Experiment Report

CHEMISTRY

EXPERIMENT REPORT

SOLUBILITY AS A FUNCTION OF TEMPERATURE

[experiment title]

Waskitarini Darmiyanti

3315133616

REGULAR CLASS OF CHEMISTRY EDUCATION 2013

FACULTY OF MATHEMATICS AND NATURAL SCIENCE

UNIVERSITAS NEGERI JAKARTA

SOLUBILITY AS A FUNCTION OF TEMPERATURE

I. PURPOSE

1. Determine the solubility of the substance at

various temperatures

2. Determine heat differential solvent

3. Determine the effect of temperature on the

solubility of a substance

4. Understanding the relationship between

temperature graph with solubility

5. Determine the average dissolution heat at

different temperature

II. THEORY

In a saturated solution of a proper balance

between the molecules of a substance soluble and

insoluble. The balance can be written as follows:

A(p) A(l)

(1)

A(l) : soluble molecules

A(p) : insoluble molecules

The equilibrium constant for the dissolution

process:

K = a2 / A2 = a2 / 1 = ys ms

(2)

a2 : activity of soluble substances

A2 : liveliness insoluble substances, which took the

price of 1 for solids in the standard state

ys : the activity coefficient of the dissolved

substance

ms : kemolalan zat yang larut yang karena larutan

jenuh disebut kelarutan

solubility is the amount of substance that can

be dissolved in a solvent to form a saturated

solution. As for how to determine the solubility of

a substance is to take a certain amount of pure

solvent, such as 1 liter. Then estimate the amount

of a substance that can form a supersaturated

solution is characterized by the presence of

insoluble solids. Having shaken or stirred

equilibrium will occur between substances that

dissolve substances insoluble (Atkins, 1994).

Equilibrium constant relationship with the process

temperature is given by van't Hoff reaction isobar:

[∂lnk∂T ]p=∆H°RT2

(3)

H° : enthalpy increase processR : ideal gas constantequation (2) and (3) provide:

[∂lnγsms∂T ]p=∆HDSRT2

(4)

∆HDS : heat differential dissolution in saturated concentration

Next (4) can be decomposed into:

∂lnγsms∂lnms

.∂lnms∂T

=∆HRT2

(5)

(∂ln∂s∂lnms+1)∂lnms∂T

=∆HDsRT2

in the case of ∂lnγsms∂lnms can be ignored, (s) can be

written as

dlnmsdT

=∆HDsRT2

(6)

or dlogms

d(1T)=

−∆H2.303R

(7)

Thus ΔH Ds can be determined from the direction

of the tangent to the curve log ms to 1 / T.

If ΔH Ds does not depend on the temperature,

the graph of log ms versus 1 / T will be linear and

integration of equation (7) between the temperatures

T1 and T2 provide:

log ms (T2 )ms(T1)

=∆Hds2.303R .

T2−T1T2.T1

(8)

III. TOOLS AND MATERIALS

Tools:

1. Beaker 1000 mL 1 piece

2. Large test tube (sheath) 1 piece

3. Large test tube 1 piece

4. Stir bar circumference 1 piece

5. Thermometer 100 ℃ 1 piece

6. Pipette volume 10 mL 1 piece

7. Erlenmeyer flask 230 mL 1 piece

8. flask 100 mL 1 piece

9. Pipette volume 25 mL 1 piece

Materials:

1. oxalic acid

2. NaOH (Sodium Hydroxide) 0.5 M

IV. WORK CHART

Make a saturated solution of oxalic acid in a way:

1. Fill water into the tube up to approximately 20

mL, heat until approximately 60 deg C,

dissolving oxalic acid, so the solution becomes

saturated.

2. Then enter the large tube containing a

saturated solution into a large sheath tube and

input into the beaker containing water at room

temperature.

3. A fill tube with a stir bar and thermometer

circumference, as shown in the picture.

Information:

A: large test tube

B: sheat tube

C: sstir bar circumference

D: Thermometer

Picture 1.1 series of experimental tool

solubility as a function of temperature

4. Continuously stir the solution in tube A, if

the temperature dropped to 40 deg C, 5 mL

pipette and dilute to 50 mL.

