7.0RESULTS

Experiment A:Propionic acid added (mL)Aqueous Layer Organic Layer X

Titrate of M/10NaOH (mL)Concentration (M)Titrate of M/10NaOH (mL)Concentration (M)

570.7044.40

348.5019.25

113.007.40

Experiment C:Flow rate of aqueous phase

Titrate of M/10NaOH (mL)Titrate of 0.025M NaOH (mL)

Feed242212.5Avg = 19.5282520Avg = 24.33

Raffinate1.00.50.5Avg = 0.672.01.51.5Avg = 1.67

Extract2.02.02.0Avg = 28.5109.7Avg = 9.4

8.0CALCULATIONS

For experiment A

Calculation of concentration: M1V1 = M2V2 M1 = concentration of NaOH M2 = concentration of propionic acid V1 = volume of NaOH To find distribution coefficient:K = Y/X Y : Concentration of solute in extract phase (Upper) X: Concentration of solute in raffinate phase (Bottom)

V2 = volume of propionic acid

For 0.1M of NaOHVolume of Propionic Acid,mLAqueous (Y)Organic (X)DistributionCoefficient,K

5.0M1V1 = M2V2(0.1M)(70.7 10-3L)=(M2)(5.0 10-3L)M2= 1.414MM1V1 = M2V2(0.1M)(44.4 10-3L)=(M2)(5.0 10-3L)M2= 0.888 MK = =

= 1.592

3.0M1V1 = M2V2(0.1M)(48.5 10-3L)=(M2)(3.0 10-3L)M2= 1.617MM1V1 = M2V2(0.1M)(19.25 10-3L)=(M2)(3.0 10-3L)M2= 0.642 MK = =

= 2.519

1.0M1V1 = M2V2(0.1M)(13 10-3L)=(M2)(1.0 10-3L)M2= 1.3MM1V1 = M2V2(0.1M)(7.4 10-3L)=(M2)(1.0 10-3L)M2= 0.74MK = =

= 1.757

For experiment C

Finding Mass Transfer Coefficient:Formulae used;

1. M1V1 = M2V2Where: M1 = concentration of NaOH M2 = concentration of propionic acid V1 = volume of NaOH V2 = volume of propionic acid

2. Rate of acid transfer = Vw (Y1- 0) Vw : Water Flow rate

3.Vo (X1-X2) = Vw (Y1 -0) Vo : Organic solvent flow rate

4. K = Y1/X*

5. Log mean driving force= (X1-X2) / ln (X1/X2) X1: Driving force at the top of the column = (X2-0) X2: Driving force at the bottom of the column = (X1-X1*)

6. Packing dimension: Length = 1.2 m Diameter = 50 mm Radius = = 0.025m

Therefore, packing volume, V= r2L = = 2.356 10-3 m3

For 0.1M of NaOH

Feed: M1V1= M2V2 (0.1M)(19.5 mL) = M2 (10 mL) M2 =0.195M of propionic acid (the X1)Raffinate: M1V1 = M2V2 (0.1M)(0.67 mL)= M2 (10 mL) M2=0.0067M of propionic acid (the X2)Extract: M1V1 = M2V2(0.1M)(2.0 mL)= M2 (10 mL) M2=0.02M of propionic acid (the Y1)

The flow rate of aqueous and organic phase = 0.0033 L/s

Mass Balance:

Propionic Acid extracted from the organic phase (raffinate)= V0 (X1 X2)= 0.0033 (0.195 0.0067)= 0.00062139 kg/sPropionic Acid extracted from the aqueous phase (extract)= Vw (Y1 Y2 ) since Y2 = 0.0= 0.0033 (0.02 0)= 0.000066 kg/s

To calculate the X1*, the average distribution coefficient from experiment A = 1.956K = ; X* = X* = = 0.0102 M

Log mean driving force =

= = 0.0559 Mass transfer coefficient = =

= 14851 = 14.851

For 0.025M of NaOH

Feed: M1V1= M2V2 (0.1M)(24.33 mL) = M2 (10 mL) M2 =0.2433M of propionic acid (the X1)Raffinate: M1V1 = M2V2 (0.1M)(1.67 mL)= M2 (10 mL) M2=0.0167M of propionic acid (the X2)Extract: M1V1 = M2V2(0.1M)(9.4 mL)= M2 (10 mL) M2=0.94M of propionic acid (the Y1)

The flow rate of aqueous and organic phase = 0.0033 L/s

Mass Balance:

Propionic Acid extracted from the organic phase (raffinate)= V0 (X1 X2)= 0.0033 (0.2433 0.0167)= 0.00074778 kg/s

Propionic Acid extracted from the aqueous phase (extract)= Vw (Y1 Y2 ) since Y2 = 0.0= 0.0033 (0.94 0)= 0.003102 kg/s

To calculate the X1*, the average distribution coefficient from experiment A = 1.956K = ; X* = X* = = 0.4806 M

Log mean driving force =

= = 0.0846Mass transfer coefficient = =

= 9813 = 9.813