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RESULTS

Experiment A
Propionic
acid

Titration

of

Titration

of

(mL)

5
3
1

Propionic

Propionic

acid

acid

(organic)

(aqueous)

concentratio

concentratio

(mL)

(mL)

n in aqueous

n in organic

25.5
12.3
7.9

46.9
30.5
20.0

layer, Y
0.938
0.610
0.400

layer, X
0.510
0.246
0.158

Experiment B

Flow

rate

of

aqueous

phase (L/min)

Flow rate of organic phase

(L/min)

Sodium

hydroxide

0.20

0.20

0.1 M

0.025M

Feed (mL)

0.40

0.20

Raffinate (mL)

0.30

1.40

Extract (mL)

0.5

3.30

concentration (M)

Propionic

acid

extracted

(mol/min)

0.008

K=Y/X

1.8392
2.480
2.532

Propionic

acid

extracted

0.007734

(mol/min)

(kg/min)

0.003732

: 0.20 L/min

Organic flow rate

: 0.20 L/min

Packing dimension

: Length

= 1.2 m

Diameter = 50 mm

CALCULATIONS

Formula for finding distribution coefficient (Experiment A):

K = Y/X
Where Y: Concentration of solute in extract phase.
X: Concentration of solute in raffinate phase.

5.0mL of propionic acid

Upper (Y)
M1V1 = M2V2
(0.1M) (46.9mL) = (M2) (5mL)

Bottom (X)
M1V1 = M2V2
(0.1M) (25.5mL) = (M2) (5mL)

M2 = 0.938 M

M2 = 0.510 M

= Y/X
= 0.938 M / 0.510 M
= 1.8392

3.0mL of propionic acid

Upper (Y)
M1V1 = M2V2
(0.1M) (30.5mL) = (M2) (5mL)
M2 = 0.610 M

Bottom (X)
M1V1 = M2V2
(0.1M) (12.3mL) = (M2) (5mL)
M2 = 0.246 M

= Y/X
= 0.610 M / 0.246 M
= 2.480

1.0mL of propionic acid

Upper (Y)
M1V1 = M2V2
(0.1M) (20.0mL) = (M2) (5mL)
M2 = 0.400 M

= Y/X
= 0.400 M / 0.158 M
= 2.532

Bottom (X)
M1V1 = M2V2
(0.1M) (7.90mL) = (M2) (5mL)
M2 = 0.158 M

Formula for finding mass transfer coefficient (Experiment B):

M1V1 = M2V2
Where:
M1 = concentration of NaOH
M2 = concentration of propionic acid
V1 = volume of NaOH
V2 = volume of propionic acid

Rate of acid transfer = Vw (Y1 0)

Vo (X1 X2) = Vw (Y1 0)
K = Y1/X*
Log mean driving force = (X1-X2) / ln (X1/X2)
Where:
X1 : Driving force at the top of the column = (X2 0)
X2 : Driving force at the bottom of the column = (X1 - X1*)

Diameter = 50mm

50 10
2

= 0.025m

Therefore, packing volume, V = r2L

= ( 0.025m)2(1.2m)
= 2.36 10-3 m3
= 2.36 L

For 0.1M of NaOH

Raffinate:
M1V1

= M2V2

(0.1)(0.30mL) = M2 (15mL)
M2

= 0.00200 M of propionic acid X1

M1V1

= M2V2

Feed:

(0.1)(0.40mL) = M2 (15mL)
M2

= 0.00267 M of propionic acid

M1V1

= M2V2

Extract:

(0.1)(0.50mL) = M2 (15mL)
= 0.00333 M of propionic acid Y1

M2

Rate of acid transfer = Vw (Y1- 0)

= (0.20 L/min) (0.00333 mol/L)
= 0.00067 mol/min

Vo (X1-X2)

= Vw (Y1 -0)

(0.20 L/min)(0.00200 mol/L - X2)

= 0.00067 mol/min

X2

= -0.00027 mol/L @ M

X1 = (X2-0)
= -0.00027 M

K = Y1 / X*
X* = Y1 / K

At equilibrium, assume K = 0.9273 (from experiment A)

= 0.00333 / 0.9273
= 3.5911 x 10-3 M
X2 = (X1-X1*)
= -0.00027 M 3.5911 x 10-3 M
= -3.8611 x 10-3 M
Log mean driving force = (X1-X2) / ln (X1/X2)
= (-0.00027 + 3.8611 x 10-3) M / ln (-0.00027 / -3.8611 x 10-3)
= -1.350 x10-3

