H ydrophilic Lipophilic Ba la nc e (H LB) Ca lcula t ions
Pharmaceutics Lecture 11.1.07
Wha t is t he HLB va lue of a surfa c t a nt syst e m c om pose d of 1 0 g Spa n 6 0
(H LB=4 .7 ) a nd 2 0 g T w e e n 6 0 (H LB=1 4 .9 )?
Q1  10 g
C1  4.7
First, recognize the HLB is equivalent to concentration. Label the variables:
Q2  20 g
C2  14.9
Q f  (10 g  20 g )  30 g C f  ??
So now we’re going to use the same equation we’ve been using for a while, rearrange it, and solve
for Cf.
C1Q1  C2Q2  C f Q f
4.7  10 g  14.9  20 g
C1Q1  C2Q2
 Cf 
 11.5
Qf
30 g
Rx
M ine ra l Oil
Wool fa t
Ce t yl a lc ohol
Em ulsifie r
Wa t e r qs a d
30g
1 .5 g
1g
5g
90mL
H LB
12
10
15
a ) Ca lc ula t e t he re quire d (fina l) H LB.
b) U sing T w e e n 6 0 (HLB=1 4 .9 ) a nd Spa n 6 0 (H LB=4 .7 ), de t e rm ine how m uc h of
e a c h you ne e d t o fill t his Rx . (se e be low )
Label our variables. Do not use the final volume as Qf. When solving HLB problems consider only
the oils (the guys with an HLB value).
Q1  30 g
C1  12
Q2  1g
C2  15
Q2  1.5 g
Q f  (30  1.5  1)  32.5 g
C1Q1  C2Q2  C3Q3  C f Q f
C2  10
C f  ??
Rearrange the equation and solve.
12  30 g  10  1.5 g  15  1g
C1Q1  C2Q2  C3Q3
 Cf 
 12
32.5 g
Qf
b) U sing T w e e n 6 0 (HLB=1 4 .9 ) a nd Spa n 6 0 (H LB=4 .7 ) a s e m ulsifie rs, de t e rm ine
how m uc h of e a c h you ne e d t o fill t his Rx .
The Rx above stated that we need 5g of emulsifier to fill this prescription. We will use the median
allegation method to solve the problem (because it’s easier). Remember we are trying to get a
final HLB of 12.
Tween
HLB
14.9
Final
7.3 parts Tween
12
Span
+
4.7
2.9 parts Span
10.2 parts HLB 12
So we need 5g of emulsifier with HLB=12. Determine the number of grams of each Tween and
Span we need based on this mass.
5 g HLB 12  7.3 parts Tween 
  3.6 g Tween
 
 10.2 parts HLB 12 
5 g HLB 12  2.9 parts Tween 
  1.4 g Span
 
10
.
2
parts
HLB
12


Or instead of doing the second calc, just subtract the number of grams of Tween from the total
mass of the emulsifier.
5 g  3.6 g  1.4 g Span
Rx
M ine ra l Oil
Ce t yl a lc ohol
Em ulsifie r
Wa t e r qs a d
5 0 % w /v
1g
5 % w /v
120mL
H LB
12
15
a ) De t e rm ine t he re quire d H LB.
b) De t e rm ine the a m ount of T w e e n 6 0 (H LB=1 4 .9 ) a nd Spa n 6 0 (4 .7 ) re quire d t o
re a c h t his H LB. (se e be low )
Before we label our variables we need to determine the mass of mineral oil used in the Rx. The Rx
consists of 50% w/v mineral oil (50g/100mL) and is written for 120mL
120mL  50 g 

  60 g
 100mL 
Q1  60 g
C1  12
Q f  (60  1)  61g
C f  ??
Label the variables.
Q2  1g
C2  15
C1Q1  C2Q2  C f Q f
Rearrange the equation and solve:
C1Q1  C2Q2
12  60 g  15  1g
 Cf 
 12.05  12
Qf
61g
b) De t e rm ine t he a m ount of Tw e e n 6 0 (H LB=1 4 .9 ) a nd Spa n 6 0 (4 .7 ) re quire d
t o re a c h t his H LB.
First determine the number of grams of emulsifier required by this Rx. Above the prescription
states that of the final 120mL 5% w/v is emulsifier.
120mL  5 g emulsifier 

