So far, our discussion has focused on simple systems in which the only pH buffers are either (1) one or more buffers or (2) . Under these circumstances, it is relatively easy to predict the effects of acidbase disturbances. Real biological systems, however, are mixtures of and many buffers. Thus, to understand the effects of respiratory and metabolic acidbase disturbances in a biological system, we must consider multiple competing equilibria—one for (described by Equation 2816) and one for each of the buffers (each described by its version of Equation 285). Obtaining a precise solution to such clinically relevant problems is impossible. One approach is to use a computer to make increasingly more precise approximations of the correct answer. Another, more intuitive approach is to use a graphical method to estimate the final pH. The Davenport diagram is the best such tool.
The Davenport diagram is a graphical tool for interpreting acidbase disturbances in blood
What happens to the pH of blood—a complex mixture that includes many buffers—when doubles? We simplify matters by lumping together the actions of all buffers, so that H^{+} + B^{(n)} ⇄ HB^{(n+1)} represents the reactions of all buffers. When we raise , almost all of the newly formed H^{+} reacts with B^{(n)} to form HB^{(n+1)} so that the free [H^{+}] rises only slightly.
(2828)
In this simplified approach, the final pH depends on two competing buffer reactions—one involving and the other involving HB^{(n+1)}/B^{(n)}. Computing the final pH requires solving two simultaneous equations, one for each buffer reaction.
The Buffer
The first of the two equations that we must solve simultaneously is a rearrangement of the HendersonHasselbalch equation (see Equation 2816):
(2829)
For a of 40 mm Hg, the equation requires that [] be 24 mM when pH is 7.40—as in normal arterial blood plasma. If pH decreases by 0.3 at this same , Equation 2829 states that [] must fall by half to 12 mM. Conversely, if pH increases by 0.3, [] must double to 48 mM. The blue column of Table 284 lists the [] values that Equation 2829 predicts for various pH values when is 40 mm Hg. Plotting these [] values against pH yields the blue curve labeled “ = 40 mm Hg” in Figure 286A. This curve, which is known as a CO_{2} isobar or isopleth (from the Greek isos [equal] + plethein [to be full]) represents all possible combinations of ] and pH at a of 40 mm Hg. Table 284 also summarizes [] values prevailing at values of 20 and 80 mm Hg. Note that at a of 20 (orange column), representing respiratory alkalosis, [] values are half those for a of 40 at the same pH. At a of 80 (green column), representing respiratory acidosis, [] values are twice those at a of 40. Each of the isopleths in Figure 286A rises exponentially with pH. The slope of each isopleth also rises exponentially with pH and represents β_{open} for . At a particular pH, an isopleth representing a higher (i.e., a higher []) has a steeper slope, as Equation 2820 predicts.
TABLE 284
Relationship between and pH at Three Fixed Levels of
[] (mM) 

pH 
= 20 mm Hg 
= 40 mm Hg 
= 80 mm Hg 
7.1 
6 mM 
12 mM 
24 mM 
7.2 
8 
15 
30 
7.3 
10 
19 
38 
7.4 
12 
24 
48 
7.5 
15 
30 
60 
7.6 
19 
38 
76 
7.7 
24 
48 
96 
FIGURE 286 Davenport diagram. In A, the three isopleths are CO_{2} titration curves. In B, the solution contains nine buffers, each at a concentration of 12.6 mM and with pK values evenly spaced 0.5 pH unit apart, as in Figure 282B. In C, the red arrow represents the transition from a normal acidbase status for blood (“Start”) to respiratory acidosis (point A) produced by raising to 80 mm Hg. Point B represents respiratory alkalosis produced by lowering to 20 mm Hg. In D, the labeled points represent the results of respiratory acidosis when is zero (point A_{1}), 25 mM/pH (point A), and infinity (point A_{2}).
