FLUID AND ELECTROLYTE BALANCE Prof. ÖZCAN GÖKÇE, MD Director of the Department of General Surgery Yeditepe University Hospital

Chemical and biologic activities of fluid Total particle count in per unit volume of solution (mmol/L) Electrically active particles in per unit volume of solution (mEql/L) Osmotically active particles in per unit volume of solution (mO/L)

Osmolality is a measure of the osmoles of solute per kilogram of solvent (osmol/kg) The osmole (Osm or osmol) is a non-SI unit of measurement that defines the number of moles of a chemical compound that contribute to a solution's osmotic pressure. Osmolarity is the measure of solute concentration (osmol/L)

Every increase 100 mg/dl glucose  1,6-3 meq/L Na+ decrease

Henderson-Hasselbach equation

pCO2= 37-42 mm Hg HCO3- = 22-28 meq/L pH=7.34-7,44

P50 it is common to use the concept of P50 to describe the affinity of a given hemoglobin for oxygen. The P50 is the PO2 at which the hemoglobin becomes 50% saturated with oxygen. As the P50 decreases, oxygen affinity increases and visa versa. Normal adult Hemoglobin A has a P50 of 26.5 mm Hg while Fetal Hemoglobin F has a P50 of 20 mm Hg and sickle cell anemia Hemoglobin S has a P50 of 34 mm Hg.

B C A A<B<C (a – v) O2

Anionic gap Measured cathions = Na,K Measured anions= HCO3- , Cl-

Anionic gap = Unmeasured anions-Unmeasured cathions Na+K + Unmeasured cathions = HCO3- + Cl- + Unmeasured anions Anionic gap = Na+K - ( HCO3- + Cl- ) Anionic gap = 140+4 - ( 28+ 104 ) Anionic gap = 12 Meq/L Correcred anionic gap = Actual anionic gap - {2,5(4,5-Albumin)}

FLUID REPLACEMENT

PRACTICAL ISSUES 1 Protein intake increases urea(Urine) and causes osmotic diuresis. 1gr protein needs 7 ml water

PRACTICAL ISSUES 2 Treatment of Hypernatremia %5 dextroz, CVP and control of electrolytes

PRACTICAL ISSUES 3 Treatment of Hyponatremia 135 - Na mEq/lt = A A*0,5*Body weight = B %3 NaCl……518 mEq/lt B*1000 = C C/2….in 2 hours 518 C/2….in other 6 hours

PRACTICAL ISSUES 4 Treatment of Acidosis 15-CO2 mEq/lt = A A*0,3*Body weight = B B*1000 = C (1/6 M Lactate) C/2….in 2 hours 167 C/2….in other 6 hours B*1000 = C (1/6 M NaHCO3) 178

PRACTICAL ISSUES 5 Treatment of Hypopotasemia 3,5 - K mEq/lt = A A*0,2*Body weight = B B is not given more than 40mEq/hour 40mEq/liter Blood K should be measured every 2 hs K solutions contains 1 mEq/cc

PRACTICAL ISSUES 6 Daily Na requirement is 2-4 mEq/kg Daily K requirement is Na/2 mEq/kg

THE END