EXPERIMENT 3THERMODYNAMIC FUNCTIONS AND SOLUBILITY PRODUCT OF BARIUM NITRATE.docx

EXPERIMENT 3 THERMODYNAMIC FUNCTIONS AND SOLUBILITY PRODUCT OF BARIUM NITRATE OBJECTIVE To study the relationship of the solubility product constant (Ksp) to the change of Gibbs free energy from the solution process of barium nitrate. INTRODUCTION The dissolution of barium nitrate, Ba(NO3)2 , in water occurs according to the reaction Ba(NO3)2 (s) The solubility product constant is given by Ksp = If M is the molar solubility of barium nitrate in pure water, then [NO3 ] = 2M. for every one barium nitrate formula unit there are 2 nitrate ions. The Ksp expression can be written as : Ksp = (M)(2M)2 = 4M3 On the other hand, if the salt is dissolved in a solution which contains an additional source of nitrate ion, such as nitric acid, then the concentration of nitrate ion arises from two sources : from the dissolution of the Ba(NO3)2 and from the nitric acid. For example : Dissolution of Ba(NO3)2 in 0.5 M HNO3 solution comprises where M’ is the 2+ solubility of Ba(NO3)2 in 0.5 M HNO3(aq). [Ba ] becomes M’ and K’sp is given by K’sp = (M’)(2M’ + 0.5)2 When an ionic compound dissociates into its ions, the equilibrium constant (K) is defined as

Since the concentration of the solid is constant, it is effectively incorporated into the equilibrium constant : When the state of the system changes at constant temperature : The equation reads with the usually applied symbols. Quantity

relates to quantity K as follow.

In part A of this experiment, the solubility of Ba(NO3)2 at room temperature in pure water and in 0.5 M HNO3 will be measured and quantities Ksp will be calculated. The values of Ksp for the two solutions should be equal since Ksp depends only on temperature. Temperature dependence of Ksp. The temperature dependence of Ksp can be used to determine the changes of enthalpy ( ) and the entropy ( ) accompanying the dissolution of 1 mole of Ba(NO3)2 in pure water as stated in the reaction.

Quantity Ksp is related to

ln Ksp = -

where the unit of

Substituting ln Ksp = -

as stated in the equation before and for the solution process by : are J/mol, T in Kelvin and R is 8.3145 J

.

into the equation ; the unit of

respectively.

The last equation shows that if Ksp is measured as a function of temperature, then a semi- logarithmic plot of Ksp versus 1/T yields ( . In part B of the experiment, Ksp as a function of temperature will be studied. Quantity of and will be estimated as well. The values of Ksp will be calculated in terms of molal rather than molar concentration. The molal concentration (m) is given by consequently the Ksp expression is now written as

CHEMICALS Barium nitrate, Ba(NO3)2 0.5 M nitric acid, HNO3 APPARATUS 50mL clear glass volumetric pipette (2) 10mL clear glass volumetric pipette (8) General purpose and mercury filled thermometer 0 to 100 , gradually every 1 100mL beaker (8) 250mL beaker (2) 500mL beaker (1) Stopwatch (1) Hotplate (1) Glass rod (1) Label (10)

(1)

PROCEDURE Part A (i) 1. 250mL of dry and clean beaker was weighed to 2. About 5 g of solid barium nitrate was placed into the beaker and weighed to 3. 50 mL of deionised water was added with volumetric pipette. 4. The mixture was stirred for 10 minutes. 5. The temperature of the solution was then measured. The saturated solution was decanted as much as possible into a waste container. 6. The beaker and content was heated on a hot plate. 7. The beaker was left to evaporate to dryness. 8. The beaker was then left to cool to room temperature and the beaker and its contents was weighed to Part A (ii) 1. The procedure in A(i) was repeated, except that 50mL of deionised water was replaced with 50mL of 0.5M nitric acid and the contents was evaporated in the fume hood. Part B 1. 8 dry and clean beakers was weighed and labelled. 2. About 40g of solid barium nitrate was placed in 500mL beaker and weighed to 3. 200mL of deionised water was added with volumetric pipettes. 4. The solution was heated on a hotplate with stirring, until the temperature is about 70 , the heating was stopped. 5. As soon as the solution cools to 45 , 10 mL of the solution was quickly decant into one of the pre-weighed 100mL beakers and the temperature was recorded. 6. The temperature of the solution was being continued monitored. 7. As the temperature reach 40 , another 10mL of the solution was decanted into 2nd preweighed 100mL beaker. The step was repeated when the solution cools to 35 , 30 , 25 , 20 , 15 and 10 . 8. The last temperature was attained by placing the solution in an ice bath. Allow each of the solution to come to room temperature before weighed to 9. Each of the decanted solutions was heated on a hotplate and evaporated to dryness. 10. The beakers were left to cool to room temperature before weighed again with the residue.