Buffers
The second of the two equations that we must solve simultaneously describes the lumped reaction of all buffers. To understand the origin of this equation, we begin with the green curve (surrounded by black curves) in Figure 286B, the titration curve of a single buffer with a pK of 7 and total buffer concentration of 12.6 mM (as for the green curve in Fig. 282B). At a pH of 10, [HB^{(n+1)}] for this single buffer is extremely low because almost all of the “B” is in the form B^{(n)}. As we lower pH by successively adding small amounts of HCl, [HB^{(n+1)}] gradually rises—most steeply when pH equals pK. Indeed, at any pH, the slope of this curve is the negative of β for this single buffer. The black curves in Figure 286B are the titration curves for eight other buffers, each present at a [TB] of 12.6 mM, with pK values evenly spaced at intervals of 0.5 pH unit on either side of 7. The red curve in Figure 286B is the sum of the titration curves for all nine buffers. N288 Its slope—the negative of the total buffering power of all nine buffers—is remarkably constant over a broad pH range. (The red curve in Figure 282B shows how β, the slope of the red curve here in Figure 286B, varies with pH.) The red curve in Figure 286B represents the second of the two equations that we must solve simultaneously.
N288
Davenport Diagram—Constructing the Buffer Line
Contributed by Emile Boulpaep, Walter Boron
The buffer line (the red line in Fig. 286B, C) describes the buffering power of all buffers other than (). In whole blood—here we refer to the line as the blood buffer line—hundreds of buffer groups may contribute to the overall buffering power. Each of these buffer groups has its own pK and concentration, and thus its own bellshaped curve—like the nine small curves in Figure 282B—describing the pH dependence of its buffering power. Each of these buffer groups also has its own titration curve, as shown in Fig. 286B, and the sum of these titration curves (the red curve in this figure) is the combined titration curve of these buffers (i.e., ). Notice that this red curve is nearly a straight line over the physiological pH range. In other words, in the physiological range, the slope of the red line is nearly constant, which is another way of stating that the buffering power is nearly constant over this pH range—as illustrated in Figure 282B.
We already know from Equation 288 (reproduced below as Equation NE 2821) that we can define buffering power in terms of the amount of strong acid added to the solution:
(NE 2821)
From the perspective of the buffers that are at work during respiratory acidosis, Δ[strong acid] is the H^{+} formed from CO_{2} and H_{2}O as a result of the increase in (see arrow from “Start” to point A in Fig. 286C). Because respiratory acidosis produces and H^{+} in a 1 : 1 ratio, the amount of formed during respiratory acidosis—that is —is the same as the amount of H^{+} added (Δ[strong acid]). Thus, we can replace Δ[strong acid] in Equation NE 2821 with and therefore somewhat paradoxically express the buffering power in terms of []:
(NE 2822)
and pH_{init} represent initial values (before increasing —that is, the point labeled “Start” in Fig. 286C), and [] and pH represent final values (after increasing —that is, point A in Fig. 286C). The equation describing the buffer line is simply a rearrangement of Equation NE 2822:
(NE 2823)
This equation describes a line whose slope is negative . For whole blood, is 25 mM/pH unit, so that the slope of the line is −25 mM/pH unit in Figure 286C. Note that because we assumed that the buffering power of the buffers is constant over the pH range of interest, and because this buffering power is represented by the slope of the buffer curve in Figure 286C, the function describing this buffering power must be a straight line (i.e., a function with a constant slope).
We are now in a position to understand why, in Figure 286C, the buffer line (see Equation NE 2823) shares the same yaxis as the CO_{2} isopleth (e.g., the blue curve in Fig. 286C), which is described by Equation 2829, reproduced here:
(NE 2824)
In both the case of the line describing the buffering power (see Equation NE 2823) and the exponential curve describing the opensystem CO_{2} buffering power (see Equation NE 2824), the dependent variable is [].
