High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Physiology Textbook > Body Fluids

Characteristics

DistributionPrevious Section
Chapter IntroNext Section
Popular Courses
Find My Course
0:00 / 0:00
The compositions of the different compartments of body fluids is not uniform: intracellular (ICF) and extracellular (ECF) have very different concentrations of various solutes.

There are many key ions dissolved in body fluids that are essential for physiological functioning. They include: Na+, K+, Cl−, HCO3−, Mg2+, PO43−, Ca2+Na^+,\ K^+,\ Cl^-,\ HCO_3^-,\ Mg^{2+},\ PO_4^{3-},\ Ca^{2+}, etc.



Review of Units of Concentration
  • Amounts of solute: moles, equivalents (Eqs), or osmoles
  • Concentrations of solute: moles/L (M), Eq/L, or osmol/L (Osm)
  • One mole is 6.02 x 10236.02\ x\ 10^{23} molecules of the substance.
  • One equivalent is used to describe the amount of ionized solute. It is equal to # moles x valence of ionized solute.
  • An osmole is the number of particles into which a solute dissociates in solution. If a solute does not dissociate into solution (e.g. glucose), then osmolarity = molarity.

Watch Out!
In some texts, the notation osm/L is used instead of just Osm to indicate osmolarity. You might see me using that in some exercise videos. However, to avoid any confusion, I will stick with Osm to indicate osmolarity (osmoles/L) in the future.


Important Formulas
Molarity = Mass of solute (g)Molecular Weight (g(mol))⋅Volume solution (L)Molarity\ =\ \frac{Mass\ of\ solute\ \left(g\right)}{Molecular\ Weight\ \left(\frac{g}{\left(mol\right)}\right)\cdot Volume\ solution\ \left(L\right)}

EquivalentL =Molarity ⋅ Valence of ionized solute\frac{Equivalent}{L}\ =Molarity\ \cdot\ Valence\ of\ ionized\ solute

Osmolarity=Molarity ⋅ (Number of particles dissociated)Osmolarity=Molarity\ \cdot\ \left(Number\ of\ particles\ dissociated\right)

% wtvol=Mass of solute (g)Volume of solution (mL)⋅ 100%\%\ \frac{wt}{vol}=\frac{Mass\ of\ solute\ \left(g\right)}{Volume\ of\ solution\ \left(mL\right)}\cdot\ 100\%
0:00 / 0:00
One mole of calcium chloride (CaCl2CaCl_2) is put into solution. How many equivalents of calcium (Ca2+Ca^{2+}) and of chloride (Cl−Cl^-) does it dissociate into?
By definition, an equivalent is the number of moles of a solute multiplied by its valence. Therefore:
One mole of calcium chloride therefore dissociates into one mole of calcium and two moles of chloride. The valence of calcium is 2 and that of chloride is 1. This gives:

For calcium = 1 mole x 2 (valence) = 2 equivalents
For chloride = 2 moles x 1 (valence) = 2 equivalents

CaCl2CaCl_2yields 2 equivalents of calcium and 2 equivalents of chloride.
0:00 / 0:00
One mole of calcium chloride (CaCl2CaCl_2) is put into one liter of water. What is the osmolarity of the resulting solution?
In solution, 1 mole of calcium chloride in 1 liter of water makes a solution of 1 mol/L or 1 M (molarity). It dissociates into 1 mole of Ca2+Ca^{2+}and 2 moles of Cl−Cl^-as follows:
By definition, osmolarity is the concentration of particles in solution (Osm/L). An osmole is the number of particles a solute dissociates into when in solution. When a solute dissociates into more than one particle in solution, its osmolarity equals its molarity times the number of particles.

Therefore, for CaCl2CaCl_2: 1 mol/L = 3 osmoles/L = 3 Osm (1 Ca2+Ca^{2+}+ 2 Cl−Cl^-).
A solution is prepared such that 48.6 g of Mg2+Mg^{2+}(MW = 24.3 g/mol) is put in 1 L of water. The concentration of this solution can be expressed as:
Relative to intracellular fluid (ICF), ECF has:
Absolute Amount of Water (L) = Weight (kg) ⋅ Body Water %100Absolute\ Amount\ of\ Water\ \left(L\right)\ =\ \frac{Weight\ \left(kg\right)\ \cdot\ Body\ Water\ \%}{100}

Drug Concentration in Body (mgmL)= Amount of Medication Given (mg)Absolute Amount of Water (L)Drug\ Concentration\ in\ Body\ \left(\frac{mg}{mL}\right)=\ \frac{Amount\ of\ Medication\ Given\ \left(mg\right)}{Absolute\ Amount\ of\ Water\ \left(L\right)}

Indicator Dilution Method:
  1. Total body water: Antipyrine, D2O or T2OD_2O\ or\ T_2O
  2. ECF: Inulin, sucrose or mannitol
  3. Plasma: Evan's Blue or I131I^{131}-Albumin
Total Body Water = ICF + ECF = ICF + (ISF + Plasma)

Important Percentages:
  1. Total body water: 60% of body weight (BW)
  2. ICF: 40% of BW (2/3 of TBW)
  3. ECF: 20% of BW (1/3 of TBW)
  4. ISF: 15% of BW (3/4 of ECF)
  5. Plasma: 5% of BW (1/4 of ECF)
  6. Transcellular fluid and Lymph: 1-2% of ECF

Molarity = Mass of solute (g)Molecular Weight (g(mol))⋅Volume solution (L)Molarity\ =\ \frac{Mass\ of\ solute\ \left(g\right)}{Molecular\ Weight\ \left(\frac{g}{\left(mol\right)}\right)\cdot Volume\ solution\ \left(L\right)}

EquivalentL =Molarity ⋅ Valence of ionized solute\frac{Equivalent}{L}\ =Molarity\ \cdot\ Valence\ of\ ionized\ solute

Osmolarity=Molarity ⋅ (Number of particles dissociated)Osmolarity=Molarity\ \cdot\ \left(Number\ of\ particles\ dissociated\right)

% wtvol=Mass of solute (g)Volume of solution (mL)⋅ 100%\%\ \frac{wt}{vol}=\frac{Mass\ of\ solute\ \left(g\right)}{Volume\ of\ solution\ \left(mL\right)}\cdot\ 100\%

Hematocrit % =height of RBC columnheight of blood⋅100Hematocrit\ \%\ =\frac{height\ of\ RBC\ column}{height\ of\ blood}\cdot100