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Fluid and Electrolyte Therapy

Fluid and Electrolyte Therapy. Introduction:. The molecules of chemical compounds in solution may remain intact, or they may dissociate into particles known as ions which carry an electrical charge. Introduction:.

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Fluid and Electrolyte Therapy

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  1. Fluid and Electrolyte Therapy

  2. Introduction: • The molecules of chemical compounds in solution may remain intact, or they may dissociate into particles known as ions which carry an electrical charge.

  3. Introduction: • Electrolyte ion in blood plasma include cations Na+ , K+ , Ca++ and Mg++ and the anions Cl- , HCO3- , HPO4-- , SO4-- , organics acids and proteins.

  4. Role of electrolytes in body fluids: • Maintaining the acid-base balance in the body. • Controlling body water volumes. • Regulate body metabolism.

  5. Electrolyte solution: • Are liquid preparations used for the treatment of disturbances in the electrolytes and fluid balance in the body.

  6. Concentration expression of electrolyte solutions:

  7. Cont’: • In preparing a solution of K+ ions, a potassium salt is dissolved in water. In addition to the K+ ions, the solution will also contain ions of opposite negative charge. These two components will be chemically equal in that the milliequivalent of one are equal to the milliequivalent of the other. • The interesting point is that if we dissolve enough potassium chloride in water to give us 40 milliequivalents of K+ per litter, we also have exactly 40 milliequivalents of Cl-, but the solution will not contain the same weight of each ion. • A milliequivalents (mEq) : represent the amount, in mg, of a solute equal to 1/1000 of its gram equivalent weight.

  8. Conversion between units: • Examples: What is the concentrations, in mg per ml, of a solution containing 2 mEq of potassium chloride per ml ? Mwt of KCl = 74.5 g Equivalent weight of KCl =74.5 1 mEq of KCl = 1/1000 X 74.5 g = 0.0745 g = 74.5 mg 2 mEq of KCl = 74.5 mg X 2 = 149 mg/ml

  9. Examples: • A solutions contains 10 mg% of Ca++ ions. Express this concentration in terms of mEq per liter ? Atomic weight of Ca++ = 40 Equivalent weight of Ca++ = 40/2= 20 1 mEq of Ca++ = 1/1000 X 20 = 0.02 g = 20 mg 10 mg % = 100 mg / L So, in 1 litter we have 5 mEq

  10. Examples: • How many milliequivalents of magnesium sulfate arerepresented in 1.0 g of anhydrous magnesium sulfate ? Mwt of MgSO4 = 120 Equivalent weight of MgSO4 = 60 1 mEq of MgSO4 = 1/1000 X 60 = 0.06 g = 60 mg 60 mg 1 mEq 1000 mg X X= 16.7 mEq

  11. Examples: • A person is to receive 2 mEq of sodium chloride per 1 kg of body weight. If the person weight 60 kg, how many milliliters of an 0.9 % sterile solution of NaCl should be administered? Mwt of NaCl = 58.5, Equivalent Weight of NaCl= 58.5 2 mEq of NaCl= (1/1000 X 58.5 ) X 2 = 0.117 g Nacl needed for this person= 0.117 X 60 = 7.02 g 100 ml (0.9% sol. ) 0.9 gm of NaCl X 7.02 g of NaCl X= 780 ml

  12. Osmotic Activity: • Electrolytes play their part in controlling body water volumes by establishing osmotic pressure. This pressure is proportional to the total number of particles in solution. • The unit that is used to measure osmotic concentration is the milliosmole ( mOsmol).

  13. Osmotic Activity: • For Non electrolytes: 1 millimole (1 formula weight in mg) = 1 mOsmol However, this relationship is not the same with electrolytes, Since the total number of particles in solution depend upon the degree of dissociation of the substance. Eg : 1 millimole of NaCl = 2 mOsmol of total particle.

  14. Osmotic Activity: • So, for separate ions : The milliosmolar value will obtained by dividing the concentration, in mg per litter, of the ion by its atomic weight. • For, Whole electrolytes solutions: The milliosmolar value is equal to the sum of the milliosmolar values of the separate ions.

  15. Examples: • A solution contain 5 % of anhydrous dextrose in water for injection. How many milliosmles per liter are represented in this concentration? Formula weight of anhydrous dextrose= 180 g 1 millimole of anhydrous dextrose (180 mg) = 1 mOsml 5 % sol. contain 50 g or 50000 mg per liter 50000 / 180 = 278 mOsmol per litter

  16. Examples: • A solution contain 10 mg% of Ca++ ions. How many milliosmoles are represented in 1 liter of the solution? Atomic weight of Ca++ = 40 1 millimole of Ca++ (40 mg) = 1 mOsmol 10 mg% of Ca++ = 100 mg/L 100 mg / 40 = 2.5 mOsmol are represented in 1 L.

  17. Examples: • How many milliosmoles are represented in a liter of an 0.9% NaCl solution? Formula weight of NaCl = 58.5 1 millimole of NaCl ( 58.5 mg) = 2 mOsmol 1000 ml of 0.9% NaCl contain 9 g or 9000 mg of NaCl 58.5 mg(NaCl) 2 mOsmol 9000 mg (NaCl) X X= 307.7 mOsmol

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