5. Do a similar decision at 30 deg C, 20 deg C, 10

deg C, to be able to reach a temperature of 20

deg C and 10 deg C. The ice is required in the

cooling water. Pipette tip needs to be wrapped

with filter paper so that solids do not enter

the pipette, when pipetting is done.

6. Fourth solution titration with NaOH 0.25 M

solution

a. Determine the solubility of oxalic acid in

the four aforementioned temperature

b. Compute the average heat dissolving the

route (10 deg C - 20 deg C), (20 deg C –

30 deg C), and (30 deg C – 40 deg C).

c. Create logarithmic graphs solubility

versus 1/T and determine the heat of

dilution chart.

V. OBSERVATIONAL DATA

Temperature

(deg C)

V.C2H2O4 + 3

drops PP

V.NaOH 0.5 M Color of

the

solution40

10 mL14.06 mL

Pink hint30 13.8 mL20 10.2 mL10 7.73 mL

Standardization NaOH 0.5 M

V.C2H2O4 (mL) V.NaOH (mL) Color of the

solution10 mL 3.2 mL Pink hint

3.0 mL

VI. CALCULATION

1. Standardization NaOH 0.5 M

H2C2O4 (aq) + 2NaOH (aq) Na2C2O4 (aq) + 2H2O

(l)

NaOH 0.5 M

M H2C2O4 = 1.26gram

1.26gram /mol× 1000mL

50mL=0.2M

Mmol H2C2O4 = 10 mL X 0.2 M = 2 mmol

V NaOH = (3.2+3.0)mL2

=3,1mL

M NaOH = 4mmol3.1mL=1.29M

2. Solubility H2C2O4 at each temperature


(l)

Temperature 40 deg C

T = 40 deg C = 313 K

V NaOH = 14,06 mL

M NaOH = 0.2 M

V H2C2O4 = 10 mL

n NaOH = M NaOH x V NaOH

= 0.2 M x 14.06 mL

= 2.812 mmol

N H2C2O4 = ½ x n NaOH

= ½ x 2.812 mmol

= 1.406 mmol

[H2C2O4 ] = nH2C2O4VH2C2O4=1.406mmol

10mL=0.1406M

(S1)


(l)


T = 30 deg C = 303 K

V NaOH = 13.8 mL

M NaOH = 0.2 M

V H2C2O4 = 10 mL


= 0.2 M x 13.8 mL

= 2.76 mmol


= ½ x 2.76 mmol

= 1.38 mmol

[H2C2O4 ] = nH2C2O4VH2C2O4=1.38mmol10mL

=0.138M

(S2)


(l)


T = 20 deg C = 293 K

V NaOH = 10.2 mL

M NaOH = 0.2 M

V H2C2O4 = 10 mL


= 0.2 M x 10.2 mL

= 20.4 mmol


= ½ x 20.4 mmol

= 10.2 mmol

[H2C2O4 ] = nH2C2O4VH2C2O4=10.2mmol10mL

=0.102M

(S3)


(l)


T = 10 deg C = 283 K

V NaOH = 7.73 mL

M NaOH = 0.2 M

V H2C2O4 = 10 mL


= 0.2 M x 7.73 mL

= 1.546 mmol


= ½ x 1.546 mmol

= 0.773 mmol

[H2C2O4 ] = nH2C2O4VH2C2O4=0.773mmol

10mL=0.077M

(S4)

3. Average heat dissolution

T1 = 10 deg C = 283 K

T2 = 20 deg C = 293 K

S3 = 0.102 M

S4 = 0.077 M

Delta Hds ?

log S4S3

=∆Hds2.303R

.T4−T3T4.T3

log 0.0770.102

=∆Hds

2.303x8.314 Jmol .k

. 283−293K283.293K2

log0.755= ∆Hds19.15

. −1082919

∆Hds=log0.755×19.15×82919−10

= (-0.122) x (-158789.88)