The buffer line passes through the point describing the initial conditions (pH_{init}, ). In Figure 286C, pH_{init} is 7.40 and is 24 mM, as indicated by the point labeled “Start.” Because these initial conditions for the buffers can change, we will see below in the text that the buffer line can translate up and down on the Davenport diagram in metabolic alkalosis and metabolic acidosis. However, as long as is constant, the slope of the line is always the same.
Solving the Problem
Figure 286C is a Davenport diagram, a combination of the three CO_{2} isopleths in Figure 286A and the linear part of the red titration curve in Figure 286B. The red titration line (representing for whole blood, 25 mM/pH unit) intersects with the CO_{2} isopleth for a of 40 at the point labeled “Start,” which represents the initial conditions for arterial blood. At this intersection, both and buffers are simultaneously in equilibrium.
We are now in a position to answer the question raised at the beginning of this section: What will be the final pH when we increase the of whole blood from 40 to 80 mm Hg? The final equilibrium conditions for this case of respiratory acidosis must be described by a point that lies simultaneously on the red titration line and the green isopleth for 80 mm Hg. Obtaining the answer using the Davenport diagram requires a threestep process:
Step 1: Identify the point at the intersection of the initial isopleth and the initial titration line (see Fig. 286C, “Start”).
Step 2: Identify the isopleth describing the final (80 mm Hg in this case).
Step 3: Follow the titration line to its intersection with the final isopleth. In Figure 286C, this intersection occurs at point A, which corresponds to a pH of 7.19 and an [] of 29.25 mM.
As discussed above, in the absence of buffers, this same doubling of causes a larger pH decrease, from 7.4 to 7.1.
By following three similar steps, we can use the Davenport diagram to predict the final pH and [] under conditions of a respiratory alkalosis. For example, what would be the effect of decreasing by half, from 40 to 20 mm Hg? In Figure 286C follow the red titration line from “Start” to its intersection with the orange isopleth for a of 20 mm Hg (point B), which corresponds to a pH of 7.60 and an [] of 19 mM. If the solution had not contained buffers, halving the would have caused a larger pH increase, from 7.4 to 7.7.
The amount of formed or consumed during “respiratory” acidbase disturbances increases with
Although it is reasonable to focus on CO_{2} during respiratory acidbase disturbances, it would be wrong to assume that [] is constant. For example, in Figure 286C, where is 25 mM/pH unit, increasing from 40 to 80 mm Hg causes [] to increase from 24 to 29.25 mM. Thus, in each liter of solution, 5.25 mmol CO_{2} combines with 5.25 mmol H_{2}O to form 5.25 mmol (which we see as the value between “Start” and point A) and 5.25 mmol H^{+}. Nearly all of this H^{+} disappears as nearly 5.25 mmol of the deprotonated buffers (B^{(n)}) consumes H^{+} to form nearly 5.25 mmol of their conjugate weak acids (HB^{(n+1)}). Thus, the flux through the reaction sequence in Equation 2828 is nearly 5.25 mmol for each liter of solution.
(2830)
Thus, the buffers drive the conversion of CO_{2} to . These buffers minimize the increase in free [H^{+}] that a given flux of CO_{2} can produce. Thus, with a high , a large amount of CO_{2}must “flux” through Equation 2830 before the free concentrations of and H^{+} rise sufficiently to satisfy the equilibrium (see Equation 2814).
varies with the hemoglobin content of blood. Thus, patients with anemia have a low , whereas patients with polycythemia have a high .
Figure 286D illustrates the importance of . If is zero, the titration line is horizontal, and doubling from 40 to 80 mm Hg does not change [] significantly (see Fig. 286D, point A_{1}); this verifies our earlier conclusion (see Fig. 285, stage 3A). Simultaneously, pH falls by 0.3, which corresponds to a neardoubling of [H^{+}]. Thus, doubling [CO_{2}] leads to a doubling of the product [][H^{+}], so that the ratio [][H^{+}]/[CO_{2}]—the equilibrium constant, K—is unchanged (Table 285, top row). Figure 286D also shows what happens if has the normal value of 25 mM/pH unit. Here, point A is the same as point A in Figure 286C, and it shows that [] increases by 5.25 mM, or 17% (see Table 285, middle row). Finally, if were ∞, the titration line would be vertical, and doubling would not change pH at all (point A_{2} in Fig. 286D). In this case, [] would double, so that the doubling of the product [][H^{+}] would match the doubling of (see Table 285, bottom row).
