Page 21 - Practical DF2 Corrected (2)

Pharm D Clinical Pharmacy program 2024-2025 Level 2, Semester 2 Pharmaceutical Dosage Forms II (PT 405)

- The amount of alkali (Borax) can be calculated as follows:

Rx/

Purified water 24 g

Borax ?g

Olive oil 50 g

White beeswax 12.5 g

White soft paraffin 12.5 g

Fiat cream. Mitte 20 g.

Knowing that:

- Molecular weight of borax (Na2B4O7.10 H2O) = 381.43 g/mol

- Molecular weight of potassium hydroxide (KOH) = 6.11 g/mol

- Molecular weight of sodium hydroxide (NaOH) = 40 g/mol

- Acid value of olive oil = 2

- Acid value of white beeswax = 20

- Acid value: is the number of (mg) of KOH required to neutralize 1 g of

oil or fat.

Calculations:

Na2B4O7 + 3 H2O Partial hydrolysis 2H3BO3 + 2NaBO2

2NaBO2 + 4 H2O 2H3BO3 + 2NaOH

i.e each mole of Na2B4O7 = 2 moles of NaOH
So, the quantity of KOH (mg) required to neutralize the free fatty acids of

the cream obtained from:
- beeswax = 20 × 12.5 = 250 mg.
- olive oil = 2 x 50 = 100 mg.

⁂ Total quantity of KOH required for neutralization = 250+100=350 mg = 0.35g

1 mole of borax → 2 moles of NaOH → 2 moles of KOH

381.43 g → → → 2 × 56.11 g

X → → → 0.35 g

- Amount of borax (X) = 0.35 ×381.43 = 1.19 g.

2 ×56.11

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