RESULTS Part A i)

Mass of empty beaker Mass of beaker + dry Ba(NO3)2 Temperature of the solution

= = =

98.4784 98.6520 27

g g

ii)

Mass of empty beaker Mass of beaker + dry Ba(NO3)2 Temperature of the solution

= = =

99.4881 101.6804 27

g g

Part B

T 45 40 35 30 25 20 15 10

Mass of empty beaker (g) 50.2243 45.0375 49.0882 49.5452 48.6129 49.8936 49.2812 49.0390

Mass of beaker + saturated solution (g) 60.6288 55.4005 58.7247 60.1352 58.3172 59.8914 58.9139 57.0260

Mass of beaker + dry Ba(NO3)2 (g) 51.7518 46.5198 50.3151 50.8052 49.6547 50.9127 50.1717 49.8360

CALCULATION AND DISCUSSION PART A (i)

Ksp = (M)(2M)2 = 4M3 Ksp = 4(0.36938)3 Ksp = 0.2016 PART A (ii)

K’sp = (M’)(2M’ + 0.5)2 K’sp = (0.2160)[2(0.2160) + 0.5]2 K’sp = 0.1876

PART B ⁄

T

KSP = 4m3

m 45

(

KSP

)

= 4 (0.6575)3 = 1.1368

)

=4 (0.6386)3 = 1.04174

)

= 4 (0.55818)3 = 0.6957

= 0.6575 m 40

( = 0.6386 m

35

( = 0.55818 m

30

3 ) = 4 (0.51669) = 0.55177

( = 0.51669m

25

3 ) = 4 (0.460135) = 0.38969

( = 0.460135m

20

(

= 4 (0.4343)3 = 0.3275

)

= 0.4343m 15

(

)

= 4 (0.389)3 = 0.2362

)

= 4 (0.3469)3 = 0.16699

= 0.38943m 10

( = 0.3469m

KSP 1.14 1.04 0.70 0.55 0.39 0.33 0.24 0.17

ln KSP 0.13 0.04 -0.36 -0.59 -0.94 -1.12 -1.44 -1.79

ln Ksp = 0.5 = J mol-1 k-1

J/mol theory

ln Ksp = -

+

=-

+

= 0.4841 ln Ksp = 0.4841 = =-1199.3705 J/mol

T (K) 318 313 308 303 298 293 288 283

1/T (K-1) 3.14x 10-3 3.19 x 10-3 3.25 x 10-3 3.30 x 10-3 3.36x 10-3 3.41x 10-3 3.47 x 10-3 3.53 x 10-3

This experiment is basically to experimentally determine how temperature influence the solubility of . An experimental determination of the solubility product constant, Ksp, at two temperatures will be employed to determine ∆Hº, ∆Sº, and ∆Gº for the solubility reaction. Ksp is the solubility product constant (or solubility product) for the equilibrium expression for the dissolution or formation of a solid. In this experiment, there are 2 parts. In part A, there are 2 different experiments that compare the value of Ksp. In the part A(i) , the is added with the deionized water while in the part A(ii), the is added with the nitric acid. Based on the calculation above, its shows that the value of KSP and K’SP are slightly different that is 0.2016 and 0.18. the reasons for this condition is strongly because there is an error occurring as we proceed throughout the experiment; as it is stated in theory that the value of c is only dependent on the effect of temperature change. The temperature at the end of both results is the same but different values are obtained, which refers to errors. One of the main reasons which might influence the most is happening as we weighed the beaker and the mass of the product. The moist from our hand might have been transferred to the beaker as we hold it with our bare hand. In part B, the is added with the deionised water at different temperature. In summarized, the value of KSP is different at different temperature. Based on this data, the graph of ln KSP is directly proportional to 1/T. Based on the graph, the equation is ln Ksp = -

+

From this equation, it shows that if Ksp is measured as a function of temperature, then a semilogarithmic plot of Ksp versus 1/T yields (slope) and (intersect). From the graph, we can get the value of the

. based on the equation,

= Y- intercept.

By using the same graph, we manage to obtained the value of The value o the is positive. Its shows that this reaction is endothermic. The heat of system is absorbed the heat from the surrounding. The of this reaction is positive. Based on the theory, when the is positive, the system is random and orderly. In this experiment, the exothermic process is occurred. Process is agreed with the Le Chatelier principle. This principle states that, when decreasing the temperature shifts a reaction in a direction that produces an exothermic (heat-releasing) change. From the above calculation, it shows that the ∆ G theory is same with the ∆G experiment that is -1199. Therefore, the value has been proven that this experiment is a non-spontaneous reaction. There might be some errors in this experiment. There are splattering occur during heat the decanted solution that effect the mass of the Barium nitrate which might reduce the real mass of barium nitrate. CONCLUSION ∆G theory is same with the ∆G experiment that is -1199 J/mol. So, this value proved that this experiment is a spontaneous reaction.

REFERENCES 1. http://depts.washington.edu/chemcrs/bulkdisk/chem152A_sum07/notes_Lecture_13_The rmoLabII.pdf 2. http://www.lahc.edu/classes/chemistry/arias/Lab6Solsp11.pdf 3. http://chemconnections.org/general/chem121/Excel%20Sol-I-07.pdf 4. http://wiki.answers.com/Q/What_is_the_different_between_solubility_and_solubility_pr oduct 5. http://wwwchem.csustan.edu/chem1112/1112bar.pdf 6. http://science.csustan.edu/perona/1112/barium.htm 7. http://webserver.dmt.upm.es/~isidoro/bk3/c07sol/Solutions.pdf 8. http://www.newworldencyclopedia.org/entry/Solubility