3 Thermodynamics & Acidity

To understand how a cell Metabolism is a defining characteristic of living systems. All living things Goal drives unfavorable reactions. carry out metabolic processes, many of which are unfavorable and require energy to proceed. Indeed, energy is required for the maintenance of life. Why does life require energy, how is it obtained, and where does it go? These Objectives questions can be addressed using thermodynamics. Thermodynamics is After this chapter, you should be able to the study of energy flow within a system. Thermodynamics can be used to predict how systems will behave under different circumstances and whether • explain the concept of Gibbs free energy. reactions are energetically favorable. • explain the relationship of ΔG° to the rxn Energy cannot be created or destroyed favorability of a reaction. The first law of thermodynamics states that energy cannot be created or • relate ΔG°rxn to Keq. destroyed; it can only be converted from one form to another or used to do • apply Le Châtelier’s Principle to chemical equilibria. work. For example, when energy is added to solid water (ice) by heating, the energy of the molecules within the solid ice increases, causing them • predict protonation state from the pH to vibrate more rapidly. As the heating continues, some water molecules of a solution and pK . a gain enough energy to break away from the other water molecules. Because water molecules interact through hydrogen bonding and because energy is needed to break those interactions, the water molecules are converting thermal energy into chemical energy when water is heated.

Chemical reactions either release or absorb energy Cells require energy to carry out metabolic reactions, including the synthesis and breakdown of molecules. To obtain energy, living systems metabolize Chapter 3 Thermodynamics & Acidity 2

Glycolysis

O O O O OH ATP ADP P O ATP ADP P HO P O OH O O O O HO O O O HO O O O O HO O O HO OH HO OH P OH OH OH O OH HO HO Glucose

OH OH O O O OH O O O O O O O O O O P P P O O O HO O O + P P O O ATP ADP O O O NADH NAD , Pi O O

+ CoA-SH, NAD OH O O O O O O O O CO , NADH P P 2 ADP ATP O H2O O O O O O O

Krebs Cycle O

NADH, CoA O + + O OH NAD H O O FADH2 O O O O O O CoA-SH, H+ FAD O O O

H2O

OH O O O O O O O O O O

O O GTP, CoA-SH OH O O O CoA O O GDP, Pi O CO , CoA-SH,O O NADH, NAD+ 2 + O O NADH NAD CO2 O O

Figure 1 Glycolysis and the Krebs cycle are central reactions in metabolism Glycolysis and the Krebs cycle are metabolic reactions that release energy that is then used by the cell to synthesize ATP. The energy stored in ATP can be used later to do work. The cell uses this energy to grow, divide, move, and carry out metabolism. We will return to these energy-generating reactions and to how the molecules in these pathways are depicted in later chapters. nutrients, such as amino acids, fats, and sugars. One of the primary reactions

that occurs during metabolism is the reaction of glucose (C6H12O6) with oxygen, which releases energy and produces carbon dioxide and water. The released energy is used in a multi-step process (the glycolytic pathway, Figure 1) to generate an energy store known as adenosine triphosphate (ATP), which can later be utilized by cellular machinery to help carry out important metabolic reactions. We will have much more to say about ATP and its myriad roles in the cell in later chapters. The reaction of glucose with oxygen depicted in Figure 2 shows glucose and oxygen (the reactants) at a higher energy level than carbon dioxide and Chapter 3 Thermodynamics & Acidity 3

Respiration an energetically favorable reaction Photosynthesis an energetically unfavorable reaction

OH OH O O HO + 6 O 6 CO + 6 H O 6 CO + 6 H O HO 6 O HO OH 2 2 2 2 2 HO OH + 2 OH OH

Glucose Oxygen Carbon dioxide Water Carbon dioxide Water Glucose Oxygen high energy high energy Glucose Glucose + Energy Energy + Oxygen Oxygen

favorable unfavorable Carbon dioxide Carbon dioxide energy is released energy is required + + Water Water low energy low energy

Figure 2 Respiration and photosynthesis interconvert glucose and O2 with CO2 and water as we will explain in more depth in a later chapter

water (the products) to capture the idea that energy is released when the reaction occurs. Reactions that release energy are energetically favorable. Conversely, reactions in which the products are higher in energy than the reactants require an input of energy to proceed; such reactions are energetically unfavorable. It is important to note that the favorability of a reaction is largely unrelated to its rate; favorable reactions can be extremely slow. The determinants of reaction rates are the subject of the next chapter on kinetics. A full understanding of the reactions that drive living systems requires an appreciation of both thermodynamics and kinetics. Reactions can proceed in both directions. For example, the conversion of glucose and oxygen to carbon dioxide and water can take place in the reverse direction to synthesize glucose, as occurs during photosynthesis. Because glucose and oxygen (the products of photosynthesis) are higher in energy than carbon dioxide and water (the reactants in photosynthesis), photosynthesis requires an input of energy (Figure 2). In the case of photosynthesis, this energy comes directly from the sun, but other energy sources, such as ATP, are commonly used by cells to drive unfavorable processes. Reactions that occur in both directions are indicated with stacked arrows in which one arrow points to the left and the other points to the right.

The concentrations of reactants and products do not change at equilibrium How do we determine whether the reactants or the products of a reaction are higher in energy? One way to do this is to measure the concentrations of the reactants and products when the reaction is at equilibrium, as we will explain. A reaction is at equilibrium when the concentrations of its reactants and products no longer change with time. The reaction does not stop at equilibrium; instead, the forward and reverse reactions occur at the same rate. As a consequence, reactants are consumed by the forward Chapter 3 Thermodynamics & Acidity 4 reaction as fast as they are generated by the reverse reaction, and products are consumed by the reverse reaction as fast as they are generated by the forward reaction. Consider the example of two hypothetical molecules, A and B, which interconvert as described in the following reaction: A B If a reaction begins with only molecule A present, the concentration of molecule A will decrease with time as the reaction proceeds, and the concentration of molecule B will increase. Because no molecules of B are present initially, the reverse reaction cannot occur at first. As time passes and molecule B accumulates, however, the rate of the reverse reaction increases, which leads to a slower net rate of molecule B accumulation because some molecules of B are being converted back into molecule A. Ultimately, the reaction reaches an equilibrium, in which the rates of the forward and reverse reactions are equal; at that point, there is no further net change in the concentration of either molecule A or molecule B.

