Acid-base fluids and electrolytes made ridiculously simple

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Answer: They will all increase the measured osmolality if added to the ex- tracellular fluid. 19. Which of the following will increase the calculated serum osmolality when added to the extracellular fluid? Urea Glucose Sodium Ethanol Methanol Isopropanol Ethylene glycol Mannitol Sorbitol Answer: Only urea, glucose and sodium are included in the formula for calculated osmolality. Therefore, only urea, glucose and sodium will add to the calculated osmolality if added to extracellular fluid. 20. Which of the following will increase the osmolal gap when added to the ex- tracellular fluid? Urea Glucose Sodium Ethanol Methanol Isopropanol Ethylene glycol Mannitol Sorbitol Answer: OSM GAP = OSM(,,,,) - OSM(ca,c) Urea, glucose, and sodium are all included in the formula for calculated osmolality. They will add to both the calculated osmolality and the mea- sured osmolality and therefore will not change the osmolal gap if added to the extracellular fluid. The other compounds will increase the measured osmolality but not the calculated osmolality and will therefore increase the osmolal gap. 21. At approximately what level of GFR would a patient have problems ex- creting the daily dietary potassium load? At this point, the patient will be- gin to develop positive potassium balance, leading to hyperkalemia. Answer: The upper limit of potassium excretion is roughly proportional to the GFR. If the GFR is 100% of normal, the maximum amount of potas- sium which could be excreted in one day is roughly 10 mEq per kg body weight. This is about 70 X 10 = 700 mEq in a 70 kg person. If the GFR is reduced to 50% of normal the maximum amount of potassium that can

be excreted in one day falls to approximately 50% X 700 = 350 mEq. This is a rough approximation of maximum potassium excretion because compensatory renal potassium secretory mechanisms will increase potas- sium excretion, and stool potassium losses also increase as the body de- fends itself against hyperkalemia. If the GFR is further reduced to 20% of normal, the maximal potassium excretion would fall to the range of about 140 mEq1day (20% of 700 mEq1day). The average diet has about 1 mEq of potassium per kg body weight, which amounts to about 70 mEqIday in a 70 kg person. For a diet con- taining 70 mEqIday, the GFR would need to be reduced to approximately 701700 = 10% of normal before hyperkalemia develops. In fact, the GFR is usually below this level when hyperkalemia develops based upon usual dietary intake. Hyperkalernia may develop at less profound levels of re- nal failure if the potassium intake is increased or if there is a hidden potas- sium load. For example, a person with a diet high in potassium would develop hyperkalemia with less impairment of the GFR. A patient with a GFR 15% of normal would develop hyperkalemia if dietary potassium is over the range of 15% X 700 = 105 mEq1day. As mentioned above: this is only a rough approximation of maximum potassium excretion. The clinical point is that ifa patient has mild to moderate renal fail- ure and hyperkalemia, the hyperkulemia should not be simply ascribed to renal failure alone. A vigorous search for other causes of hyper- kulemia is needed. 22. How much potassium is there in the ECFV of a 70 kg man? Answer: The very delicate nature of the transcellular distribution of potassium is illustrated by the following calculation: TBW = .6X 70kg=42L ECFV = 113 X 42 L = 14 L Potassium concentration in ECFV 4.0 mEqL Total potassium in ECFV: 4.0 mEqL X 14 L = 56 mEq The calculated amount of potassium in the entire ECFV (56 mEq) is less than that contained in three routine supplemental 20 mEq doses of KC1 or the potassium in four glasses of orange juice! Even a small increase in the amount of extracellular potassium could cause a large increase in the ECF potassium concentration. Adding 56 mEq to the ECFV would result in an increase of potassium concentration from 4.0 mEqL to 8.0 mEqL! Thankfully, we do not double our potassium concentration after four glasses of orange juice because homeostatic mechanisms maintain the striking difference between intracellular and extracellular potassium con- centrations and, therefore, the ECFV potassium concentration. 23. How much potassium is there in the ECFV of a 40 kg woman? Answer: TBW = .5 X 40 kg = 20 L ECFV = 113 X 20 L = 6.7 L

