1a. 90 mL of sterile water is added in order to create a 125 mL bottle with a co

Question

1a. 90 mL of sterile water is added in order to create a 125 mL bottle with a concentration of 250 mg per 5 mL. If 65 mL of water were mistakenly
added instead of 90 mL when reconstituting the product, what is the new concentration of the bottle in milligrams per teaspoonful?

1b. Instructions for a 500 mg vial for injection state to reconstitute the vial with 4.8 mL sterile water in
order to make a 100 mg/mL concentration. What would be the concentration (in mg/mL) if you accidentally
reconstituted this vial with 5.8 mL sterile water instead? Round your answer to the nearest 1/10th.

Explanation / Answer

1a) Determine the initial concentration of the stock solution. 90 mL of sterile water was added to the stock solution to prepare a 125 mL solution. Assuming that volumes are additive, we have, volume of the stock solution taken = (125 – 90) mL = 35 mL.

The final concentration of the analyte in the stock is 250 mg/5 mL = (250/5) mg/mL = 50 mg/mL.

We need to determine the concentration of the analyte in the stock solution by using the dilution equation. The dilution equation is

M1*V1 = M2*V2

where V1 = 35 mL; V2 = 125 mL and M2 = 50 mg/mL. Therefore,

M1*(35 mL) = (50 mg/mL)*(125 mL)

====> M2 = (50 mg/mL)*(125 mL)/(35 mL) = 178.57 mg/mL.

However, only 65 mL sterile water was added so that the final volume is (35 + 65) mL = 100 mL. Again use the dilution equation as

M1*V1 = M2*V2

=====> (178.57 mg/mL)*(35 mL) = M2*(100 mL)

=====> M2 = (178.57 mg/mL)*(35 mL)/(100 mL) = 62.4995 mg/mL.

The concentration is (62.4995 mg/mL)*(5 mL) = 312.4975 mg/5 mL 312.5 mg/5 mL (ans).

1b) The student added 5.8 Ml sterile water instead of 4.8 mL to the 500 mg sample. The concentration is (500 mg)/(5.8 mL) = 86.206 mg/mL 86.2 mg/mL (ans).

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