The equilibrium constant is the ratio of products to reactants at equilibrium

All reactions have a characteristic equilibrium constant (Keq) under a given

set of conditions. Keq is defined as the ratio of the mathematical product of the equilibrium concentrations of the species on the right (i.e., the concentrations of the chemical products multiplied together) divided by the mathematical product of the equilibrium concentrations of the species on the left (i.e., the reactants). A hypothetical reaction in which one molecule of A reacts with one molecule of B to reversibly yield one molecule of C and one molecule of D would have an equilibrium constant equal to ([C][D])/ ([A][B]), as shown below. The brackets (“[]”) surrounding each molecule signify the concentration for that reactant or product in units of molarity (i.e., moles per liter). [C] [D] A + B C + D K = eq [A] [B]

If Keq is less than one, then the concentrations of the products are lower

than the concentrations of the reactants at equilibrium. Conversely, if Keq is greater than one, then the concentrations of products are greater than

the concentrations of the reactants at equilibrium. If Keq is exactly one, then there is an equal mixture of reactants and products at equilibrium. At

one extreme, a Keq that approaches zero means that there will be almost no

product at equilibrium. At the other extreme, a Keq that approaches infinity means that there will be nearly no reactant at equilibrium. Each chemical

reaction has its own value of Keq that is constant and only changes with temperature. The value of the equilibrium constant indicates whether or not a reaction is

favorable. Reactions for which Keq is greater than one are favorable, meaning that the reactants are higher in energy than the products and signifying that energy is released as the reaction proceeds. Similarly, reactions for which

Keq is less than one are unfavorable, meaning that the reactants are lower in energy than the products and signifying that the reaction absorbs energy Chapter 3 Thermodynamics & Acidity 5

Example 1 Writing an equilibrium constant Write an equation that expresses the equilibrium constant for the reaction shown below in terms of reactant (water and pyrophosphate) and product (hydrogen phosphate) concentrations.

-4 -2 H2O + P2O7 2 HPO4

As we saw earlier, Keq is the mathematical product of the species on the right divided by the mathematical −2 product of the species on the left. As there are two molecules of HPO4 on the right side of the equation, −2 −2 the numerator of the Keq expression will be [HPO4 ] [HPO4 ]. The denominator of the expression will be −4 −2 2 [H2O][P2O7 ]. We can simplify the numerator to [HPO4 ] because two identical terms are being multiplied together, producing the expression shown below: [HPO -2][HPO -2] [HPO -2]2 K = 4 4 = 4 eq -4 -4 [H2O][P2O7 ] [H2O][P2O7 ] We can write the equilibrium constants for reactions that involve than one molecule of the same reactant or product using the general equation shown below. In this equation, the lowercase letter a-d that precedes each molecule represents the stoichiometric coefficient of that molecule (i.e., the number of molecules of −2 that species that are involved in the reaction). In this example, two molecules of HPO4 appear in the −2 products of the reaction. Because its stoichiometric coefficient is 2, the concentration of HPO4 has an exponent of 2 in the equilibrium constant expression. [C]c [D]d a A + b B c C + d D K = eq [A]a [B]b

as it proceeds. Reactions for which Keq equals one feature products and reactants of equal energy. Let us consider the example of the hydration of carbon dioxide to produce carbonic acid, a reaction that is critical for life (Figure 3). The conversion of carbon dioxide to carbonic acid is necessary for photosynthesis in many plants, and the reverse reaction is needed for respiration. The capture of carbon dioxide, a greenhouse gas, also plays a pivotal role in the Earth’s environment. Carbon dioxide produced by industrial activity absorbs radiation reflected from the Earth’s surface, trapping it in the atmosphere and warming the planet. A significant portion of carbon dioxide that is formed is absorbed into the oceans, where some of it combines with water to form carbonic acid. Although this reaction allows the ocean to slow global warming by acting as a reservoir for carbon dioxide, the concomitant production of carbonic acid also leads to the acidification of the ocean, which can be detrimental to marine life.

Figure 3 Aqueous CO exists in 2 CO + H O H CO equilibrium with carbonic acid 2 (aq) 2 (l) 2 3 (aq)

Carbon dioxide Water Carbonic acid

O H O C O H C H O H O O Chapter 3 Thermodynamics & Acidity 6 Hydration occurs by the reaction of carbon dioxide dissolved in water

(aqueous carbon dioxide) with liquid water, represented as “CO2 (aq)” and

“H2O (l),” respectively (Figure 3). The “(aq)” and “(l)” labels indicate that the reactants are aqueous and liquid, respectively. Such labels are known as state symbols. Other state symbols include “(s)” for solid and “(g)” for gas. Note that the carbon atoms in both carbon dioxide and carbonic acid form four bonds, maintaining the octet configuration. It follows that the equilibrium constant for the hydration of carbon dioxide is: [H CO ] = 2 3 Keq [CO2] [H2O] The value of this equilibrium constant is 3.1 x 10−5 at 25°C, a number far

smaller than 1. This means that the reactants (CO2 and H2O) are strongly

favored over the product (carbonic acid, H2CO3), and when carbon dioxide is dissolved in water, only a small fraction of it is converted to carbonic acid at equilibrium. However, the capture of carbon dioxide does not end with its conversion to carbonic acid. Rather, carbonic acid undergoes subsequent reactions that acidify the oceans and ultimately harm marine organisms, as we will see later in this chapter.

The composition of a chemical system at equilibrium is independent of its initial state A key feature of a chemical equilibrium is that it is independent of the initial state. A given reaction will proceed to the same equilibrium ratio of products and reactants, even if the reaction begins with a predominance of

The composition Figure 4 100% H2CO3 of a reaction at equilibrium Equilibrium is independent of its initial CO composition 2 Shown are graphs of the change in composition of two carbon dioxide hydration reactions over time. The concentration of water is high enough to concentration H CO assume that it does not change significantly 2 3 during the course of the reactions, and time for that reason it is not shown on either graph. In the top graph the reaction started 100% CO2 from pure carbonic acid (on the left), and as time passed, the reaction reached equilibrium at a state composed almost CO2 entirely of carbon dioxide (on the right). In the bottom graph the reaction started from pure carbon dioxide (on the left), but as time passed, the reaction again equilibrated to a state composed almost entirely of concentration H CO carbon dioxide. In both cases, the reactions 2 3 approach a ratio of products to reactants time that equals Keq. Chapter 3 Thermodynamics & Acidity 7 reactants or a predominance of products. For example, a hydration reaction

that begins with 100% CO2 will reach the same ratio of products to reactants

at equilibrium as a reaction that starts with 100% H2CO3, as illustrated in Figure 4.