= 19372.36 J/mol

= 19.372 KJ/mol

T2 = 20 deg C = 293 K

T3 = 30 deg C = 303 K

S3 = 0.102 M

S2 = 0.138 M

Delta Hds ?

log S3S2

=∆Hds2.303R

.T3−T2T3.T2

log 0.1020.138

= ∆Hds

2.303x8.314 Jmol .k

. 293−303K293.303K2

log0.739= ∆Hds

19.15 Jmol .k

. −10K88779K2

∆Hds=log0.739×19.15 J

mol.K×88779K2

−10K = (-0.13) x (-170011.785)

= 22101.53 J/mol

= 22.101 KJ/mol

T3 = 30 deg C = 303 K

T4 = 40 deg C = 313 K

S2 = 0.138 M

S1 = 0.1406 M

Delta Hds ?

log S2S1

=∆Hds2.303R

.T2−T1T2.T1

log 0.1380.1406

=∆Hds

2.303x8.314 Jmol .k

. 303−313K303.313K2

log0.982=∆Hds

19.15 Jmol .k

. −10K94839K2

∆Hds=log0.982×19.15 J

mol.K×94839K2

−10K = (-7.89 x 10-3) x (-181616.685) J/mol

= 1432.95 J/mol

= 1.432 KJ/mol

T ( deg

C)

T (K) Ms Log ms 1/T

10 283 0.077 - 1.1135 3.5 x

10-3

20 293 0.102 - 0.991 3.4 x

10-3

30 303 0.138 - 0.860 3.3 x

10-3

40 313 0.1406 - 0.852 3.2 x

10-3

4. Heat of dissolution and graph log M1 and 1/T

Stretch 10 deg C – 20 deg C

M = −∆Hdsgraph2.303R

y2−y1x2−x1

=−∆Hdsgraph2.303×8.314

−0.991−(−1.1135)

(3.4−3.5 )10rank−3= −∆Hds2.303×8.134

−∆Hds=−1225×19.15∆Hds=23.45Kj /mol



y2−y1x2−x1

=−∆Hds

2.303×8.314−0.860−(−0.991)

(3.3−3.4 )10rank−3=

−∆Hds2.303×8.314

∆Hds=1310×19.15∆Hds=25.08Kj /mol



y2−y1x2−x1

= −∆Hdsgraph2.303×8.314

−0.852−(−0.860)

(3.2−3.3 )10rank−3=

−∆Hds2.303×8.314

∆Hds=80×19.15∆Hds=1.53Kj/mol

Then ∆Hds graph =(23.45+25.08+1.53)Kj/mol

3=16.68Kj/mol

Determine ∆Hds based calculation

∆Hds (10 deg C – 20 deg C) = 19.372 Kj/mol∆Hds (20 deg C - 30 deg C ) = 22.101 Kj/mol∆Hds (30 deg C – 40 deg C) = 1.432 Kj/mol∆Hds total =(19.372+22.101+1.432 )Kj /mol

3=14.30Kj/mol

VII. GRAPH

T (k) Ms (M)

1/T (x10-3)

k-1 Log Ms313 0.595 3.195 -0.8518303 0.485 3.3003 -0.86293 0.32 3.413 -0.9914283 0.28 3.533 -1.1135

3.15 3.2 3.25 3.3 3.35 3.4 3.45 3.5 3.55 3.6

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0Hubungan Log Ms terhadap

1/T

Y-values

1/T (x10-3)

Log

Ms

VIII. ANALYSIS

In this experiment aims to prove the solubility

as a function of temperature. And this is obtained

by determine the solubility of the substance at

various temperatures and determine the differential

dissolution heat. In this experiment used oxalic

acid for dissolving oxalic acid in water at a

temperature of 20 deg C is quite large, 111 g/liter.

