Medical Physiology, 3rd Edition

Acid-Base Chemistry When image Is the Only Buffer

In this section, we consider buffering by CO2 and image when these are the only buffers present in the solution. We defer to the following section the more complex example in which both image and image buffers are present in the same solution.

In the absence of other buffers, doubling image causes pH to fall by 0.3 but causes almost no change in [image]

Figure 28-5 represents 1 L of a solution with the same image composition as arterial blood plasma but no buffers other than image. What are the consequences of increasing image in the gas phase? The resulting increase in [CO2]Dis causes the image equilibrium to shift toward formation of H+ and image. This disturbance is an example of a CO2 titration, because we initiated it by altering image. More specifically, it is a respiratory acidosis—“acidosis” because pH falls, and “respiratory” because pulmonary problems (Table 28-3) are the most common causes of an increase in the image of arterial blood (see p. 680).


FIGURE 28-5 Doubling of CO2 or image concentrations.

TABLE 28-3

The Four Major Acid-Base Disorders





Respiratory acidosis

↑ image

↓ Alveolar ventilation (e.g., drug overdose)

↓ Lung-diffusing capacity (e.g., pulmonary edema)

Ventilation-perfusion mismatch

pH: ↓

[image]: ↑

image: ↑

Respiratory alkalosis

↓ image

↑ Alveolar ventilation caused by:

Hypoxia (e.g., acclimatization to high altitude)


Aspirin intoxication

pH: ↑

[image]: ↓

image: ↓

Metabolic acidosis

Addition of acids other than CO2 or H2CO3

Removal of alkali (fixed image)

↓ Urinary secretion of H+ (e.g., renal failure)

Ketoacidosis (e.g., diabetes mellitus)

Lactic acidosis (e.g., shock)

image loss (e.g., severe diarrhea)

pH: ↓

[image]: ↓

image: No change

Metabolic alkalosis

Addition of alkali

Removal of acids other than CO2 or H2CO3 (fixed image)

image load (e.g., NaHCO3 therapy)

Loss of H+ (e.g., severe vomiting)

pH: ↑

image: ↑

image: No change

In the absence of image buffers, how far pH falls during respiratory acidosis depends on the initial pH and image, as well as on the final image. For example, doubling image from 40 mm Hg (see Fig. 28-5, stage 1) to 80 mm Hg causes [CO2]Dis to double to 2.4 mM (see Fig. 28-5, stage 2A). At this point, the 1-L system is far out of equilibrium and can return to equilibrium only if some CO2 (X mmol) combines with X H2O to form X H+ and X image. Thus, according to Equation 28-14, the following must be true at equilibrium:



Solving this equation for X yields an extremely small value, nearly 40 nmol, or 0.000,040 mmol. X represents the flux of CO2 that passes through the reaction sequence CO2 + H2O → H2CO3 → H+ + image to re-establish the equilibrium. CO2 from the atmosphere replenishes the CO2 consumed in this reaction, so that [CO2]Dis remains at 2.4 mM after the new equilibrium is achieved (see Fig. 28-5, stage 3A). The reason that the flux X is so small is that, with no other buffers present, every H+ formed remains free in solution. Thus, only a minuscule amount of H+ need be formed before [H+] nearly doubles from 40 to nearly 80 nM. However, [image] undergoes only a tiny fractional increase, from 24 to 24.000,040 mM. The doubling of the denominator in Equation 28-21 is matched by a doubling of the numerator, nearly all of which is due to the near-doubling of [H+]. The final pH in this example of respiratory acidosis is



Another way of arriving at the same answer is to insert the final values for [image] and image into the Henderson-Hasselbalch equation:



The opposite acid-base disturbance, in which image falls, is respiratory alkalosis. In a solution containing no buffers other than image reducing image by half, from 40 to 20 mm Hg, would cause all of the aforementioned reactions to shift in the opposite direction, so that pH would rise by 0.3, from 7.4 to 7.7. We could produce respiratory alkalosis in a beaker by lowering the image in the gas phase. In humans, hyperventilation (see Table 28-3) lowers alveolar and thus arterial image (see p. 680).

Thus, in the absence of image buffers, doubling [CO2] causes pH to fall by 0.3, whereas halving [CO2] causes pH to rise by 0.3 (see Table 28-1). Remember, the log of 2 is 0.3.