Gibbs free energy is a measure of the energy available to drive a chemical reaction The energy that is available to drive a chemical reaction is referred to as the Gibbs free energy (G), named after the great chemist Willard Gibbs. We determine the Gibbs free energy for a reaction (rxn) by comparing the difference in energy (∆G) between a state with 100% products and a state

with 100% reactants. This is known as ∆Grxn, representing the difference in free energy between the two states:

ΔGrxn = Gproduct – Greactant Because the value of G changes with temperature, pressure, and

concentration, we define a set of conditions under which ΔGrxn values can be compared to one another. Under these conditions, called the standard state, concentrations are 1 molar (M), pressure is 1 atmosphere (atm), and the temperature is constant (usually 25°C, or 298 K). The difference in the Gibbs free energy between two states under standard-state conditions is

denoted as ΔG°rxn. Like Keq, the value of ΔG°rxn is a property of a reaction

and does not change as that reaction proceeds. The concept of ΔG°rxn will be of critical importance as we examine the chemical reactions that drive living systems. Consider the hydration of carbon dioxide (Figure 3) in a hypothetical reaction in which carbon dioxide is completely converted to carbonic acid. Given the position of the equilibrium, which strongly favors the reactants,

we know that the state of 100% CO2 (aq) is more stable (i.e., has a lower Gibbs

free energy) than the state of 100% H2CO3 (aq). ΔG°rxn for the hydration of carbon dioxide would thus be a positive value because the energy of the product is greater, indeed much greater, than the energy of the reactants. Figure 5 is a plot of free energy over a range of carbon dioxide and carbonic acid concentrations for this reaction. Notice that equilibrium is not at the

extreme left, when the reaction consists of 100% CO2 (aq). Rather, as we noted in Figure 4 above, the system achieves equilibrium when both the reactants and products are present as a mixture (though with much more carbon dioxide than carbonic acid). Given that carbonic acid is much higher in energy than carbon dioxide, you might wonder why any carbonic acid

exists at equilibrium. The answer is that at equilibrium (when [H2CO3]/

[CO2] = Keq[H2O]), the system is at an even lower level of Gibbs free energy

than at a state consisting of 100% CO2 (aq). Indeed, equilibrium is defined as the state at which the Gibbs free energy is minimized. The reason that free energy is at a minimum when both carbon dioxide and carbonic acid are present is that additional energy is released when chemical systems are disordered and contain a mixture of products and reactants. We will learn more about how disorder influences Gibbs free energy in Chapter 6, when we introduce the concept of entropy. Chapter 3 Thermodynamics & Acidity 8

100% H2CO3

100% CO2

(C) ∆G°rxn G°

(A) (B) [H2CO3] = Keq [H2O] [CO2]

equilibrium

100% CO2 composition of system 100% H2CO3 Figure 5 Equilibrium is the lowest-energy state for a system Shown is a Gibbs free energy diagram for the hydration of carbon dioxide. The curve plots the Gibbs free energy of the reaction at varying relative amounts of CO2 and H2CO3. The minimum point on the curve presents the lowest-energy state for the system. At this minimum point, the concentrations of CO2 and H2CO3 equal those present at equilibrium. For clarity, the axes are not drawn to scale. The line tangent to the curve is indicated (black line) at three points, (A), (B), and (C). The slope of the tangent line reveals the sign of ΔG at each point. At points (A), (B), and (C), the slopes of the tangent lines, and thus the signs of ΔG, are negative, zero, and positive, respectively.

The Gibbs free energy diagram also allows us to predict the direction of the reaction at different ratios of reactants and products. To do this, we determine the slope of a line tangent to a point along the curve, which represents the change in free energy (ΔG) at that particular point. When

the product-to-reactant ratio is less than Keq [point (A) in Figure 5], ΔG has a negative sign moving from left to right (the tangent slopes down), meaning that the reaction is favorable in this direction and consequently that the reaction proceeds to produce more products. Conversely, when the

product-to-reactant ratio is greater than Keq [point (C)], ΔG has a positive sign moving from left to right (the tangent slopes up), meaning that the reaction is unfavorable in this direction. That is, it would require an increase

in energy to convert more CO2 to carbonic acid. On the other hand, the reverse reaction at point (C) (that is, moving from right to left) is favorable, meaning that it results in a decrease in energy (ΔG is negative). Finally,

when the product-to-reactant ratio is equal to Keq [i.e., at equilibrium; point (B)], ΔG equals zero (the tangent has no slope). At this point there is no net reaction, and the concentrations of products and reactants do not change

with time. Unlike Keq and ΔG°rxn, which are fixed properties of the reaction, the value of ΔG changes as the reaction proceeds; its value approaches zero as the reaction approaches equilibrium.

Finally, the value of ΔG is directly related to ΔG°rxn and to the product- to-reactant ratio by a simple equation. In this equation, which is shown below, “R” is the ideal gas constant (2.0 x 10−3 kcal∙mol−1∙K−1), “T” is the temperature in Kelvin, and “ln” is the natural logarithm. [A], [B], [C], and [D] are the concentrations of reactants and products for a hypothetical reaction in which A and B reversibly react to form C and D. This equation can be used to calculate ΔG under a given set of conditions in order to Chapter 3 Thermodynamics & Acidity 9

Consider the following reaction for which ΔG° is negative: Breakout rxn

A B

Which of these curves best describes this equilibrium?

(A) (B)

G° G°

100% A [A] = [B] 100% B 100% A [A] = [B] 100% B (C) composition of system (D) composition of system

G° G°

100% A [A] = [B] 100% B 100% A [A] = [B] 100% B composition of system composition of system

determine the direction in which a reaction will proceed, as shown in the example in Example 2. [C][D] ΔG = ΔG° + RT ln rxn [A][B] A particularly useful feature of this equation is that it enables us to express the equilibrium constant in units of energy. In the special case when the reaction is at equilibrium, ΔG = 0, and the above equation can be re-written as:

ΔG°rxn = –RT ln Keq

From this important relationship , we see that reactions in which ΔG°rxn is positive are unfavorable, require an input of energy to proceed, and have

an equilibrium that favors the reactants (Keq < 1). Conversely, reactions

in which ΔG°rxn is negative are favorable, release energy as they proceed,

and have an equilibrium that favors the products (Keq > 1). Reactions in

which ΔG°rxn is zero are energetically neutral and have an equilibrium that

favors neither products nor reactants (Keq = 1). These relationships are summarized in Figure 6. Chapter 3 Thermodynamics & Acidity 10

Example 2 Calculating and interpreting ΔG°rxn values A reversible reaction interconverts the sugars glucose and fructose:

glucose fructose

An experiment was conducted in which the reaction was allowed to reach equilibrium at 25°C (298 K). Upon reaching equilibrium, the concentrations of glucose and fructose were measured be 1.4 mM and 1.2

mM, respectively. Use these data to calculate ΔG°rxn for the conversion of glucose to fructose. Based on your answer, which molecule, glucose or fructose, is more stable?

