HLB Practise Flashcards
The formula for a cosmetic cream calls for 5% of an emulsifier blend consisting of Span60 and Tween20. If the “Required HLB” of the oil phase is 14.0, how many grams of each emulsifier should be used in preparing 500grams of the cream?
HLB of Span60 = 4.7
HLB of Tween20 = 16.7
By Alligation,
14
Span60: 4.7 2.7
Tween20: 16.7 9.3
Relative Amounts
Tween20 : Span60
9.3 : 2.7 (9.3+2.7 = 12)
9.3/12 x 100 = 77.5% : 2.7/12 x 100 = 22.5%
Since 5% of emulsifying blend is used in the preparation of 500 grams of cream, therefore,
5/100 x 500 grams = 25 grams of mixture should be used in the preparation of 500 grams of cream.
Span60 = 25 grams x 22.5/100 = 5.625 grams.
Tween20 = 25 grams x 77.5/100 = 19.375 grams
Answer: So, 5.625 grams of Span60 and 19.375 grams of Tween20 should be used in the preparation of 500 grams of cream.
a) Stearic Acid = 8%
b) Cetyl Alcohol = 1%
c) Lanolin(Anhydrous) = 1%
d) Emulsifier = 4%
e) Glycerin = 10%
f) Preserved Water quantity sufficient to produce = 100%
1. Calculate the “Required HLB” of the oil phase?
2. How many grams of Span80 and how many grams of Tween60 should be used in formulating 1000 grams of the product?
(1).
Oil Phase: a)+b)+c) =8%+1%+1% = 10%
Required HLB x Fraction of oil phase
Stearic Acid 15 x 8/10 = 12
Cetyl Alcohol 15 x 1/10 = 1.5
Lanolin(Anhydrous) 10 x 1/10 = 1
12+1.5+1 = 14.5
Answer(1): The “Required HLB” of oil phase is 14.5.
(2).
HLB of Span80 = 4.3
HLB of Tween60 = 14.9
14.5
Span80 4.3 0.4
Tween60 14.9 10.2
Relative Amounts
Span80 : Tween60
0.4 : 10.2 0.4+10.2 = 10.6
0.4/10.6 x 100 = 3.8% : 10.2/10.6 x 100 = 96.2%
100 grams of formulation contains = 4 grams of emulsifier
1 grams of formulation contains = 4/100 grams of emulsifier
1000 grams of formulation contains = 4/100 x 1000 grams of emulsifier
= 40 grams of emulsifier
Span80 40 x 3.8% = 1.52 grams
Tween60 40 x 96.2% = 38.48 grams
Answer(2): So, in making 1000 grams of formulation, 1.52 grams of Span80 and 38.48 grams of Tween60 should be used.
Calculate the HLB of a mixture of 45 grams of Span80 and 55 grams of Polysorbate80.
HLB of Span80 = 4.3
HLB of Polysorbate (Tween80)= 15
Total quantity of mixture = 45 grams +55 grams = 100grams
%Age of Span80 = 45/100
= 45%
%Age of Tween80 = 55/100
= 55%
HLB x % of mixture
Span80: 4.3 x 45% = 1.9
Tween80: 15 x 55% = 8.3
1.9+8.3 = 10.2
Answer: HLB of mixture is 10.2.
What is the HLB of an emulsifier blend consisting of 20% of Span60, 20% of Span80 and 60% of Tween60?
HLB of Span60 = 4.7
HLB of Span80 = 4.3
HLB of Tween60 = 14.9
HLB x % of mixture
Span60: 4.7 x 20% = 0.94
Span80: 4.3 x 20% = 0.86
Tween60: 14.9 x 60% = 8.94
0.94+0.86+8.94 = 10.74
Answer: HLB of emulsifying blend is 10.74.
Calculate the “Required HLB” for the oil phase of the following O/W type lotion?
Stearyl Alcohol: 250grams.
White Petrolatum: 250grams.
Propylene Glycol: 120grams.
Emulsifier quantity sufficient.
Preserved Water quantity sufficient to produce 1000 grams.
Representation of the oil phase in the formulation = 250+250+120 = 620grams.
The oil phase represents 620 parts of entire formula.
Representation of different components in the oil phase:
Stearyl Alcohol = 250/620 = 0.403 parts of oil phase.
White Petrolatum = 250/620 = 0.403 parts of oil phase.
Propylene Glycol = 120/620 = 0.194 parts of oil phase.
HLB of Stearyl Alcohol = 14
HLB of White Petrolatum = 12
HLB of Propylene Glycol = 11.6
HLB x Fraction of the oil phase
Stearyl Alcohol: 14 x 0.403 = 5.642
White Petrolatum: 12 x 0.403 = 4.836
Propylene Glycol: 11.6 x 0.194 = 2.250
Required HLB =5.642+4.836+2.250 = 12.73
Answer: Required HLB of the oil phase is 12.73.
Calculate the Required HLB of the oil phase of the following O/W type lotion?
Mineral Oil = 30%
Lanolin (Anhydrous) =2%
Cetyl Alcohol = 3%
Emulsifier quantity sufficient.
Preserved water quantity sufficient to produce 100%
Representation of the oil phase in the entire formula = 30%+2%+3% = 35%
Representation of different components in the oil phase:
Mineral Oil = 30/35 = 0.86 parts of the oil phase.
Lanolin (Anhydrous) =2/35 = 0.06 parts of the oil phase.
Cetyl Alcohol = 3/35 =0.086 parts of the oil phase.
HLB of Mineral Oil = 12
HLB of Lanolin (Anhydrous) = 10
HLB of Cetyl Alcohol = 15
Required HLB x Fraction of oil Phase.
Mineral Oil = 12 x 0.86 = 10.3
Lanolin (Anhydrous) = 10 x 0.06 = 0.6
Cetyl Alcohol = 15 x 0.086 = 1.3
Required HLB of the oil phase = 10.3+0.6+1.3 = 12.2
Answer: Required HLB of the oil phase is 12.2.
: In what proportion should Tween60 and Arlacel83 be blended to obtain a “Required HLB” of 11.5?
HLB of Tween60 = 14.9
HLB of Arlacel83 = 3.7
By Alligation,
11.5
Tween60 14.9 7.8
Arlacel83 3.7 3.4
Tween60 : Arlacel83
7.8 : 3.4
7.8+3.4 = 11.2
Tween60 = 7.8/11.2 x 100 = 69.6%
Arlacel83 = 3.4/11.2 x 100 = 30.4%
Answer: Tween60 and Arlacel83 should be blended in a proportion of 69.6% and 30.4% respectively to obtain a “Required HLB” of 11.5.
The “Required HLB” of and oil phase is 13.2. What %Age of Tween40 and of Span40 should be used to give the “Required HLB”?
HLB of Tween40 = 15.6
HLB of Span40 = 6.7
By Alligation,
13.2
Tween40 15.6 6.5
Span40 6.7 2.4
Relative Amount
Tween40 : Span40
6.5 : 2.4
6.5+2.4 = 8.9
Tween40 = 6.5/8.9 x 100 = 73%
Span40 = 2.4/8.9 x 100 = 27%
Answer: Tween40 and Span40 should be mixed in a proportion of 73% and 27% to get a “Required HLB of 13.2.