In the absence of other buffers, doubling [image] causes pH to rise by 0.3

What would happen if we doubled [image], rather than image? If we start with 1 L of solution that has the ionic composition of arterial blood (see Fig. 28-5, stage 1), adding 24 mmol of image (e.g., NaHCO3) drives the image equilibrium toward CO2 (see Fig. 28-5, stage 2B). The new equilibrium is achieved when X mmol of image combines with X H+ to produce X CO2 and X H2O. Because the system is open to CO2, the generation of X mmol CO2 causes no change in [CO2]Dis; the newly formed CO2 simply evolves into the atmosphere. Thus, according to Equation 28-14, the following must be true at equilibrium:



Solving this equation yields an X of nearly 0.000,020 mM. Thus, in the absence of other buffers, an initial doubling of [image] from 24 to 48 mM causes [H+] to fall by nearly half, from 40 to 20 nM. Although [image] also decreases by 20 nM, its fractional change from 48 to 47.999,980 mM is insignificant. The final pH is:



The Henderson-Hasselbalch equation yields the same result:



Thus, in the absence of image buffers, doubling [image] causes pH to increase by ~0.3. Rather than adding image, we could produce an identical effect by adding the same amount of strong base (e.g., NaOH), as we did in Figure 28-3, stages 2B and 3B, or by removing the same amount of strong acid (e.g., HCl). Although the removal of HCl from a solution in a beaker may seem artificial, the removal of gastric HCl from the body occurs during vomiting (see Table 28-3). imageN28-7 We also could add a weak base (e.g., NH3) or remove any weak acid (e.g., lactic acid) other than CO2 or H2CO3. However, to produce the same alkali load, we would have to add/remove more of a weak base/acid than for a strong base/acid. All of these maneuvers, carried out at a fixed image, are examples of metabolic alkalosis—“alkalosis” because the pH increases, and “metabolic” because derangements in metabolism are common clinical causes.


Metabolic Alkalosis Caused by Vomiting

Contributed by Walter Boron

Because the stomach secretes HCl into the lumen of the stomach, vomiting necessarily results in a net loss of HCl, imageN28-9 and this by itself causes some degree of metabolic alkalosis (see p. 635). In addition, the loss of Cl leads to volume contraction, that is, a decrease in effective circulating volume (see pp. 554–555). The body's multipronged response to this volume contraction (see p. 838) includes a stimulation of the renin-angiotensin-aldosterone axis, the end point of which is increased retention of NaCl and the osmotically obligated H2O. However, at the level of the renal proximal tubule, the increased levels of systemic [angiotensin II] lead not only to an increase in NaCl reabsorption, but also to an increase in NaHCO3 reabsorption. Moreover, at the level of the distal nephron, the increased levels of aldosterone lead not only to an increase in NaCl reabsorption, but also to an increase in NaHCO3 reabsorption. The result is that the hormonal stimuli that represent the appropriate response to volume depletion lead, as a side effect, to a metabolic alkalosis. The proper treatment of this contraction alkalosis is not to infuse the patient with NaHCO3, but rather to replenish the lost volume by delivering “normal saline” intravenously. With effective circulating volume returned to normal, the stimulus to the renin-angiotensin-aldosterone axis is removed, and the acid-base disturbance resolves.

It is interesting to note that although volume depletion causes the kidney not only to retain NaCl but also to retain NaHCO3 inappropriately, the converse is not true. That is, respiratory or metabolic acidosis cause the kidney to retain NaHCO3 but to decrease the reabsorption of NaCl, so as to maintain a constant effective circulating volume.

What this all teaches us is that the defense of effective circulating volume trumps everything (including acid-base homeostasis). However, the body's mechanisms for regulating acid-base balance are far more sophisticated than those for regulating effective circulating volume. Thus, the response to acidosis or alkalosis is precise and appropriate—and does not involve inappropriate changes in effective circulating volume that would otherwise lead to hyper- or hypotension!


Equivalency of Adding image, Adding OH, and Removing H+

Contributed by Walter Boron

Imagine that we start with 1 L of an aqueous solution at 37°C. This solution mimics blood plasma, 5% CO2/24 mM image at pH 7.4, and the system is “open” for CO2 (i.e., CO2 can freely enter and leave the aqueous solution so as to keep [CO2] constant). Moreover, this solution contains a image buffer (HB(n+1) ⇄ H+ + B(n)) with a buffering power of 25 mM/pH unit.

Add NaHCO3

If we were to add 1 mmol of NaHCO3 to this 1 L of solution, then [image] would instantly increase by 1 mM. Of course, over time, subsequent reactions cause [image] to gradually fall as the added image equilibrates with the pre-existing image buffer according to the overall reactions:


(NE 28-25)

Furthermore, because the consumption of image goes hand in hand with the consumption of H+, the pH will gradually rise, which in turn upsets the equilibrium of image buffers:


(NE 28-26)

Thus, the rise in pH causes the HB(n+1)/B(n) equilibrium to shift to the right, thereby increasing [B(n)]. As a result, the final image would be <1 mM. However, the changes in [image] and [B(n)] would sum to very nearly 1 mM (i.e., the amount of NaHCO3 originally added):


(NE 28-27)

The reason why we use “approximately equal” in the above equation is that a tiny amount of the added image causes the pH to rise (i.e., it is not reflected in the Δ[B(n)]).