Explanation: We can begin by calculating the value of Keq. Keq equals the ratio of products over reactants present at equilibrium:

−3 −3 Keq = [fructose]/[glucose] = 1.2 x 10 M / 1.4 x 10 M = 0.86

Now we can use the general formula that relates Keq to ΔG°rxn:

ΔG°rxn = –RT ln Keq −1 −1 ΔG°rxn = –(0.002 kcal∙mol ∙K )∙(298 K)∙ln(0.86) −1 −1 ΔG°rxn = –(0.002 kcal∙mol ∙K )∙(298 K)∙(–0.151) −1 ΔG°rxn = 0.090 kcal∙mol

Because the sign of ΔG°rxn is positive, we know that the conversion of glucose to fructose requires an input of energy; therefore, glucose is lower in energy and more stable than fructose.

Figure 6 The favorability of a Favorable reaction is indicated by both Keq + Energy Keq > 1 ∆G°rxn < 0 A B and ΔG°rxn

Unfavorable

Energy + Keq < 1 ∆G°rxn > 0 A B

Energetically Neutral

Keq = 1 ∆G°rxn = 0 A B Chapter 3 Thermodynamics & Acidity 11

(A) (B)

Energy + A B ∆G°rxn > 0 Glutamine synthesis ∆G°rxn = +3.3 kcal/mol

+ + Energy ∆G° << 0 ∆G° = -8.6 kcal/mol + ATP ADP Pi rxn + Cleavage of ATP rxn

+ + + + Energy ∆G° < 0 ∆G° = -5.3 kcal/mol A ATP ADP Pi B rxn ATP-driven glutamine synthesis rxn

Figure 7 ATP hydrolysis can drive unfavorable reactions (A) ATP hydrolysis releases a large amount of energy that can be used to drive unfavorable reactions, such as the hypothetical conversion of “A” to “B”. In this example, the amount of energy released during ATP hydrolysis is greater than the amount of energy absorbed during the conversion of A to B, resulting in a favorable net reaction. (B) Shown is a specific example in which energy released from the cleavage of ATP is used to drive the synthesis of glutamine, a process that is otherwise unfavorable.

Cells use ATP to drive otherwise unfavorable reactions Cells are complex systems that make and break down molecules, and in doing so, they routinely carry out unfavorable reactions. It requires an input of energy to drive reactions in the disfavored direction, and in the cell, the source of this energy is most commonly ATP. The ATP molecule is high in energy relative to adenosine diphosphate (ADP) and phosphate −3 (PO4 ); consequently, breaking ATP into ADP and phosphate features a

large, negative ΔG°rxn. The energy released from this reaction can be used to drive other, unfavorable reactions toward completion (Figure 7A). Because

energy is conserved, ΔG°rxn for an overall reaction can be expressed as the

sum of the ΔG°rxn values of its component reactions, as follows:

ΔG°rxn (overall) = ΔG°rxn (reaction 1) + ΔG°rxn (reaction 2) + ... + ΔG°rxn (reaction “n”)

For example, consider the production of the amino acid glutamine from ammonia and another amino acid, glutamate. The reaction of ammonia

with glutamate is unfavorable, with a ΔG°rxn of +3.3 kcal/mol. However,

cells break apart ATP and use the energy that is released (ΔG°rxn of −8.6 kcal/mol) to drive the unfavorable synthesis of glutamine. The overall

ΔG°rxn value of the ATP-driven synthesis of glutamine equals the sum of

the ΔG°rxn values of its component reactions (Figure 7B). In later chapters we will consider the mechanisms by which energy from ATP is harnessed to drive otherwise unfavorable reactions. Cells exist far from equilibrium and hence require constant inputs of energy. They use energy to grow, divide, and respond to environmental stresses in order to survive. Cells are constantly changing, and by now, you will appreciate that a system in a constant state of flux cannot be at equilibrium. Cells that reach equilibrium are dead. Thus, cells take in energy from outside sources and harness it via molecules such as ATP to do work in order to avoid reaching equilibrium. Chapter 3 Thermodynamics & Acidity 12 Cells also use Le Châtelier’s principle to drive unfavorable reactions Chemical systems at equilibrium react to perturbations (e.g., changes in concentration, temperature, pressure, etc.) by adopting a different equilibrium state, a concept known as Le Châtelier’s principle. Le Châtelier’s principle states that a chemical system that is displaced from equilibrium will adjust so that it can return to a state of equilibrium. As we will see, this phenomenon is commonly used by cells to drive unfavorable chemical processes. As an example, consider a hypothetical reaction in which molecules A and B react to produce molecules C and D, as shown in Figure 8. When this reaction is at equilibrium, the ratio of products over reactants, [C][D]/([A][B]), equals the reaction’s equilibrium constant. Notice, however, that there are an infinite number of [A], [B], [C], and [D] values

for which [C][D]/([A][B]) equals Keq. Consequently, the actual values of [A], [B], [C], and [D] at equilibrium are not fixed and instead depend on the initial composition of the reaction. In effect, perturbing an equilibrated reaction by adding or removing a reactant or product is the equivalent of starting the reaction again, but with a different initial composition. Consequently, this perturbation leads to a new equilibrium state in which the concentrations of products and reactants are different, even though the

ratio of products to reactants still equals Keq. Figure 8A applies this concept to the hypothetical reaction of A and B producing C and D by showing an example in which extra reactant (molecule A) is added to the equilibrated reaction. This increase in [A] causes the equilibrium to shift to the right (toward the products) in order to

