680
SECTION VIII
Renal Physiology
the urine that is 1000 times the concentration in plasma. pH 4.5
is thus the
limiting pH.
This is normally reached in the collect-
ing ducts. If there were no buffers that “tied up” H
- in the urine,
this pH would be reached rapidly, and H
secretion would stop.
However, three important reactions in the tubular fluid remove
free H
, permitting more acid to be secreted (Figure 40–2).
These are the reactions with HCO
3
- to form CO
2
and H
2
O, with
HPO
4
2–
to form H
2
PO
4- , and with NH
3
to form NH
4
+
.REACTION WITH BUFFERS
The dynamics of buffering are discussed in Chapter 1 and be-
low. The pK' of the bicarbonate system is 6.1, that of the diba-
sic phosphate system is 6.8, and that of the ammonia system is
9.0. The concentration of HCO
3
- in the plasma, and conse-
quently in the glomerular filtrate, is normally about 24 mEq/
L, whereas that of phosphate is only 1.5 mEq/L. Therefore, in
the proximal tubule, most of the secreted H
- reacts with
HCO
3
- to form H
2
CO
3
(Figure 40–2). The H
2
CO
3
breaks
down to form CO
2
and H
2
O. In the proximal (but not in the
distal) tubule, there is carbonic anhydrase in the brush border
of the cells; this facilitates the formation of CO
2
and H
2
O in
the tubular fluid. The CO
2
, which diffuses readily across all bi-
ological membranes, enters the tubular cells, where it adds to
the pool of CO
2
available to form H
2
CO
3
. Because most of the
H
- is removed from the tubule, the pH of the fluid is changed
very little. This is the mechanism by which HCO
3
- is reab-
sorbed; for each mole of HCO
3
removed from the tubular flu-
id, 1 mol of HCO
3
diffuses from the tubular cells into the
blood, even though it is not the same mole that disappeared
from the tubular fluid.
Secreted H
- also reacts with dibasic phosphate (HPO
4
2–
) to
form monobasic phosphate (H
2
PO
4
- ). This happens to the
greatest extent in the distal tubules and collecting ducts,
because it is here that the phosphate that escapes proximal
reabsorption is greatly concentrated by the reabsorption of
water. The reaction with NH
3
occurs in the proximal and dis-
tal tubules. H
+
also combines to a minor degree with other
buffer anions.
Each H
+
ion that reacts with the buffers contributes to the
urinary
titratable acidity,
which is measured by determining
the amount of alkali that must be added to the urine to return
its pH to 7.4, the pH of the glomerular filtrate. However, the
titratable acidity obviously measures only a fraction of the
acid secreted, since it does not account for the H
2
CO
3
that has
been converted to H
2
O and CO
2. In addition, the pK' of the
ammonia system is 9.0, and the ammonia system is titrated
only from the pH of the urine to pH 7.4, so it contributes very
little to the titratable acidity.
AMMONIA SECRETION
Reactions in the renal tubular cells produce NH
4+
and HCO
3- .
NH
4
+
is in equilibrium with NH
3
and H
+
in the cells. Because
the pK' of this reaction is 9.0, the ratio of NH
3
to NH
4+
at pH
7.0 is 1:100 (Figure 40–3). However, NH
3
is lipid-soluble and
diffuses across the cell membranes down its concentration
gradient into the interstitial fluid and tubular urine. In the
urine it reacts with H
+
to form NH
4+
, and the NH
4+
remains
in the urine.FIGURE 40–1
Secretion of acid by proximal tubular cells in
the kidney.
H
- is transported into the tubular lumen by an antiport in
exchange for Na
. Active transport by Na, K ATPase is indicated by ar-
rows in the circle. Dashed arrows indicate diffusion.
Interstitial
fluidTubular
lumenH+Na+K+Na+K+Renal tubule cellHCO 3 − HCO 3 −CO 2 +H 2 OH 2 CO 3+Carbonic
anhydraseH+FIGURE 40–2
Fate of H
+
secreted into a tubule in exchange
for Na
+. Top:
Reabsorption of filtered bicarbonate via CO
2
.
Middle:
Formation of monobasic phosphate.
Bottom:
Ammonium formation.
Note that in each instance one Na
- ion and one HCO
3
- ion enter the
bloodstream for each H
- ion secreted. A
- , anion.
Interstitial
fluidRenal
tubule cellTubular
lumenNa+Na+A−Na+HCO 3 −HCO 3 −HCO 3 −HCO 3 −H+H+Na+HCO 3 − HCO 3 −Na++ HCO 3 −
H++ HCO 3 −CO 2 + H 2 ONa+Na+HPO 42 −Na+H 2 PO 4 −+NH 3 NH 3NH 4 +A−H+H+ H+