Potassium concentration in ECFV 4.0 mEqn Total potassium in ECFV 4.0 mEqn X 6.7 L = 26.8 mEq The calculated amount of potassium in the entire ECFV is about one sup- plemental 20 mEq dose of KCl! 24. Calculate the total amount of HC03- present in the ECFV of a 50 kg woman with an ECF HC03- concentration of 25 mEqL. Answer: Total body water: .5 X 50 = 25 liters. The ECFV is approxi- mately 113 of total body water: 2513 = 8.3 liters. The normal ECF stores of HC03- are 25 mEqL X 8.3 L = 207.5 mEq! This corresponds to about four standard ampules of sodium bicarbonate. How much HC03- is being reabsorbed each day by the proximal tubule assuming a GFR of 100 mymin? Answer: Total amount filtered = total amount reabsorbed by the proximal tubule: 100 mVmin X 1440 midday X 25 mEqL = 3600 mEq/day ! This is about 17 times the total amount of bicarbonate in the ECF. 25. To illustrate one aspect of the importance of the urinary buffers the fol- lowing is a calculation of what the urine pH would be if there were no uri- nary buffers. I don't expect you to know how do this calculation. It is included for illustration only. Normally, the daily excretion of hydrogen ion is approximately 50-100 mmoVday and is equal to the amount of fixed acid produced by the me- tabolism of the diet. Assuming a hydrogen ion excretion of 100 mmol in a 24-hour urine volume of, say, 1 L, this would result in a urinary pH of pH = - log(H+) = - log(100 mmolAL) = -log(. 100 mmoVrnl) = 1 It hurts just to imagine urine with a pH of 1 ! Compare this pH of 1 to the normal minimum urinary pH of 4.5. The urinary buffers allow for large increases in hydrogen ion excretion (200-300 mmoVday) in states of in- creased hydrogen ion load without appreciable decreases in urine pH. The primary means by which the kidney rids the body of excess hydrogen ion is by increasing renal amrnoniagenesis. In situations when excess hydro- gen ion is added to the body, the kidney responds by increasing produc- tion and excretion of NH4+.

FIGURE 9-1. Three Step Approach to Acid-Base Disorders Step 1: Identify the most apparent disorder. Disorder PH Pcoz HCO3- Metabolic acidosis Decreased Decreased (secondary) Decreased (primary) Metabolic alkalosis Increased Increased (secondary) Increased (primary) Respiratory acidosis Decreased Increased (primary) Increased (secondary) Respiratory alkalosis Increased Decreased (primary) Decreased (secondary) Step 2: Apply the formulas to determine if compensation is appropriate. If not, a second disorder co-exists. Metabolic acidosis: PCOZ = 1.5 X [HC03-] + 8 Metabolic alkalosis: PCOZ = 40 + .7 X ( [ H C O S - ( ~ ~ ~ ~ ~ ~ ) ] - [HCO3-(normal)] ) Respiratory acidosis: Acute: [HCO3-] increases by 1 mEqn for every 10 mm Hg increase in PCOZ Chronic: [HC03-] increases by 3.5 mEqL for every 10 mm Hg increase in PC02 Respiratory alkalosis: Acute: [HCOs-] decreases by 2 mEqn for every 10 mm Hg decrease in PCOZ Chronic: [HC03-] decreases by 5 mEqL for every 10 mm Hg decrease in PC02 Step 3: Calculate the anion gap. AG = [Na+] - ([Cl-] + [HCOs-I) The normal AG is 9-16 mEq/L. If AG > 20 mEqn, high AG acidosis is probably present. If AG > 30 mEq/L, high AG acidosis is almost certainly present. For lactic acidosis, the ratio of the increase in the anion gap to the decrease in the HCO3- averages approximately 1.5. In ketoacidosis, the ratio of the increase in the anion gap to the decrease in the HC03- averages approximately 1 .O. In respiratory disorders, the PCO~ is abnormal and we want to see if there is also a coexisting metabolic disorder. We ask: What should the [HC03-] be after compensation? If the [HC03-] differs significantly from that predicted by the formula for compensation, then a coexisting metabolic disorder is present. Apply the formula for the disorder you have identified to see if the compensation is correct. If the compensation is not what is predicted by the formula, then an additional disorder is present. Step 3: Calculate the anion gap. The normal value of the anion gap used in this book is 9-16 mEq/L, although many hospitals may prefer to use the smaller range 10-14. If the calculated anion gap is normal, you are finished. The presence of an increased anion gap is a powerful clue to the diagnosis of metabolic acidosis. If the anion gap is increased above 20 mEq/L, then an an- ion gap metabolic acidosis is probably present. If the anion gap is increased above 30 mEq/L, then an anion gap metabolic acidosis is almost certainly pre- sent, regardless of the pH and [HC03-1.