counter the perturbation of the equilibrium and reduce [A] again until Keq is reached. As you can see from Figure 8B, the ratio of products to reactants equals the equilibrium constant when the reaction is at equilibrium, even though the concentrations of products and reactants have changed. Now we are ready to return to the question of why an increase in carbon dioxide in the atmosphere leads to the acidification of the world’s oceans. The answer lies in the fact that carbon dioxide hydration is part of a series

of sequential reactions, as described in Figure 9. First, atmospheric CO2 (g)

dissolves in water to produce CO2 (aq), which then reacts with water to

produce H2CO3 (aq). H2CO3 (aq), in turn, dissociates via reaction with water + − to yield hydronium (H3O (aq)) and HCO3 (aq). Since acidity is simply a + + measurement of H3O concentration, any increase in [H3O (aq)] results in an increase in acidity (we discuss acidity in more detail later in this chapter). The production of carbon dioxide from industrial activity increases its concentration in the atmosphere. As you would expect given Le Châtelier’s

principle, this increase in [CO2 (g)] causes the dissolution equilibrium

(Figure 9) to shift toward products, resulting in an increase in [CO2 (aq)]. Since aqueous carbon dioxide is also a reactant in the hydration reaction,

the increase in [CO2 (aq)] causes the hydration equilibrium to shift toward

products, causing an increase in [H2CO3 (aq)]. Finally, since carbonic acid

is a reactant in the acid dissociation reaction, the increase in [H2CO3 (aq)] causes the acid dissociation equilibrium to shift toward products, causing an + increase in [H3O (aq)]. As you can see, the concentration of carbon dioxide in the atmosphere is connected to the production of hydronium through a

series of coupled equilibria, and increases in [CO2 (g)] cause a perturbation Chapter 3 Thermodynamics & Acidity 13

(A) (B) A + B C + D [C] [D] (5.0 x 10-3)(5.0 x 10-3) 1 2 1 Keq = = = 25 [A] [B] (1.0 x 10-3)(1.0 x 10-3)

Molecules C & D 5 [C] [D] (5.5 x 10-3)(5.5 x 10-3) 2 Keq = = = 25 [A] [B] (2.4 x 10-3)(5.1 x 10-4) Molecule A Concentration (mM) Concentration 1 Molecule B

Time More Molecule A added

(C) Perturbation E ect on equilibrium

More Molecule A is added. The equilibrium shifts towards products. Additional C and D will be produced until a new equilibrium is reached. At the new equilibrium, [A], [C], and [D] will increase relative to the previous equilibrium, whereas [B] will decrease.

More Molecule C is added. The equilibrium shifts towards reactants. Additional A and B will be produced until a new equilibrium is reached. At the new equilibrium, [A], [B], and [C] will increase relative to the previous equilibrium, whereas [D] will decrease.

Molecule A is removed. The equilibrium shifts towards reactants. Additional A and B will be produced until a new equilibrium is reached. At the new equilibrium, [A], [C], and [D] will decrease relative to the previous equilibrium, whereas [B] will increase.

Molecule C is removed. The equilibrium shifts towards products. Additional C and D will be produced until a new equilibrium is reached. At the new equilibrium, [A], [B], and [C] will decrease relative to the previous equilibrium, whereas [D] will increase.

Figure 8 Le Châtelier’s principle describes how chemical systems at equilibrium respond to perturbation by moving toward an equilibrium state with a different composition Shown is a hypothetical reaction between A and B to produce C and D whose equilibrium constant is 25. (A) Shown is a graph of [A], [B], [C], and [D] for a reaction beginning with equal amounts of A and B and no C or D. The reaction proceeds until it achieves equilibrium (state 1), where ([C][D])/([A][B]) equals Keq. At that point, additional A is added to the system, which suddenly increases

[A] and perturbs the equilibrium. The reaction produces more product until the ratio ([C][D])/([A][B]) again equals eqK (state 2). (B) The concentrations of A, B, C, and D are different at state 2 compared to state 1, but at both states ([C][D])/([A][B]) equals eqK . Notice that [A], [C], and [D] are higher at state 2 than at state 1, whereas [B] is lower. (C) Listed are four examples of perturbations and their effects on this equilibrium. The first row lists the perturbation described in (A). None of these perturbations affectseq K .

K = 0.83 Figure 9 Atmospheric carbon eq Dissolution: CO CO dioxide affects ocean acidity 2 (g) 2 (aq) through a multistep series of coupled equilibria -5 Keq = 3.1 x 10 CO2 in the atmosphere is connected to the Hydration: CO + H O H CO production of hydronium through a series 2 (aq) 2 (l) 2 3 (aq) of coupled equilibria. As expected from Le Châtelier’s principle, an increase in [CO ] 2 (g) -6 Keq = 4.0 x 10 causes a perturbation in each equilibrium - + + Acid Dissociation: H CO + H O HCO + H O that ultimately increases [H3O (aq)], thus 2 3 (aq) 2 (l) 3 (aq) 3 (aq) increasing the ocean’s acidity (see details in text). Chapter 3 Thermodynamics & Acidity 14 that propagates through each individual equilibrium, ultimately causing an increase in hydronium concentration and thus an increase in ocean acidity. An increase in the acidity of the ocean is consequential because of its harmful effects on marine life. Acidification is especially problematic for organisms like corals and mollusks, whose mineralized shells dissolve under acidic conditions.

Acid-base reactions involve the exchange of a hydrogen ion

The dissociation of 2H CO3 belongs to a class of reactions that are particularly important in living systems, acid-base reactions. Acid-base reactions are a special type of equilibrium that involves the movement of a hydrogen ion (H+), or proton, between two molecules. The molecule that gives up the proton is called an acid, and the molecule that receives the proton is called a base. The product of the reaction that consists of the acid without its proton is called the conjugate base. Similarly, the product that consists of the base with an added proton is called the conjugate acid. Whether a molecule is an acid or a conjugate acid is only an issue of reaction direction; the conjugate acid in any given reaction is the acid in the reverse reaction. We often refer to the transfer of a proton as protonation and deprotonation. Once an acid has lost its proton, it is said to be deprotonated, and once a base has gained a proton, it is said to be protonated. A molecule is said to be acidic when it readily gives up a proton. The stronger the acid, the more acidic it is, and the more easily it gives up its proton. Acids often contain highly polar X-H bonds, where X represents an electronegative atom. This is because electronegative atoms in polar bonds have a tendency to acquire the electrons from the bond and release a free hydrogen ion, making the acid’s bond to the hydrogen atom more labile. A molecule is said to be basic when it readily accepts a proton. The stronger the base, the more basic it is, and the more easily it abstracts a proton from an acid. Bases tend to contain regions of high electron density, almost always in the form of a lone pair of electrons. Note that acids can have one (e.g., hydrochloric acid), two (e.g., carbonic acid) or even three labile X-H (e.g., phosphoric acid) bonds.