TABLE 285
Relationship between and the Amount of Formed in Response to a Doubling of
BUFFERING POWER 
ΔpH 
FORMED (mM) 
FRACTIONAL Δ[H^{+}] 
FRACTIONAL 
FRACTIONAL × Δ[H^{+}] 
0 
−0.30 
0.000,040 
1.999,997 
1.000,002 
2.00 
25 
−0.21 
5.25 
1.71 
1.17 
2.00 
∞ 
0 
24.0 
1.00 
2.00 
2.00 
Adding or removing an acid or base—at a constant —produces a “metabolic” acidbase disturbance
Above, we examined the effect of adding HCl in the absence of other buffers (see Fig. 283). Predicting the pH change for such an acidbase disturbance was straightforward because the titration of to CO_{2}consumed all buffered H^{+}. The situation is more complex when the solution also contains buffers (Fig. 287, stage 1). When we add 10 mmol of HCl to 1 L of solution, the opensystem buffer pair neutralizes most of the added H^{+}, buffers handle some, and a minute amount of added H^{+} remains free and lowers pH (see Fig. 287, stage 2A). Because the system is open, the CO_{2} formed during buffering of the added H^{+} escapes to the atmosphere. If this buffering reaction were occurring in the blood, the newly formed CO_{2} would escape first into the alveolar air, and then into the atmosphere.
FIGURE 287 Metabolic acidosis and alkalosis in the presence of buffers. The red arrows in the Davenport diagrams represent the transition to metabolic acidosis (point C) and metabolic alkalosis (point D). This example differs from Figure 283, in which the solution contained no buffers other than .
How much of the added H^{+} (Δ[strong acid] = 10 mM) follows each of the three pathways in this example of metabolic acidosis? Answering this question requires dealing with two competing equilibria, the and the buffering reactions. As for respiratory acidbase disturbances, we cannot precisely solve the equations governing metabolic acidbase disturbances. However, we can use the Davenport diagram on the left of Figure 287 to obtain a graphical estimate of the final pH and [] through a fourstep process:
Step 1: Identify the point describing the initial conditions (see Fig. 287, “Start” in graph on left).
Step 2: Following the black arrow labeled “2,” move downward (in the direction of decreased []) by 10 mM—the concentration of added H^{+}—to the point labeled with an asterisk. In this maneuver, we assume that the reaction + H^{+} → CO_{2} + H_{2}O has initially consumed all of the added H^{+}. Of course, if this were true, the reaction would be far out of equilibrium and the pH would not have changed at all. Also, the buffers would not have had a chance to participate in the buffering of H^{+}.
Step 3: Through the asterisk, draw a line (see Fig. 287, black line in left graph) that is parallel to the titration line.
Step 4: Following the black arrow labeled “4,” move to the intersection of the new black line and the original isopleth. This intersection occurs at point C, which corresponds to a pH of 7.26 and an [] of 17.4 mM. This maneuver tracks the reaction CO_{2} + H_{2}O → + H^{+}. buffers consume nearly all of this H^{+} in the reaction H^{+} + B^{(n)} → HB^{(n+1)}. As a result, [], [H^{+}], and [HB^{(n+1)}] all rise and the system equilibrates.
As a shortcut, we could bypass the two black arrows and simply follow the red arrow along the CO_{2} isopleth from “Start” to point C.