If a high anion gap acidosis due to lactic acidosis or ketoacidosis is pre- sent, then it may be helpful to compare the change in the anion gap to the change in the bicarbonate concentration. By doing this, one may identify an additional "hidden" metabolic disorder, either a metabolic alkalosis or a nor- mal anion gap metabolic acidosis. Step 1: Identify One Disorder. Look at the pH, PCOZ and [HC03-] to identify the most apparent acid-base disorder. In general: If the pH is low (<7.35), either a metabolic acidosis or a respiratory acido- sis is present; if the [HC03-] is low: metabolic acidosis; if the PCOZ is high: respiratory acidosis. If the pH is high (>7.45), either a metabolic alkalosis or a respiratory al- kalosis is present; if the [HC03-] is high: metabolic alkalosis; if the PCOZ is low: respiratory alkalosis. If the pH is normal, but either the [HC03-] or the PCOZ, (or both) is abnor- mal, then pick the most abnormal of the [HC03-] or PCOZ. For example: pH 7.40, PCOZ 60 mm Hg, HC03 36 mEq/L. Both the PCOZ and the [HC03-] are abnormal. Because the pH is normal in this case, you could start by diag- nosing either a metabolic alkalosis ([HC03-] 36 mEqL ) or a respiratory acidosis (PCOZ 60 mm Hg). This method will allow you to start either way. Step 2: Apply the Formulas to See If Compensation Is Correct. Apply the formulas for expected compensation to determine if a second disorder is present. This section deals with what the formulas for expected compensation to simple disorders mean and how to use them. Once you iden- tify a disorder, the general question is: Is the compensation close to that pre- dicted by the formula for expected compensation? Once you have made a diagnosis of one disorder, then apply the formula for that specific disorder to see if the compensation is, appropriate. For metabolic disorders ask: What should the PCO~ be after compensation? For respiratory disorders ask: What should the [HC03-] be after compensation? The formulas give approxima- tions for the expected compensation for acid-base disorders. If the compensa- tion is not consistent with the given formula, then a second disorder is present. Remember to use the values of both the PCOZ and the [HC03-] from the arterial blood gas (ABG) for purposes of determining if compensation is appropriate (Step 2). Also, remember to use serum values to calculate the anion gap (Step 3). In this book, the serum bicarbonate and the calculated bicarbonate from the ABG are almost always equal, but this is not always the case in clinical practice.

Metabolic Acidosis The hydrogen ion concentration of ECF is determined by the ratio of the PCO~ (which is controlled by the lungs) to the [HC03-] (which is controlled by the kidneys) according to the relation: A metabolic acidosis is a process that causes a primary decrease in [HC03-1. The respiratory compensation for a metabolic acidosis is increased ventilation, which produces a secondary decrease in Pco~. This returns the Pcod[HCO3-] ratio (and therefore the hydrogen ion concentration) toward the normal range. Typically, the lungs do not return the hydrogen ion concentration all the way into the normal range, but only toward the normal range. What should the PCO~ be after compensation for a metabolic acidosis? The quantitative answer to this question is obtained by using the formula for expected respiratory compensa- tion for a metabolic acidosis. That is, the PCO~ should be equal to: What if the measured P C O ~ differs from this value? A significant difference means that there is also a respiratory disorder in addition to the metabolic aci- dosis, because the PCO~ is not behaving as we would expect. If the measured PCO~ is higher than predicted by the formula, there is a coexisting respiratory acidosis. If the measured P C O ~ is lower than predicted, there is a coexisting respiratory alkalosis. This formula is approximate, and we should allow the measured PCO~ to be 2 2 mm Hg off from that predicted by the formula. A more significant deviation in either direction from the value predicted by the for- mula, however, indicates that in addition to a metabolic acidosis, there is also a respiratory disorder present. Metabolic Alkalosis The hydrogen ion concentration of the ECF, as mentioned, is determined by the ratio of the Pco~, which is controlled by the lungs, to the [HC03-1, which is controlled by the kidneys, according to the relation: A metabolic alkalosis is a process which causes a primary increase in [HCO3-1. The respiratory compensation for a metabolic alkalosis is decreased ventilation, which produces a secondary increase in P C O ~ This returns the Pcoz/[HCO3-] ratio (and therefore the hydrogen ion concentration) toward the normal range. The lungs do not bring the hydrogen ion concentration into the normal range, but only toward the normal range. By how much should the PCO~ increase in compensation for a metabolic alkalosis? The quantitative answer

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