pH describes the acidity of a solution Free protons are unstable on their own, and in aqueous solutions they + readily combine with water to form hydronium ions (H3O ): H H O O H H H H

+ The total concentration of H3O in a solution is a measure of the overall + acidity of the solution. The acidity of the solution increases as [H3O ] + increases. Because [H3O ] often varies by many orders of magnitude, it is expressed logarithmically as pH, which equals the negative base-10 + + logarithm of [H3O ] (i.e., pH = −log10[H3O ]). Solutions with low pH values are the most acidic, and solutions with high pH values are the most basic. Solutions with pH values near 7.0 are considered roughly neutral, with higher pH values being basic and lower pH values being acidic. Earlier Chapter 3 Thermodynamics & Acidity 15

Box 1 Le Châtelier’s principle in photosynthesis

Many plants, including many important food crops like corn and sugar cane, use a specialized form of − photosynthesis that uses bicarbonate (HCO3 ) as a substrate. As we saw in Figure 9, however, the series of steps that transform atmospheric carbon dioxide into bicarbonate are all unfavorable, especially the hydration and acid dissociation steps. Nonetheless, plants are able to produce significant concentrations of bicarbonate through the use of Le Châtelier’s principle. Plants do this by keeping the pH (a measure of + [H3O (aq)]) constant in the cell through a complex process whose details we will not discuss here. Suffice + + it to say that when extra H3O is produced in the cell, it is immediately removed, ensuring that [H3O (aq)] + does not change. In this way, the plant couples the dissolution of carbon dioxide to removal of H3O , thereby driving the conversion of atmospheric carbon dioxide to bicarbonate.

+ To understand why keeping [H3O (aq)] constant allows plants to sequester enough bicarbonate to carry out photosynthesis, we must look at the dissociation of carbonic acid. Despite the fact that the reaction is highly unfavorable, nearly all carbonic acid in the cell dissociates into hydronium and bicarbonate. To understand why this occurs, consider the equilibrium constant for this reaction: [HCO -][H O+] = 3 3 Keq [H2CO3][H2O]

Remember that Keq is constant for a reaction at a given temperature; it is always the same, and the cell cannot − change it. Notice, however, that [H2CO3] and [HCO3 ] are not the only terms present in the equilibrium + constant for this reaction. It also includes [H2O] and [H3O ]. Therefore, the cell could alter the equilibrium − + ratio of [HCO3 ] to [H2CO3] by adjusting either [H2O] or [H3O ]. The concentration of water in the cell is + extremely high (approximately 55 M) and effectively unchangeable, but [H3O ] is held constant at a relatively −8 + low concentration of about 4.0 x 10 M. When additional H3O is produced in the cell by some chemical + reaction, such as by the dissociation of H2CO3, those H3O molecules are removed from the cell such that + −6 [H3O ] remains constant. Knowing that Keq must always equal 4.0 x 10 , [H2O] equals 55 M, and that + −8 [H3O ] is held constant at 4.0 x 10 M, we can substitute these values into the formula for Keq to solve for the − equilibrium ratio of [HCO3 ] to [H2CO3] that would exist in a cell: [HCO -][H O+] 3 3 = Keq [H2CO3][H2O]

[HCO -][4.0 x 10-8] 3 = 4.0 x 10-6

[H2CO3][55]

[HCO -] (55) (4.0 x 10-6) 3 = = 5500 -8 [H2CO3] (4.0 x 10 )

− + As this calculation shows, the cell is able to keep the ratio of [HCO3 ] to [H2CO3] high by keeping [H3O ] + artificially low. This situation is an example of Le Châtelier’s principle; the product 3(H O ) is continually removed from the equilibrium, which causes the equilibrium to shift toward products, thereby increasing − the concentration of HCO3 and decreasing the concentration of H2CO3. This perturbation is equivalent to the example shown on the last line of Figure 8C, in which removal of one product increases the equilibrium concentration of the other product. Because the dissociation equilibrium shifts toward products, the Chapter 3 Thermodynamics & Acidity 16

concentration of H2CO3 (a reactant) decreases. The decrease in [H2CO3], in turn, perturbs the hydration

equilibrium because H2CO3 is a product of that reaction. Therefore, a decrease in [H2CO3] causes the

hydration equilibrium to shift toward products, which in turn decreases the concentration of aqueous CO2

(a reactant). The resulting decrease in [CO2 (aq)] in turn shifts the dissolution equilibrium toward products, causing the dissolution of additional carbon dioxide from the atmosphere. As you can see, cells are able to capture carbon dioxide from the atmosphere and produce bicarbonate by applying Le Châtelier’s principle to + a series of coupled equilibria. By removing H3O , cells initiate a perturbation in the dissociation equilibrium that propagates through each equilibrium in the series, ultimately resulting in the flux of material through a series of otherwise unfavorable chemical reactions. Living systems commonly exploit Le Châtelier’s principle in this way to drive chemical reactions; indeed, we will see other examples in later chapters.

+ −8 we saw that [H3O ] in most cells is approximately 4 x 10 , equating to a pH of 7.4. Therefore, the pH of a cell is nearly neutral. Figure 10 shows the pH values of a few commonly encountered substances. Because of the base-10 logarithmic scale, each pH unit represents a factor of 10. For example, a solution with a pH of 6 is 10 times more acidic than a solution with a pH of 7 and 100 times more acidic than a solution with a pH of 8.

Ka and pKa describe the strength of an individual acid molecule pH measures the acidity of a solution as a whole, which may contain several different acids of different strengths. To quantify the strength of an individual acid, we use the equilibrium constant of its acid dissociation reaction, that is, the reaction of the acid with water to produce its conjugate base and hydronium. Because the concentration of water is essentially constant, it is

combined with Keq, as shown in Figure 11, to define a new term called the

acid dissociation constant, or Ka.