Equation oxalic acid solubility in water is:

(COOH)2 . 2H2O (COOH)2 + 2H2O

Based on the theory, according to van’t Hoff,

the higher the temperature, the more the dissolved

substances (hot dissolution positive: endothermic)

heat while dissolving negative: exothermic, occurs

when dissolved substances on the wane when the

temperature is higher. Based on the calculation of

the experimental solubility data obtained as

follows:

Temperature (T) Solubility10 deg C 0.077 M20 deg C 0.102 M30 deg C 0.138 M40 deg C 0.1406 M

From above the data, it can be concluded that

the higher the temperature, the higher solubility.

And is consistent with the literature.

In this experiment, a saturated solution of

oxalic acid was prepared by dissolving crystals of

oxalic acid slowly into the water that has been

heated to 60 deg C to excessive oxalic acid. As for

heating water up to 60 deg C is done because if the

heating is more than 60 deg C, oxalic acid will

break down into CO2 and H2O. so that the amount of

oxalic acid in solution will be reduced, and this

may affect the volume of NaOH during the titration.

And if oxalic acid is reduced, then the number of

moles of NaOH equivalent equal to the number of

moles of oxalic acid equivalent will be reduced as

well.

After reaching a temperature of 60 deg C and

becomes saturated or not dissolved yet, be cooling

until the temperature reaches 40 deg C, 30 deg C, 20

deg C, 10 deg C. cooling of the solution is done by

adding ice cubes. Any decrease in temperature of 10

deg C, 10 mL diluted in 100 mL volumetric flask,

then titrated with NaOH, after drops of indicator

solution pp. As for the volume of NaOH from each

titration is:

Temperature (T) V NaOH (mL)10 deg C 14.0620 deg C 13.830 deg C 10.240 deg C 7.73

From the table above can be analyzed that the

lower the temperature, the amount or volume of NaOH

required to titrate the less. This is due to the

lower temperature, the concentration of oxalic acid

solubility becomes smaller so that the volume of

NaOH used less.

The method used to determine the delta H sector

in the reaction can be done by law Van Haff by

equation:

dlnmsdT

=∆HDsRT2 , which later on became an integral

logs=−∆H

2.303RT+C

Based on the calculation of the data obtained,

the value of delta H ds obtained by calculation and

based on the graph, that is:

Stretch

temperature

Delta Hds

(calculation)

Delta Hds

(graph)10-20 deg C 19.372 Kj/mol 23.45 Kj/mol20-30 deg C 22.101 Kj/mol 25.08 Kj/mol30-40 deg C 1.432 Kj/mol 1.53 Kj/mol

Average delta Hds (calculation) = 14.30 Kj/mol

Average delta Hds (graph) = 16.68 Kj/mol

From the following data, obtained delta Hds

calculation results and experimental experience the

difference, the difference is caused by an error

when titration (NaOH volume excess) and the solution

temperature is not exactly as prescribed

(temperature lower/higher than it should be).

IX. CONCLUSION

1. Solubility is the maximum amount of dissolved

substance/can be dissolved in a solvent to form

a saturated solution.

2. Ms value at reach temperature is:

10 deg C = 0.077 M

20 deg C = 0.102 M

30 deg C = 0.138 M

40 deg C = 0.1406 M

3. The higher temperature, the greater solubility

of the substance. Solubility is propotional to

the temperature.

4. The higher temperature, the volume of NaOH

required to titrate the less.

5. The titration aims to determine the equivalent

point when added H2C2O4 pp indicator turned

purple hint.

6. Differential heating value proportional to the

temperature, the greater the differential

heating value greater solubility of the

substance.

7. Delta H ds based calculation is 14.30 Kj/mol,

while the delta H ds is based on the graph is

16.68 Kj/mol.

X. BIBLIOGRAPHY

Alberty, Robert A. 1991. Kimia Fisik. Jakarta:

Erlangga.

Atkins, PW. 1999. Kimia Fisik Jilid I, Edisi Lima. Jakarta:

Erlangga.

Daniel, F. 1992. Kimia Fisik I. Jakarta: Erlangga.

Sutardjo. 1989. Kimia Fisik. Jakarta: Bina Aksara

Vogel. 1990. Buku Teks Anorganik Kualitatif. Jakarta:

PT.Kalman Media Pustaka

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