - 4.8 = 9.2.
(C) is very close to 9.2, so it must be the right answer.