Finding the mass transfer coefficient:

Mass transfer coefficient =

Rate of acid transfer

Volume of packing Mean driving force
0.00067 mol/min
2.36 L x 0.9084

= 3.1253 x 10-4 mol/L.min

= 3.1253 x 10-4 M/min

Raffinate:
M1V1

= M2V2

(0.025)(1.40mL)

= M2 (15mL)

M2

M1V1

= M2V2

Feed:

(0.025)(0.20mL)

= M2 (15mL)

M2

= 0.00033 M of propoinic acid

M1V1

= M2V2

Extract:

(0.025)(3.30mL)
M2

= M2 (15mL)
= 0.0055 M of propoinic acid Y1

Rate of acid transfer = Vw (Y1- 0)

= (0.20 L/min) ( 0.0055 mol/L)

= 0.0011 mol/min

Vo (X1-X2)

= Vw (Y1 -0)

(0.20 L/min)(0.00233 mol/L - X2)

= 0.0011 mol/min

X2

= - 0.00063 mol/L @ M

X1 = (X2-0)
= - 0.00063 M

K = Y1 / X*
X* = Y1 / K

At equilibrium, assume K = 0.9273 (from experiment B)

= 0.0055 / 0.9273
= 0.0059 M
X2 = (X1-X1*)
= - 0.00063 M 0.0059 M
= -0.0065 M
Log mean driving force = (X1-X2) / ln (X1/X2)
= (-0.00063 + 0.0065) M / ln (-0.00063 / -0.0065)
= -0.0025

Finding the mass transfer coefficient:

Mass transfer coefficient =

Rate of acid transfer

Volume of packing Mean driving force
0.0011 mol /min
2.36 L x0.05438

= -0.0086 mol/L.min
= -0.0086 M/min
= -0.0086 kg/min

DISCUSSION

In this experiment, we want to identify the distribution of coefficient and to identify the mass
transfer coefficient. This experiment is referred to the solubility which is the solvent is also
soluble with a specific solute contained in the solution and also about the separation of a
substance from a mixture by preferentially dissolving the substance in a suitable solvent.
For the first experiment, which is to determine the distribution coefficient, we used the titration
method from the upper (Y) and bottom (X) layer sample. By titration with 0.1 M, the values are
6.5333 in 1.0 ml of propionic acid, 2.7143 in 3.0 ml of propionic acid and 1.8913 in 5.0 ml of
propionic acid. From the result in this experiment, if titrated with 0.1 M NaOH, it shows that the
value of distribution coefficient decrease as the volume of propionic acid increase. This
experiment is totally successful because those results followed the actual results which is the
value of distribution coefficient must be decrease when the volume of propionic acid is
increasing.

For the second experiment, which is to determine the mass transfer coefficient, we used the
liquid-liquid extraction column to get the feed, raffinate and extract solution to be the samples.
The samples then are titrated with 0.1 M NaOH and 0.025 M NaOH. The value of mass transfer
coefficient when titrated with 0.1 M NaOH is 0.003732 kg/min while titrated with 0.025 M, the
value of mass transfer coefficient is-0.06026 kg/min. Result for this experiment shows that the
experiment is failed because the values of mass transfer coefficient supposedly in positive.
Besides that, we cannot show that the relation between mass transfer rate with the concentration
of NaOH. For the actual result, this experiment can shows the increment of mass transfer rate
when the concentration of NaOH is decreasing.

Several errors occurred during this experiment progress that totally affects the result of second
experiment which is for finding the mass transfer coefficient. The most common error is the
position of the eye during taking the volume value at the burette. As the solution, the eye position
should be straight to the scale and must be perpendicular to the meniscus. Besides that, the error
also occurred while using the apparatus that is not properly clean. We should use the clean
apparatus to avoid the oil emission and impurities at the beaker, conical flask or burette. If the
apparatus not clean, we should use the distilled water to clean up those apparatus.
Besides that, the experiment should be repeated at least 2 or 3 times to get the accurate values
and the mistake during the experiment progress can be identified. However, the first experiment
only done with one trial those cause the experiment being failed. Furthermore, we choose light
pink color for the color indicator during titration, but to get all the same color might be hard
enough to identify. As the color is not same in each titration, the value of mass transfer
coefficient will be different from the actual and made the experiment failed. This is because the
color indicates that the NaOH is at equilibrium with the sample solution.
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