Acid-Base Balance Seminar No. 11 Parameters of acid base balance Measured in arterial blood • pH = 7.40 ± 0.04 = 7.36 – 7.44 • pCO[2] = 4.8 – 5.8 kPa • supporting data: pO[2], tHb, sO[2], HbO[2], COHb, MetHb Calculated • [HCO[3]^-] = 24 ± 3 mmol/l (from H.-H. eq.) • BE = 0 ± 3 mmol/l (from S.-A. nomogram, see physilogy) • BB[s] = 42 ± 3 mmol/l • BB[b] = 48 ± 3 mmol/l Q. 1 Buffer bases in (arterial) plasma Q. 2 A. 2 BB[s] = 42 ± 3 mmol/l BB[b] = 48 ± 3 mmol/l hemoglobin in erythrocytes increases BB[b] by 6-8 mmol/l Q. 3 Oxygen parameters and hemoglobin derivatives Oxygen parameters and hemoglobin derivatives Q. 4 A. 4 7.4 = 6.1 + log [HCO[3]^-] / 0.22 × 5.3 1.3 = log [HCO[3]^-] / 1.2 10^1.3 =^ [HCO[3]^-] / 1.2 20 = [HCO[3]^-] / 1.2 [HCO[3]^-] = 24 mmol/l Four types of acid-base disorders Maintanance of constant pH in body Responses to acute change • compensation • correction Q. 6 A. 6 Metabolic acidosis is the most common condition Metabolic alkalosis is the most dangerous condition Q. 8 Q. 9 A. 9 Excessive infusions of NaCl isotonic solution lead to metabolic acidosis Q. 10 Hyperchloremic MAc • excessive infusions of NaCl solution • the loss of HCO[3]^- + Na^+ + water (diarrhoea, renal disorders) TH relative higher concentration of chlorides in plasma Q. 11 How is AG calculated? AG A. 11 MAc with increased AG • Hypoxia of tissues – insufficient supply of O[2] TH anaerobic glycolysis: glucose  2 lactate • elevated AG – lactoacidosis • Starvation, diabetes • TAG  FA (β-oxidation in liver) acetyl-CoA (excess, over the capacity of CAC)  KB production • elevated AG - ketoacidosis • Renal insufficiency – elevated phosphates, sulfates • Various intoxications Q. 12 A. 12 • AG – normal values • SID – buffer bases (mainly HCO[3]^-) – decreased • compare Q. 8a) Q. 13 Metabolic acidosis Q. 15 Methanol intoxication Metabolic oxidation of methanol provides a rather strong formic acid Compare two acids ethanol acetic acid pK[A] = 4.75 K[A] = 1.8 ´ 10^-5 methanol formic acid pK[A] = 3.75 K[A] = 1.8 ´ 10^-4 ethylene glycol intoxication Intoxication by ethylene glycol Calcium oxalate is insoluble chelate Calcium oxalate is insoluble chelate Why MAc occurs in anemia? Not enough hemoglobin TH insufficient supply of O[2] TH hypoxia  anaerobic glycolysis to lactate elevated AG – lactoacidosis Q. 16 Metabolic oxidation of ethanol leads to excess of NADH Metabolic consequences of EtOH biotransformation Q. 17 • thiamine is the cofactor of aerobic decarboxylation of pyruvate • thiamine deficit TH pyruvate cannot be converted to acetyl-CoA • therefore pyruvate is hydrogenated to lactate • even in aerobic conditions: glucose  lactate • increased plasma lactate  elevated AG  lactoacidosis Q. 18 A. 18 Causes of metabolic alkalosis • Repeated vomiting – the loss of chloride (Cl^-) anion TH hypochloremic alkalosis • Direct administration of buffer base HCO[3]^- per os: baking soda, some mineral waters intravenous infusions of sodium bicarbonate • Hypoalbuminemia severe malnutrition liver damage, kidney damage What is baking soda? A. NaHCO[3 ] sodium hydrogen carbonate (sodium bicarbonate) sold in pharmacy Q. 19 SID corresponds to buffer bases of plasma Q. 20 Q. 21 Metabolic alkalosis Q. 23 A. 23 Q. 26 Respiratory acidosis Q. 27 Describe the scheme on p. 5 • Excess of CO[2] in the body produces more H[2]CO[3] in blood • Carbonic acid in buffering reaction with proteins gives HCO[3]^- ion • Hydrogen carbonate ion is driven to ICF • Therefore the level of HCO[3]^- in ECF is normal or slightly elevated Q. 29 A. 29 pH 7.32 ........................................- pCO[2] 9.3 kPa ............................. [HCO[3]^-] = 39 mmol/l .................. [Na^+] = 136 mmol/l ......................OK [K^+] = 4.5 mmol/l .........................OK [Cl^-] = 92 mmol/l ..........................- AG = 136 + 4.5 - 39 – 92 = 9.5 mmol/l .......  no MAc Conclusion: compensated RAc Q. 30 Respiratory alkalosis Combined disorders Q. 33 A. 33 pH 7. 4 ............................. OK pCO[2] 5.13 kPa .......................OK BE 1 mmol/l ............................OK TH HCO[3]^- = 25 mmol/l ....... OK Na^+ 140 mmol/l ......................OK K^+ 4.6 mmol/l ........................OK Cl^- 89 mmol/l ......................... - AG = 140 + 4.6 – 25 – 89 = 30.6 mmol/l ............... SID = 140 + 4.6 – 89 = 55.6 mmol/l ....................... Conclusion: MAc + MAlk Q. 34 A. 34