We now can return to the question of how much of the added H^{+} follows each of the three pathways in Figure 287, stage 2A. Because [] decreased by 24 − 17.4 = 6.6 mM, the amount of H^{+} buffered by must have been 6.6 mmol in each liter. Almost all of the remaining H^{+} that we added, nearly 3.6 mmol, must have been buffered by buffers. A tiny amount of the added H^{+}, ~0.000,015 mmol, must have remained unbuffered and was responsible for decreasing pH from 7.40 to 7.26 (see Fig. 287, stage 3A).
Figure 287 also shows what would happen if we added 10 mmol of a strong base such as NaOH to our 1L solution. The opensystem buffer pair neutralizes most of the added OH^{−}, buffers handle some, and a minute amount of added OH^{−} remains unbuffered, thus raising pH (see Fig. 287, stage 2B). The Davenport diagram on the right of Figure 287 shows how much of the added OH^{−}follows each of the three pathways in this example of metabolic alkalosis. The approach is similar to the one we used above for metabolic acidosis, except that in this case, we generate a new black line that is displaced 10 mM above the titration line. We follow the isopleth to its intersection with this black line at point D, which corresponds to a final pH of 7.51 and an [] of 31.1 mM. Thus, [] rose by 31.1 − 24.0 = 7.1 mM. This is the amount of added OH^{−} that buffered. buffers must have buffered almost all of the remaining OH^{−} that we added, ~2.9 mmol in each liter. The unbuffered OH^{−}, which was responsible for the pH increase, must have been in the nanomolar range (see Fig. 287, stage 3B, and Box 281).
Box 281
Strong Ion Difference
Some clinicians assess acidbase disorders in terms of a virtual parameter termed strongion difference (SID). Unlike weak acids and bases, the socalled strong cations and strong anions are fully dissociated at physiological pH. In blood,
Three major limitations to the SID approach are that (1) proteins and physiological processes generally depend on pH, not SID; (2) cells and the body closely regulate pH but have no known mechanism for directly sensing or regulating SID; and (3) SID neither is uniquely related to pH nor has a causal role in changing pH. Because the SID approach provides no new mechanistic insight, we focus on the classical pH/buffer approach.
During metabolic disturbances, makes a greater contribution to total buffering when pH and are high and when is low
In Figure 287, neutralized 7.1 mmol of the 10 mmol of OH^{−} added to 1 L (ΔpH = +0.11) but only 6.6 mmol of the added 10 mmol of H^{+} (ΔpH = −0.14). The reason for this difference is that the buffering power of in an open system increases exponentially with pH (see Fig. 284, blue curve). In the Davenport diagram, this pH dependence of β_{open} appears as the slope of the CO_{2} isopleth, which increases exponentially with pH. Thus, adding alkali will always cause a smaller pH change than adding an equivalent amount of acid. In our example of metabolic alkalosis (see right side of Fig. 287), the mean β_{open} is (7.1 mM)/(0.11 pH unit) = 65 mM/pH unit in the pH range 7.40 to 7.51 (Table 286). On the other hand, in our example of metabolic acidosis (see left side of Fig. 287), the mean β_{open} is (6.6 mM)/(0.14 pH unit) = 47 mM/pH unit in the pH range 7.26 to 7.40, substantially less than in the more alkaline range. Because is 25 mM/pH unit over the entire pH range, β_{total} is greater in the alkaline pH range.
TABLE 286
Buffering Produced by and Buffers*
ADDITION 
ΔpH 


Δ[B^{(n)}] 

β_{total} 
+10 mmol H^{+} 
~ −0.14 
~6.6 mM 
47 mM/pH unit 
~3.4 mM 
25 mM/pH unit 
72 mM/pH 
+10 mmol OH^{−} 
~ +0.11 
~7.1 mM 
65 mM/pH unit 
~2.9 mM 
25 mM/pH unit 
89 mM/pH 
^{*}The additions are made to 1 L of solution. Initial pH = 7.40, = 40 mm Hg, = 24 mM. assumed to be 25 mM/pH unit.
