Tear fluid is isotonic with normal saline and shows a freezing point of -0.52°C.
Points:
(1) 1 osmole per kg water reduces the freezing point to -1.858°C.
(2) The molecular weight of sodium chloride is 58.45 grams.
(3) A freezing point depression of -0.52°C caused by sodium chloride would reflect 280 mOsm per kg water. Since sodium chloride dissociates into 2 osmotically active atoms, this indicates 140 mmol/L sodium chloride, or 8.18 grams per liter, or 0.82% solution.
(4) To simplify calculations the percent sodium chloride is rounded to 0.85% or 0.9%.
(5) The eye can tolerate a tonicity ranging from 0.5% to 1.6% sodium chloride equivalents (page 345, Turco). So the minor changes in salt concentration associated with the rounding of the percent sodium chloride are within tolerance limits.
(6) The osmolarity of a final solution is additive of the osmolarity of the solutes added.
To prepare an ophthalmic solution isotonic with tears:
(1) Prepare the solution according to directions.
(2) Measure its freezing point depression.
(3) Determine the amount of sodium chloride to add to make the solution isotonic (target 280 mOsm per kg water).
Method 1 to calculate amount of sodium chloride to add:
(1) The amount of freezing point depression required from sodium chloride is (-0.52°C) - (freezing point of prepared solution in °C).
(2) 0.52 / (0.9% sodium chloride) = (freezing point depression needed from sodium chloride) / (percent sodium chloride needed), or rearranged:
percent sodium chloride needed in final solution =
= (-0.52 - (freezing point of prepared solution)) * 0.9% / (-0.52) =
(3) The percent sodium chloride needed is grams per deciliter (100 mL). This needs to be multiplied by 10 for grams per liter.
Method 2 to calculate the amount of sodium chloride to add:
(1) Use the freezing point determination to determine the number of mOsm present in the solution as prepared, as
number of mOsm per kg water =
= (freezing point depression in °C) / 1.858
(2) 280 - (mOsm per kg water) = mOsm of sodium chloride to add
(3) Since sodium chloride dissociates, (mOsm required of sodium chloride) / 2 = mmoles sodium chloride to add.
(4) The number of grams sodium chloride to add per liter prepared solution = (58.45) * (mmoles sodium chloride to add) / 1000