  6 g emulsifier

 100mL
Remember the final HLB we calculated above is 12. Again we’ll use median allegation method.
Tween
HLB
14.9
Final
7.3 parts Tween
12
Span
4.7
+
2.9 parts Span
10.2 parts HLB 12
So we need 6g of emulsifier with HLB=12. Determine the number of grams of each Tween and
Span we need based on this mass.
6 g HLB 12  7.3 parts Tween 
  4.3g Tween
 
parts
HLB
10
.
2
12


6 g HLB 12  2.9 parts Tween 
  1.7 g Span
 
parts
HLB
10
.
2
12


H LB Work she e t from c la ss
1 . Wha t is t he H LB of a n e m ulsion ble nd of 3 0 % spa n 4 0 (H LB=6 .7 ) a nd 7 0 %
t w e e n 4 0 (H LB=1 5 .6 )?
Assume a total mass of 100g for ease of calculations.
Label the variables.
Q1  30 g
C1  6.7
Q f  (30  70)  100 g
C f  ??
Q2  70 g
C2  15.6
C1Q1  C2Q2  C f Q f
Rearrange the equation and solve:
6.7  30 g  15.6  70 g
C1Q1  C2Q2
 Cf 
 12.9
Qf
100 g
2 . a ) Wha t is the H LB of t he follow ing e m ulsion?
b) De t e rm ine t he a m ount of Tw e e n 6 0 (H LB=1 4 .9 ) a nd Spa n 6 0 (4 .7 )
re quire d t o re a c h this H LB. (se e be low )
Rx
Ce t yl a lc ohol
Whit e w a x
Whit e pe t rola t um
Em ulsifie r
Wa t e r qs a d
14g
1g
10g
6 % w /w
100g
Q1  14 g
H LB
15
12
12
Label the variables.
Q2  1g
Q2  10 g
Q f  (14  10  1)  25 g
C1Q1  C2 Q2  C3Q3  C f Q f
C1  15
C2  12
C2  12
C f  ??
Rearrange the equation and solve:
C1Q1  C 2 Q2  C3Q3
15  14 g  12 1g  12  10 g
 Cf 
 13.7
25 g
Qf
b) De t e rm ine t he a m ount of Tw e e n 6 0 (H LB=1 4 .9 ) a nd Spa n 6 0 (4 .7 ) re quire d
t o re a c h t his HLB. (se e be low )
First determine the number of grams of emulsifier required by this Rx. Above the prescription
states that of the final 100g 6% w/w is emulsifier.
100 g  6 g emulsifier 
  6 g emulsifier
 
100
g


Remember the HLB we calculated above was 13.7.
Tween
HLB
14.9
Final
9 parts Tween
13.7
Span
4.7
+
1.2 parts Span
10.2 parts HLB 13.7
So we need 6g of emulsifier with HLB=13.7. Determine the number of grams of each Tween and
Span we need based on this mass.
6 g HLB 13.7  9 parts Tween 
  5.29 g Tween
 
 10.2 parts HLB 13.7 
6 g HLB 13.7  1.2 parts Tween 
  0.71g Span
 
 10.2 parts HLB 13.7 
3 . Ca lc ula t e t he re quire d H LB for the oil pha se of t he follow ing oil pha se .
Rx
Ce t yl e st e rs w a x
Whit e w a x
M ine ra l oil
PART S
1 2 .5
1 2 .0
5 6 .0
H LB
15
12
12
Q1  12.5 parts
C1  15
Q2  56 parts
C2  12
Label the variables.
Q2  12 parts
C2  12
Q f  (12.5  12  56)  80.5 parts
C1Q1  C 2 Q2  C3Q3  C f Q f
C f  ??
Rearrange the equation and solve:
C1Q1  C 2 Q2  C3Q3
15  12.5 parts  12  12 parts  12  56 parts
 Cf 
 12.5
80.5 parts
Qf