We have already seen that β_{open} is proportional to [] (see Equation 2820) and that—at a fixed pH—[] is proportional to (see Equation 2829). Thus, other things being equal, the contribution of β_{open} to total buffering increases with . Patients with a high due to respiratory failure will, therefore, have a higher buffering power than normal individuals at the same pH.
Because and buffers compete for added OH^{−} or H^{+}, the contribution of to total buffering also depends on . Figure 288 illustrates the effects of adding 10 mmol NaOH to our standard 1L solution at values of 0, 25 mM/pH unit, and ∞. We saw in Figure 283, stage 3B, that adding 10 mmol of NaOH to a 1L solution containing but no other buffers causes [] to increase from 24 to 34 mM, and pH to increase from 7.40 to 7.55. To use a Davenport diagram to solve the same problem (see Fig. 288A), we generate a titration line with slope of zero (because = 0) through the “Start” point. We also draw a black line with the same slope, but displaced 10 mM higher. Following the blue = 40 isopleth from “Start” to the point where the isopleth intersects with the black line at point D_{1}, we see that—as expected—the final pH is 7.55 and the final [] is 34 mM. Thus, must buffer virtually all 10 mmol of OH^{−} added to 1 L. Because the total buffering power is simply β_{open}, the pH increase must be rather large, 0.15.
FIGURE 288 Effect of on the pH increase caused by metabolic alkalosis.
When is 25 mM/pH unit, as in the Davenport diagram in the upper right of Figure 287 (replotted in Fig. 288B), can buffer only 7.1 mmol because the buffers neutralize 2.9 mmol. With the increased total buffering power, the pH increase is only 0.11.
Finally, when is ∞, the black line lies right on top of the vertical titration line (see Fig. 288C). Thus, neither [] nor pH changes at all. However, [HB^{(n+1)}] increases by 10 mM because the infinitely powerful buffers do all of the buffering.
This set of three examples—for values of 0, 25, and ∞—is comparable to the set that we discussed above for respiratory acidosis in Figure 286D.
A metabolic change can compensate for a respiratory disturbance
Thus far, we have considered what happens during primary respiratory and metabolic acidbase disturbances. When challenged by such acid or alkaline loads in the blood plasma, the body compensates by altering [] or , returning pH toward its initial value and minimizing the magnitude of the overall pH change.
In Figure 289A we revisit an example, originally introduced in Figure 286C, in which we produced a primary respiratory acidosis by increasing from 40 to 80 mm Hg (red arrow in Fig. 289A between “Start” and point A at pH 7.19). If the high persists, the only way we can restore pH toward its initial value of 7.40 is to add an alkali (e.g., or OH^{−}) or remove an acid (e.g., H^{+}), all of which are equivalent. N289 Adding 10 mmol of OH^{−} to 1 L, for example, superimposes a metabolic alkalosis on the primary respiratory acidosis—a metabolic compensation to respiratory acidosis.
FIGURE 289 Metabolic compensation to primary respiratory acidbase disturbances.
The Davenport diagram predicts the consequences of adding 10 mmol of OH^{−} to 1 L. We start by generating a black line that is parallel to the red titration line, but displaced upward by 10 mM. Point A_{1}, the intersection of the black line and the isopleth for 80 mm Hg, represents a final pH of 7.29, still lower than the normal 7.40, but much higher than the 7.19 prevailing before compensation.