Acids with large Ka values are stronger than acids with small Ka values. Like + [H3O ], Ka values vary across many orders of magnitude, and are therefore

often expressed logarithmically as pKa, which equals the negative base-10

logarithm of the Ka (i.e., pKa = –log10 Ka). Acids with low pKa values are the

milk (6.5) lemon juice (2.2) physiological 1 M HCl (0) soda (2.5) pH (7.4) soapy water (10.0) bleach (12.6)

pH 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14

gastric acid co ee (5.0) seawater (7.5 to 8.4) household ammonia (1.5 to 3.5) (11.8)

Figure 10 The acidity of common solutions varies across many orders of magnitude Chapter 3 Thermodynamics & Acidity 17

Generic acid dissociation Carbonic acid dissociation

Acid dissociation H-A + H O A- + H O+ H CO + H O HCO - + H O+ reaction (aq) 2 (l) (aq) 3 (aq) 2 3 (aq) 2 (l) 3 (aq) 3 (aq)

- + - + Equilibrium [A ] [H3O ] [HCO3 ] [H3O ] Keq = Keq = constant (Keq) [H-A] [H2O] [H2CO3] [H2O]

- + - + Acid dissociation [A ] [H3O ] [HCO3 ] [H3O ] Ka = [H2O] Keq = Ka = [H2O] Keq = constant (Ka) [H-A] [H2CO3]

Figure 11 Ka quantifies the strength of an individual acid molecule

Shown here are the acid dissociation reaction, Keq expression, and Ka expression for a generic acid dissociation and the dissociation of carbonic acid where H-A is a generic acid and A− is the resulting conjugate base.

most acidic. As with pH, each pKa unit represents a factor of 10. Therefore,

an acid with a pKa of 1 is 10 times more acidic than an acid with a pKa of

2 and 100 times more acidic than an acid with a pKa of 3. For acids with two or more labile hydrogens, each individual proton within an acid has its

own unique Ka and pKa values. Ka and pKa values are a property of a given acid; because they are equilibrium constants, they do not change during the

course of the reaction. Figure 12 shows the pKa values of a few commonly encountered acids.

A molecule’s protonation state depends on the pH of the surrounding solution

The protonation state of a molecule depends on both its pKa and the pH of the surrounding solution. The protonation state of a molecule can be related

to the pKa and the pH of the solution using the Henderson-Hasselbalch

equation, which can be derived from the Ka expression as shown in Figure 13. The Henderson-Hasselbalch equation is important for understanding the molecules of life because it makes it possible to determine whether a molecule is predominately protonated or deprotonated at a particular pH. Since the protonation state determines whether or not a molecule is charged, determining protonation state is crucial for understanding the properties of molecules and how they will interact with other molecules.

Summary Chemical reactions are driven by the release of free energy. Reactions are favorable only if they move to a lower energy state, releasing energy in the process. A reverse reaction of a favorable reaction absorbs energy instead of releasing it, making such processes unfavorable. Equilibrium is a state in which the concentrations of the reactants and products of a chemical reaction no longer change. The reaction does not stop at equilibrium; instead, the forward and reverse reactions take place at the same rate such that no net production of products or reactants Chapter 3 Thermodynamics & Acidity 18

Acid Conjugate Base Ka pKa

H H Hydronium ion O H + O 50 -1.7 H H H

H O O Phosphoric acid O P OH H + O P OH 7.1 x 10-3 2.2 OH OH

H Increasing acidity O O O O O O O O Citric acid H + 7.4 x 10-4 3.1 HO OH HO OH OH OH

O O Acetic acid H H + 1.7 x 10-5 4.8 O O

H N Ammonium ion N + H H -10 H H H H 5.6 x 10 9.3 H

H -16 Water O H + O 1.8 x 10 15.7 H H

Figure 12 Ka values vary over many orders of magnitude

Shown here are the pKa values of a few acids that are relevant to biological systems. Acids with small pKa values are stronger than acids − with large pKa values. Notice that some H-A acids (e.g., phosphoric and acetic acids) are neutral and have negative conjugate bases (A ), whereas other H-A acids (e.g., hydronium and ammonium) are positively charged and have neutral conjugate bases.

[A−] [H O+] Figure 13 The Henderson- K = 3 Hasselbalch equation derives from a [HA] the acid dissociation constant [HA] + The figure shows the derivation of the K = [H O ] a [A−] 3 Henderson-Hasselbalch equation for a generic acid dissociation reaction (Figure [HA] 11). The acid dissociation constant top( −log K = −log ([H O+]) row) is rearranged to give the equation ( a [A−] ) 3 shown in the second row. Next, the negative logarithm is taken of both sides (third row), [HA] −log K - log = −log ([H O+]) and the logarithm on the left side of the a [A−] 3 equation is simplified using the rule of logs (log AB = log A + log B) (fourth row). To [HA] complete the derivation, pK and pH are pK −log = pH a a − substituted into the equation in place of [A ] + “−log Ka” and “−log [H3O ]”, respectively (fifth row), producing the Henderson- Henderson-Hasselbalch [HA] pK = pH + log Hasselbalch equation (bottom row). Equation a [A−] Chapter 3 Thermodynamics & Acidity 19

Box 2 Assessing a molecule’s protonation state

The Henderson-Hasselbalch equation can be used to calculate the molar ratio of the protonated acid to the deprotonated conjugate base that exists at a particular pH. For example, consider a molecule with a

pKa of 6.0 that is dissolved in a solution with a pH of 7.0. If we substitute these values into the Henderson- Hasselbalch equation, we see that the fraction of protonated to deprotonated molecule is 1:10, as follows:

[HA] pK = pH + log a 10 [A−]

[HA] 6.0 = 7.0 + log 10 [A−]

[HA] −1 = log 10 [A−]

1 [HA] = 10 [A-]

takes place. Equilibrium is the lowest energy state for a chemical system.

