PROBLEM SET 7.2: FLUID AND ELECTROLYTE BALANCE AND ACID-BASE BALANCE ANSWER KEY,VERIFIED ANSWERS.100% SCORE

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1. The following were measured in a 24-hour urine collection: urine volume = 1.2 L titratable acid = 33 mmol L-1 urine [HCO3-] = 4 mmol L-1 urine [NH4+] = 90 mmol L-1 Calculate the amount of HCO ... 3 - gained or lost from the body through renal mechanisms. The net amount of HCO 3 - gained or lost from the body is calculated as: net HCO 3 - = titratable acid + ammonium - HCO 3 - excreted These are in amounts or rates. In the above example, we deal with amounts. The amount of titratable acid is its concentration times the urine volume; the same applies to the ammonium and HCO 3 -. So we have net HCO 3 - = 1.2 L x 33 mmol L-1 + 1.2 L x 90 mmol L-1 - 1.2 L x 4 mmol L-1 net HCO 3 - = 39.6 mmol + 108 mmol - 4.8 mmol = 142.8 mmol 2. A patient with a history of bronchopulmonary disease was admitted to the hospital and the following laboratory data were obtained for arterial blood: Na 143 mM K 4.8 mM pH 7.34 P O2 47 mm Hg Cl 78 mM HCO 3 - 38 mM P CO2 73 mm Hg Hemoglobin 17.1 g% Classify the acid-base status of this patient in terms of acidosis/alkalosis, respiratory or metabolic, compensated or uncompensated. Fig. 7.PS2.1 Acid-Base status, problem #2 The acid-base status can be represented by point X on the pH-HCO3- diagram shown. This corresponds to a metabolic (renal) compensation for a respiratory acidosis. You could tell that because P CO2 is high and the blood is acidic; thus it cannot be metabolic acidosis, because the respiratory system would respond by hypocapnia. But the decrease in pH is only slight compared to what would happen if it were just hypercapnia - there is compensation and the compensation is brought about by increasing the plasma [HCO3-].7.PS2.2 3. The concept of free water clearance is sometimes used to assess the kidneys’ ability to handle a water load. The free water clearance is defined as: C H2O = QU - [(QU Uosmol)/Posmol] Here C H2O is the free water clearance, QU is the flow of urine, in mL min-1, Uosmol is the total concentration of materials in the urine, in osmol/L, and Posmol is the osmolarity of plasma. The term in brackets is the clearance of osmolarity, and it represents the volume of plasma that contains the total osmolarity that is excreted. Calculate the free water clearance for the following conditions: Condition QU (mL min-1) Uosmol (mOsM) Posmol (mOsM) CH2O A 0.8 900 300 B 1.5 300 300 C 6.0 70 300 What does a negative free water clearance mean? The free water clearance is calculated as specified, as QU - [QU Uosmol / Posmol] as indicated in the filled-in table. Condition QU (mL min-1) Uosmol (mOsM) Posmol (mOsM) CH2O (mL min-1) A 0.8 900 300 -1.6 B 1.5 300 300 0 C 6.0 70 300 4.6 The free water clearance is the amount of excess water is eliminated or gained by the body by the formation of urine. If urine is more concentrated than plasma, it’s like placing pure water into the plasma and the gain of water is a negative free water clearance. If urine is isosmolar with plasma, there is no net gain or loss of water in excess of salt. When a dilute urine is excreted, the urine contains water in excess of salt and so the free water clearance is positive. 4. The acid NH 4 + dissociates to its conjugate base (NH3) and H+, with a dissociation constant KD = 6.3 x 10-10 M: NH 4 + X NH 3 + H+ Suppose that NH4+ is originally present in two compartments separated by a cellular membrane. The membrane is freely permeable to NH3 but it is impermeable to NH4+ . The pH on one side is 9.0 and on the other side it is 7.4. Let T1 and T2 be the total concentration of NH 3 + NH4+ on side 1 and 2, respectively. Calculate T1/T2.7.PS2.3 Here we have the dissociation reaction: NH 4 + X NH 3 + H+ (1) with a K d = 6.3 x 10-10 M. Only the NH3 can travel across the membrane because it isn’t charged. We assume that the movement is rapid and that NH3 reaches an equilibrium across the membrane where [NH3]1 = [NH3]2. The ratio of the total of NH4+ and [NH3] on both sides of the membrane is what we want. This is T 1 / T2 = ([NH4+]1 + [NH3 ]1) / ([NH4+]2 + [NH3 ]2) (2) Since at equilibrium [NH3]1 = [NH3]2, we can divide the numerator by [NH3]1 and the denominator by [NH3]2 and still have an equation: T 1 / T2 = ([NH4+]1 / [NH3 ]1 + 1) / ([NH4+]2 / [NH3 ]2 + 1) (3) The dissociation reaction obeys kinetic laws and the Henderson-Hasselbalch equation: pH1 = pK + log ([NH3]1 / [NH4+]1) (4a) pH2 = pK + log ([NH3]2 / [NH4+]2) (4a) We solve for the ratios in the numerator and denominator of eq. (3) [Show More]

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