If we add an additional 14 mmol OH^{−}, for a total addition of 24 mmol OH^{−} to 1 L, pH returns to exactly its initial value of 7.40 (point A_{2} in Fig. 289A). In other words, we can perfectly compensate for doubling from 40 (“Start”) to 80 mm Hg (point A) by adding an amount of OH^{−} equivalent to the amount of that was present (i.e., 24 mM) at “Start.” Perfect compensation of respiratory acidosis is an example of isohydric hypercapnia (i.e., same pH at a higher ):
(2831)
The kidneys are responsible for the metabolic compensation to a primary respiratory acidosis. They acutely sense high and may also chronically sense low blood pH. The response is to increase both the secretion of acid into the urine and the transport of into the blood (see pp. 832–833), thereby raising plasma pH—a compensatory metabolic alkalosis. The renal compensation to a substantial respiratory acidosis is not perfect, so that pH remains below the normal value of 7.40.
The kidneys can also perform a metabolic compensation to a respiratory alkalosis. In Figure 289B we revisit a second example, originally introduced in Figure 286C, in which we produced a primary respiratory alkalosis by decreasing from 40 to 20 mm Hg (red arrow in Fig. 289B between “Start” and point B at pH 7.60). We can compensate for most of this respiratory alkalosis by adding 10 mmol of H^{+}to each liter of solution or by removing 10 mmol NaHCO_{3} or NaOH, all of which produce the same effect. We generate a black line that is parallel to the red titration line but is displaced downward by 10 mM. Point B_{1} (pH 7.44), at the intersection of the black line and the isopleth for 20 mm Hg, represents a partial compensation.
If we add an additional 2 mmol H^{+}, for a total addition of 12 mmol of H^{+} to each liter, pH returns to exactly its initial value of 7.40 (point B_{2} in Fig. 289B). Why is this amount so much less than the 24 mmol of OH^{−} that we added to each liter to compensate perfectly for a doubling of ? When we halve , we need only add an amount of H^{+} equivalent to half the amount of that was present (24/2 = 12 mM) at “Start.”
In response to a primary respiratory alkalosis, the kidneys secrete less acid into the urine and transport less into the blood (see pp. 833–834), thereby lowering plasma pH—a compensatory metabolic acidosis.
A respiratory change can compensate for a metabolic disturbance
Just as metabolic changes can compensate for respiratory disturbances, respiratory changes can compensate for metabolic ones. In Figure 2810A we return to an example originally introduced on the left side of Figure 287, in which we produced a primary metabolic acidosis by adding 10 mmol of HCl to 1 L of arterial blood (red arrow in Fig. 2810A between “Start” and point C at pH 7.26). Other than removing the H^{+}(or adding OH^{−} to neutralize the H^{+}), the only way we can restore pH toward the initial value of 7.40 is to lower . That is, we can superimpose a respiratory alkalosis on the primary metabolic acidosis—a respiratory compensation to a metabolic acidosis. For example, we can compensate for most of the metabolic acidosis by reducing from 40 to 30 mm Hg. Starting at C, follow the black line from the isopleth for 40 mm Hg to the new isopleth for 30 mm Hg at point C_{1} (pH 7.34). This maneuver represents a partial compensation.
FIGURE 2810 Respiratory compensation to primary metabolic acidbase disturbances.
If we reduce by an additional 6.6 mm Hg to 23.4 mm Hg, we produce a perfect compensation, which we represent in Figure 2810A by following the black line from C until we reach pH 7.40 at point C_{2}, which is on the isopleth for 23.4 mm Hg. Perfect compensation requires that we reduce by a fraction (i.e., [40 − 23.4]/40 ≅ 0.42) that is identical to the ratio of added HCl to the original [] (i.e., 10/24 ≅ 0.42).
Not surprisingly, the respiratory system is responsible for the respiratory compensation to a primary metabolic acidosis. The low plasma pH or [] stimulates the peripheral chemoreceptors (see p. 710) and, if sufficiently longstanding, also the central chemoreceptors (see pp. 713–715). The result is an increase in alveolar ventilation, which lowers (see pp. 679–680) and thus provides the respiratory compensation.