Add NaOH

Imagine that instead of adding NaHCO3, we add 1 mmol of NaOH to our 1 L of solution. We can imagine that virtually all of the added OH would react with the CO2 to form an equivalent amount of image:


(NE 28-28)

(Note: this reaction is equivalent to CO2 + H2O → image + H+.) Thus, [image] would rise by very nearly 1 mM. The reason we say “very nearly” is that a tiny fraction of the added OH would equilibrate with H+ and H2O, causing pH to rise:


(NE 28-29)

This consumption of OH would cause pH to rise, which in turn would cause the HB(n+1)/B(n) equilibrium to shift to the right:


(NE 28-30)

As a result [B(n)] would rise by a tiny amount. At the same time, however, the reaction CO2 + H2O ⇄ image + H+ is out of equilibrium because the image that we formed in Equation NE 28-28 yields an [image] far too high for the prevailing [H+]. Thus, Equation NE 28-28 would partially reverse, releasing more OH to react with H+, and leading to the formation of still more B(n). This reversal of Equation NE 28-28 (the secondary fall in [image]) would continue until eventually the reactions in both Equation NE 28-28 and Equation NE 28-30 are in equilibrium. This process is illustrated by the panels on the right side of Figure 28-7. As a result, the final image would be less than the ~1 mM that we predicted in Equation NE 28-28. However, the changes in [image] and [B(n)] would sum to very nearly 1 mM (i.e., the amount of NaOH added):


(NE 28-31)

In fact, the values of image and Δ[B(n)] in Equation NE 28-31 (i.e., the example in which we added 1 mmol NaOH) are precisely the same as in Equation NE 28-27 (i.e., the example in which we added 1 mmol NaHCO3).

Two provisos here. First, the reaction in Equation NE 28-28 would be very slow in the absence of added carbonic anhydrase.imageN18-3 Nevertheless, even in the absence of carbonic anhydrase, this reaction would eventually come to equilibrium, and we could imagine that [image] would eventually rise by very nearly 1 mM. Second, in this thought experiment, we are “imagining” that the added OH initially reacts only with CO2 to form ~1 mM image, and that some of this image then dissociates to yield some CO2 and OH. In fact, a sizeable amount of the OH that we originally added will also react instantly with H+ (according to Equation NE 28-29), in turn leading to the formation of B(n) in Equation NE 28-30.

Remove HCl

Finally, imagine that we magically remove 1 mmol of HCl from a liter of our image solution. In real life, we could remove this H+ either by consuming it in a chemical (or biochemical) reaction or by transporting it out of some biological compartment (e.g., the cytosol). We could imagine that the HCl that we remove causes the image equilibrium to shift, thereby causing [image] to rise by very nearly 1 mM:


(NE 28-32)

The consumed CO2 would come from the atmosphere. A small amount of the H+ that we removed would lower pH:


(NE 28-33)

However, the reaction in Equation NE 28-32 would be badly out of equilibrium because the rise in [image] would be out of proportion to the fall in [H+]. Thus, the reaction in Equation NE 28-32 would begin to reverse, consuming some of the ~1 mM image that we initially formed. Moreover, the HB(n+1)/B(n) reaction would also be out of equilibrium because [H+] has fallen without any change in [B(n)] or [HB(n+1)]. As a result, the following will occur:


(NE 28-34)

In fact, the reaction in Equation NE 28-32 would continue to reverse, releasing some image as well as an equal amount of H+ for consumption in Equation NE 28-34. Eventually, the reactions in Equations NE 28-32 through NE 28-34 would be in equilibrium. At this point,


(NE 28-35)

In fact, the values of image and Δ[B(n)] in Equation NE 28-35 (i.e., this example in which we removed 1 mmol HCl) are precisely the same as in Equation NE 28-27 (i.e., the example in which we added 1 mmol NaHCO3) and in Equation NE 28-31 (i.e., the example in which we added 1 mmol NaOH).


Thus, regardless of whether we add 1 mmol of NaHCO3, add 1 mmol of NaOH, or remove 1 mmol of HCl … in the end, the result is the same. The image and the HB(n+1)/B(n) buffer systems will both wind up being in equilibrium, and in each example, [image] will increase by the same amount, [B(n)] will increase by the same amount, and pH will increase by the same amount.

The opposite acid-base disturbance—in which we remove image or another alkali, or add an acid other than CO2 or H2CO3—is metabolic acidosis. If we start with a solution having the ionic composition of arterial blood, removing half of the initial image or removing 12 mM NaOH or adding 12 mM HCl would cause all of the aforementioned reactions to shift in the opposite direction, so that pH would fall by 0.3, from 7.4 to 7.1. Common causes of metabolic acidosis in humans are renal failure and metabolic problems (see Table 28-3).

We saw in Equation 28-26 that the pH of a image solution does not depend on [image] or image per se, but on their ratio. Because it is the kidney that controls [image] in the blood plasma (see Chapter 39), and because it is the lung that controls image (see Chapter 31), the pH of blood plasma is under the dual control of both organ systems, a concept embodied by a whimsical variant of the Henderson-Hasselbalch equation: