1. VOLUME PER VOLUME [Hct]

i) effect of crystalloid dilution on Hct

Vol1 x Conc1 = Vol2 x Conc2

eg initial blood volume 5000ml with Hct 43.9% —> addition 500ml fluid

What is the resultant Hct?

(5000 x 43.9%) = ((5000+500) x C2

—>C2 = 39.9%

note: 65 ml blood per Kg (adult)

ii) effect of blood infusion on Hct

Vol1 x Conc1 + (Hct x vol blood added) = Vol2 x Conc2

eg initial blood volume 5000ml with Hct 25% —> addition 500ml blood with Hct 35%

What is the resultant Hct?

(5000 x 25% + (500ml x 35%)) = (5000+500) x C2

—>C2 = 142500 ¸ 5500= 25.9%

iii) desired Hct on CPB

eg initial blood volume 5000ml with Hct 25% —> addition 2000ml prime vol with desired on CPB Hct of 25%

How many units of 60% Hct packed cells are required?

a) total vol (pt vol + priming vol) x desired Hct = total red cell vol req

7000ml x 0.25 = 1750 ml rbc

b) pt will provide 5000 x .25 = 1250 ml rbc vol

—> 1750 - 1250 = 500ml rbc vol req

c) whole blood used (Hct 35%)

—> vol whole blood req = red cell vol req ¸ whole blood Hct

= 500 ¸ .35= 1429 ml whole blood req

packed cells used (Hct 60%)

—> vol packed cells req = red cell vol req ¸ packed cells Hct

= 500 ¸ .60 = 833 ml packed cells req

iv) hemoconcentration

eg initial on CPB blood volume 7000ml with Hct 25% —> inadvertent addition of fluid giving a final Hct of 15% — using a hemoconcentrator which producers 250 ml units of 60% Hct: how many 250 ml 60% units of packed cells need to be prepared to return Hct to 25%?

a) Determine total vol of fluid inadvertently added

Vol1 x Conc1 = Vol2 x Conc2

Vol2 is the unknown

Vol2 = (Vol1 x Conc1) ¸ Conc2

Vol2 = (7000 x 25%)¸15%

Vol2 = 11667 ml

vol fluid added = Vol2 (new volume) - Vol1 (pt blood vol + prime)

11667 - (5000 + 2000) = 4667 ml

b) 4667 ml fluid will have to be removed

Now determine how much fluid is discarded in the production of each 250ml 60% Hct unit of blood

Vol1 x Conc1 = Vol2 x Conc2 Vol1 is the unknown

Vol1 = (Vol2 x Conc2)¸ Conc1

Vol1 = (250 x 60%) ¸15%

Vol1 = 1000 ml

Therefore for each 250 ml packed cells produced, 750 ml noncellular fluid is discarded

c) As 4667 ml fluid will have to be removed, 4667/750 or approx 6 processing steps are required

2. WEIGHT PER VOLUME

i) e.g. 500 ml 0.9% saline + 500 ml 5% dextrose

—> what are the final concentrations of the saline & dextrose?

0.9% saline = 0.9 grams per 100ml solution

5% dextrose = 5 grams per 100 ml solution

a) for saline:V1 x C1 = V2 x C2

C2 = V1 x C1¸ V2

C2 = 500 x 0.9% NaCl¸1000ml

C2 = 0.45% = 0.45 g/dL

b) for dextrose:V1 x C1 = V2 x C2

C2 = (V1 x C1)¸ V2

C2 = (500 x 5% NaCl) ¸ 1000ml

C2 = 2.5% = 2.5 g/dL

ii) e.g. How much 25% albumin must be added in a total vol of 2500 ml prime to give a final 5% albumin concentration?

V1 x C1 = V2 x C2 V1 is the unknown

V1 = (V2 x C2) ¸ C1

V1 = (2500 x 5%)¸25%

V1 = 500 ml

3. WEIGHT PER VOLUME/ VOLUME PER VOLUME COMBINATION

i) e.g. 680 ml blood (child) with fibrinogen 220 mg/dl & 60% Hct; 1000 ml prime

—> what is the on pump fibrinogen concentration?

a) First calculate plasma volume:

680 x (100 - 60%) = 272 ml plasma

b) V1 x C1 = V2 x C2

C2 = (V1 x C1) ¸ V2

C2 = (272 x 220 mg/dl) ¸ (total noncellular vol)1272 ml

C2 = 47 mg/dL

ii) Therefore req fibrinogen (coagulation is impaired with fibrinogen < 100 g/dL) in the form of FFP —> if give 1 ≈ FFP (250 ml with 200 mg/dL fibrinogen) to replace 250 ml prime what will be the new fibrinogen concentration?

C2 = (V1 x C1)¸ V2

C2 = (272 x 220 mg/dl + 250 ml x 200 mg/dL) ¸ (total noncellular vol)1272 ml

C2 = 86 mg/dL

4. CONVERSION of WEIGHT PER VOLUME UNITS to ACTIVITY UNITS (Equivalents or Osmoles)

i) conversion of 0.9% NaCl (weight per volume) to Eq/Osmol

a) convert weight/dL into weight/Litre

§ 0.9 grams/dL —> 9 grams/L

ii) conversion of weight/L into molar concentration

[note: a molar solution = 1 mole of solute per Litre]

§ molar solution of NaCl = 58.5 gram per litre solution

iii)58.5 gram/L NaCl: 9 gram/L NaCl

1 mole/L:X gram/L

X= (9gm/L x 1mole/L) ¸ 58.5 gm/L

X = 0.154 moles per litre NaCl

—> 0.9% solution NaCl in water contains 154 mmol/L NaCl

iv) As each NaCl molecule splits into 2 parts —> osmolarity of the 0.9% NaCl solution will be 2 times the molarity = 2 X 154 mM = 308 milliosmoles/L