The body achieves a respiratory compensation to a primary metabolic alkalosis in just the opposite manner. In Figure 2810B, we revisit an example originally presented on the right side of Figure 287, in which we produce a primary metabolic alkalosis by adding 10 mmol of NaOH to 1 L (red arrow in Fig. 2810B between “Start” and point D at pH 7.51). If we now increase from 40 to 50 mm Hg, we partially compensate for the metabolic alkalosis by moving from D to D_{1} (pH 7.44). If we raise by an additional 6.7 mm Hg to ~56.7 mm Hg, the compensation is perfect, and pH returns to 7.40 (D_{2}).
In response to a primary metabolic alkalosis, the respiratory system decreases alveolar ventilation and thereby raises . The compensatory respiratory acidosis is the least “perfect” of the four types of compensation we have discussed. The reason is that one can decrease alveolar ventilation—and thus oxygenation—only so far before compromising one's very existence.
In a real person, each of the four primary acidbase disturbances—points A, B, C, and D in Figures 289 and 2810—occurs in the extracellular fluid. In each case, the body's first and almost instantaneous response is to use extracellular buffers to neutralize part of the acid or alkaline load. In addition, cells rapidly take up some of the acid or alkaline load and thus participate in the buffering, as discussed below. Furthermore, renal tubule cells may respond to a metabolic acidosis of extrarenal origin (e.g., diabetic ketoacidosis) by increasing acid secretion into the urine, or to a metabolic alkalosis (e.g., NaHCO_{3} therapy) by decreasing acid secretion—examples of metabolic compensation to a metabolic disturbance. Points C and D in Figure 2810 include the effects of the actual acid or alkaline load, extracellular and intracellular buffering, and any renal (i.e., “metabolic”) response.
Position on a Davenport diagram defines the nature of an acidbase disturbance
The only tools needed for characterizing acidbase status (i.e., pH, [], and ) are a Davenport diagram and electrodes for measuring pH and . From these last two parameters, we can compute [] using the HendersonHasselbalch equation. The acidbase status of blood plasma—or of any aqueous solution—must fall into one of five major categories:
1. Normal. For arterial blood, pH is 7.40, [] is 24 mM, and is 40 mm Hg. This point was labeled “Start” in the previous figures and is the central point in Figure 2811A through D.
FIGURE 2811 Acidbase states represented by position on a Davenport diagram.
2. One of the four primary (or uncompensated) acidbase disturbances. The coordinates for pH and [] fall either (a) on the titration line, but off the 40mm Hg isopleth for uncompensated respiratory acidosis (point A, Fig. 2811A) or alkalosis (point B); or (b) on the 40mm Hg isopleth, but off the titration line for an uncompensated metabolic acidosis (point C) or alkalosis (point D).
3. A partially compensated disturbance. The coordinates for pH and [] fall in any of the four colored regions of Figure 2811B. For example, point A_{1} in the blue region represents a partially compensated respiratory acidosis. In a subject for whom a respiratory acidosis has existed for longer than a few hours, increased renal H^{+} excretion has probably already begun producing a metabolic compensation. Depending on the extent of the original acidosis and the degree of compensation, point A_{1} could lie anywhere in the blue region. We can reach similar conclusions for the other three disturbances/compensations.
4. A perfectly compensated disturbance. The point falls on the vertical line through pH 7.4. As indicated by point A_{2}D_{2} in Figure 2811C, a perfect compensation following respiratory acidosis is indistinguishable from that following metabolic alkalosis. In either case, the final outcome is isohydric hypercapnia. Similarly, point B_{2}C_{2} represents a perfect compensation following either metabolic acidosis or respiratory alkalosis (isohydric hypocapnia).
5. A compound respiratory/metabolic disturbance. The point falls in one of the two colored regions in Figure 2811D. For example, points A_{3} and C_{3} represent a respiratory acidosis complicated by metabolic acidosis, and vice versa. The second disturbance causes the pH to move even further away from “normal,” as might occur in an individual with both respiratory and renal failure (see Table 283). Of course, it is impossible to determine if one of the two compounding disturbances developed before the other.