Equilibrium is described using the equilibrium constant (Keq), which equals the ratio of product concentrations to reactant concentrations. The

value of Keq is specific to each reaction and does not change as the reaction

proceeds. A reaction’s Keq value describes whether the reactants or products

are favored. When Keq is greater than one, the products are lower in energy

than the reactants, and the reaction is favorable. When Keq is less than one, the products are higher in energy than the reactants, and the reaction is unfavorable. Changes in Gibbs free energy are a measure of the amount of energy that is

absorbed or released as a reaction proceeds. ΔG°rxn describes the difference in Gibbs free energy between a state of pure reactants and the alternative

state of pure products. Reactions for which ΔG°rxn is negative are favorable,

and reactions for which ΔG°rxn is positive are unfavorable. Like Keq, ΔG°rxn is a property of a given reaction, and its value does not change as the reaction proceeds. Conversely, ΔG describes the Gibbs free energy available in a system at a given ratio of reactants and products. The value of ΔG does change as the reaction proceeds, and its sign predicts the net direction in which a reaction with a given ratio of reactants and products will proceed. Reactions proceed toward products when ΔG is negative; conversely, reactions proceed toward reactants when ΔG is positive. ΔG equals zero

when the system is at equilibrium. When ΔG is zero, ΔG°rxn is related to Keq by the following simple equation:

ΔG°rxn = –RT ln Keq Living systems utilize energy released during metabolism to drive unfavorable reactions. Favorable processes, such as ATP hydrolysis, often drive unfavorable processes so that the overall net process is favorable. Cells also drive unfavorable processes using Le Châtelier’s principle, the concept Chapter 3 Thermodynamics & Acidity 20

Breakout Shown below is the dissociation of acetic acid in water.

H3C H3C C O H2O C O H3O O H O

pKa = 5

What do you think will happen if the pH of the solution is raised to 7 and the reaction is allowed to reach equilibrium at this new pH?

A. As the pH increases, the magnitude of ΔG°rxn decreases. B. As the pH increases, ΔG for the forward reaction becomes positive. C. As the pH increases, more reactants are converted to products. D. Both B and C are correct.

that chemical systems respond to perturbations (e.g., removal of product or addition of reactant) by adopting a new equilibrium state with a different composition of reactants and products. Acid-base reactions are common in living systems. Such reactions involve the movement of a hydrogen ion, or proton, from an acid to a base. The acid becomes deprotonated during the course of the reaction to give rise to the conjugate base. Similarly, the base becomes protonated during the reaction to give rise to the conjugate acid. The strength of an acid is indicated by the

acid dissociation constant (Ka), which is related to the equilibrium constant

of the acid’s dissociation reaction. Ka is usually expressed logarithmically

as pKa. Acids with low pKa values are more acidic than acids with higher

pKa values. The acidity of an overall solution is indicated by pH. Solutions with low pH values are more acidic than solutions with higher pH values. The Henderson-Hasselbalch equation is used to determine the protonation state of an acid in a solution at a particular pH: [HA] pK = pH + log a [A-] Chapter 3 Thermodynamics & Acidity 21

Practice problems

1. The following questions deal with the conversion of glyceraldhyde-3-phosphate (G3P) to dihydroxyacetone phosphate (DHAP), a step in the breakdown of glucose (‘glycolysis’) shown below. G3P DHAP a. Write an equation that expresses the equilibrium constant for the conversion of G3P to DHAP in terms of [G3P] and [DHAP]. b. The equilibrium constant for this reaction has been measured to be 22. Which molecule, G3P or DHAP, will be more abundant when the reaction reaches equilibrium? c. Given an equilibrium constant of 22, complete each of the following sentences by circling the correct term. i. The conversion of G3P to DHAP is thermodynamically ______(favorable/unfavorable). ii. The conversion of DHAP to G3P is thermodynamically ______(favorable/unfavorable). 2. Rank the following the following molecules from weakest to strongest acid:

Hydrofluoric acid, pKa = 3.14

Sodium bisulfite, pKa = 6.91 −5 Citric acid, Ka = 1.8 x 10

Hydrochloric acid, pKa < 1 3. Indicate whether the following statements are true or false. a. The conjugate base is the deprotonated form of an acid. b. The protonation of a base raises the pH of a solution. c. pH describes the acidity of a solution and not the acidity of an individual acid.

d. The pKa of a compound varies with changing pH.

e. Acids with large pKa values are weaker than acids with small pKa values.

f. The strength of an individual acid is described by its pKa.

4. The pKa of formic acid is 3.75. At what pH would equal concentrations of its con- jugate acid and conjugate base be present? At physiological pH of 7.4, what charge would most of the molecules of formic acid have? Formic acid 5. When you add vinegar (acetic acid in water) to baking soda (sodium bicarbonate,) the chemicals react to produce carbon dioxide. In fact, this reaction goes to completion, using up all reactants to form products. Given what you know about equilibria, how is this possible?

6. The amino acid glycine has two pKa values and is frequently drawn as shown. Draw the predominant structure of glycine: a. when it is in your stomach at pH 2 b. in your blood at pH 7.4 c. in your small intestine at pH 8.5 d. in sodium carbonate solution at pH 10

7. Oxygen has an electronegativity of 3.44, while that of Chlorine is 3.16. However, HCl is more acidic than water. Explain. (Solutions are located on the next page) Chapter 3 Thermodynamics & Acidity 22

Solutions to practice problems

Question 1:

a. Keq = [DHAP]/[G3P] b. DHAP c. i., favorable; ii., unfavorable

Question 2: Weak to strong: sodium bisulfite, citric acid, hydrofluoric acid, hydrochloric acid

Question 3: a., true; b., true; c., true; d., false; e., true; f., true.

Question 4: pH of 3.75; negative

Question 5: We must consider two different reactions: Rxn 1 Rxn 2 Adding vinegar (acid) to bicarbonate causes the equilibrium of reaction 1 to shift to the right, as the added acid is partly consumed by conversion of the bicarbonate ion to carbonic acid. The greater con-

centration of carbonic acid also causes reaction 2 to shift to the right, producing CO 2. Each of these shifts as a response to an increase in concentration of a reactant is an example of Le Chatalier’s principle.

Question 6: a. b. c. d.

Question 7: While electronegativity measures propensity for negative charge accumulation, as does acidity, the com- parisons made are different. Acidity is a measure of the stability of the anion (A-) compared to its acidic proton-bound form (AH). Oxygen atoms are smaller than chlorine atoms, so the covalent bond to a hydrogen atom, which depends on overlap of orbitals, will be considerably stronger. Similarly, chlorine orbitals are much larger and more diffuse than those of hydrogen, so the covalent bond will be weaker. We can clarify the distinction by considering deprotonation of an acid (HA -> H+ + A-) in two steps: (1) cleavage of the bond, assigning one of the bonding electrons to each portion (making an H atom and an unstable free radical A) and (2) transfer of the electron to form A- and H+. The second step, accepting an electron, is more favorable for the more electronegative oxygen. However, the first step, breaking the bond to the hydrogen atom, is much more difficult for the oxygen, since its smaller size leads to a stronger bond to H. For this reason, acidity does not strictly follow electronega-

tivity and HCl is more acidic than H2O