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Ion Transport Numbers

By:   •  December 20, 2015  •  Lab Report  •  2,150 Words (9 Pages)  •  1,451 Views

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PHYSICAL

CHEMISTRY II

SESSION 2015/2016

EXP 4: ION TRANSPORT NUMBERS

NAME:NOR ATIQAH BINTI ZAINAL ABIDIN

MATRIC NO.: SEZ140033

PARTNER : AIZAT LIYANA BINTI OTHMAN

MATRIC NO.: SEW140001

GROUP: S

DATE OF EXPERIMENT: 9TH DECEMBER 2015

DATE OF SUBMISSION:  17TH DECEMBER 2015

LECTURER: DR. TEO YIN YI

Objective:

To determine the ionic transport number of H+ in aqueous solution of 0.0990 M HCl.

Procedure:

[pic 1]

1) The following solutions are prepared.

Leading solution : 0.5 cm3 of methanolic bromophenol blue solution is added to 120 cm3  0.0990 M HCl solution.

Indicator solution : 0.5 cm3 of methanolic bromophenol blue solution is added to 100 cm3 0.0361 M LiCl solution.

2) The electrolysis cell is cleaned first with dilute sulphuric acid  followed by distilled water.

3) The cathode cell compartment is washed with HCl twice Then it is filled with the leading solution until rising up above the cell tap. The cell tap is closed and the cell compartment is filled up with HCl until full.

4) The Ag/AgCl reference electrode is put carefully into the cell compartment so that no air bubbles trapped in the cell compartment. Throughout the experiment, the cell compartment should not contain any air bubbles. Any excess HCl should be removed from the top of the cell tap with a dropper.

5) The anode cell compartment is washed with the LiCl solution several times. Next, it is filled with the indicator solution and the Platinum electrode is put inside it.

6) The cell is placed in the water bath.

7) The cell electrodes are connected to the DC power supply.

8) The cell tap is opened before the DC power supply is switched on. First a current of 4-5 mA is delivered through the cell until boundary reaches the first graduated scale in the graduated tube.

9) Then the current is held steady at 3 mA throughout the experiment.

10) The time taken for the boundary to reach each graduated scale is recorded.

11) When the boundary has reached the final scale, these qualitative tests are performed:

(i)   The electrical current is switched off for a while and the boundary is observed.   The electrical supply is switched on again.

(ii)   The wire connection to the electrodes is reversed so that the electrical current flow tp the opposite direction.  The boundary is observed. Then the wires is reconnected to the electrodes as before.

12) All the observations are recorded.

Results and Calculations:

Concentration of HCl: (0.0990 ± 0.0004) M

Concentration of LiCl: (0.0361 ± 0.0009) M

Temperature: (30.00 ± 0.01) °C

Time (minutes)

Volume (mL)

Time (minutes)

Volume (mL)

0

0.10

396

0.36

15

0.11

412

0.37

35

0.12

429

0.38

53

0.13

440

0.39

75

0.14

461

0.40

93

0.15

479

0.41

111

0.16

495

0.42

126

0.17

514

0.43

143

0.18

529

0.44

154

0.19

549

0.45

168

0.20

566

0.46

185

0.21

584

0.47

191

0.22

602

0.48

202

0.23

620

0.49

222

0.24

642

0.50

239

0.25

659

0.51

252

0.26

680

0.52

265

0.27

697

0.53

278

0.28

715

0.54

291

0.29

735

0.55

306

0.30

755

0.56

322

0.31

777

0.57

335

0.32

797

0.58

350

0.33

815

0.59

364

0.34

835

0.60

379

0.35

[pic 2]

From regression:

 

Coefficients

Standard Error

t Stat

P-value

Lower 95%

Upper 95%

Intercept

0.104330466

0.00265887

39.23864303

3.95358E-38

0.0989844

0.1096765

X variable

0.000616289

5.66815E-06

108.7285583

4.19723E-59

0.0006049

0.0006277

From the graphed plotted,

Y= 0.0006x + 0.104

the gradient obtained

= 0.0006

= (0.000600 ± 0. 000006)

From the equation: [pic 3]

where          [pic 4]        =         gradient of the graph

           F           =         Faraday’s constant

                =         (9.64870 ± 0.00016) x 104 C mol-1

        CAX         =         concentration of AX

          V            =         volume in cm3

           I            =         electrical current

                =         (3.00 ± 0.05) mA

           t            =         time in seconds

          A            =         H+

        AX         =         HCl

Because [pic 5]= m,

Hence,        [pic 6]                                  = CHCl F.m / I

                    =[pic 7]   

                           =[pic 8]                                                 

...

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