Question
Posted on: November 8 2016Ciclosporin eye drops 0.05% in oil
Answer
Info
Sandimmun contains ciclosporin as an active substance. One ml of the concentrate for solution for infusion contains 50 mg Ciclosporin. The other ingredients in Sandimmun are: Ethanol anhydrous, Polyoxylricinusolie. Sandimmun concentrate for solution for infusion Contains approximately 278 mg/ml ethanol ( 34.4% v/V) en 650 mg/ml macrogolglycerolricinoleate ( https://www.medicines.org.uk/emc/medicine/1317) P. p1 {margin: 0.0 px 0.0 px 0.0 px 0.0 px; font: 13.0 px Arial;-webkit-Text-stroke: #000000} span. s1 {font-kerning: none}
. A dose of 100 mg of Sandimmun contains 556 mg ethanol.
Material: 2 sterile eye drop vials and 2 sterile syringes to measure 1 ml and 1 sterile syringe of 10 ml
Preparation: The used oil must be strengthened. We Make a small beaker clean with purified water and rinse after with Gedenat. alcohol. The beaker is covered with all foil and we place it on a soft heating source (radiator) to allow the alcohol to evaporate. In this bekergla we weigh 25 g olive oil, cover the beaker back with the already foil and place the whole in a warm air oven at 150 ° C for 75 minutes. Then let's cool the oil.
Preparation of the solution
The concentration of the prescribed ampoule is 5%. You should dilute it to 0.05%. I suspect we should do this in two steps.
From the 5 ml ampoule, with a sterile syringe, we bring 1 ml (= 50 mg C) into a sterile collyrevial (1) and add 9 ml of sterile oil and mix to homogenous. 1 ml of this solution (1) contains 5 mg C. From Solution 1, with a sterile syringe, we take 1 ml of solution, transfer it in Collyrevial 2 and return 9 ml of sterilized olive oil. The collyrevial (1) is closed. 1 ml of the solution in Collyrevial (2) contains 0.5 mg C or 0.05%. The Collyre Bottle 2 is closed and provided with the required labels.
Problem of alcohol in the delivered solution
100 mg Sandimmun contains 556 mg ethanol. The density of this solution is not specified and we assume = +/-1 so 1 ml of Sandimmun contains 5560 mg ethanol. This 1 ml is diluted to 10 ml (solution 1) so that 1 ml contains 556 mg of ethanol. From this 10 ml of solution 1 we take 1 ml and dilute to 10 ml solution (2). 1 ml of solution (2) contains 55.6 mg ethanol. 1 ml of olive oil gives 45 drops so we have a little more than 1 mg ethanol per drop.