thermodynamic functions and solubility product of barium nitrate
ABSTRACT The objective of this experiment is to study the relationship of the solubility product constant (Ksp) to the change of Gibbs free energy (ΔG) from the solution process of barium nitrate. There are two parts in this experiment where part 1 the Ksp value was calculate based on the room temperature. The value for Ksp calculated is 1.4292. Hence, in part 2 is the calculation based on the function of temperature that vary to the Ksp value. So the value of Ksp was calculated is 0.1822. INTRODUCTION Le Chatelier’s principle said when a chemical system at equilibrium is disturbed, it retains equilibrium by undergoing a net reaction that reduces the effect of disturbance (Silberberg: principles of general chemistry). Standard thermodynamic functions of reaction is know how to use thermodynamic data to find ΔH, ΔS, ΔG when a system of constant composition undergoes changes in temperature and pressure, (ira N,levine: physical chemistry 3th edition). So here, in order to know standard state we must know how to calculates the change in H,S, and G for a process that start with separated reactant, each in a specified state at temperature T, called its standard state. So for the barium nitrate the solubility product constant is defined by Ksp. When barium nitrate in pure water so the value of molar solubility = 2S, which is it’s for one barium nitrate complex there are 2 nitrate ions. So, the Ksp value can be written as: Ksp = (S) (2S) 2 = 4S3 When the salt was dissolved in the solution that have an additional source of nitrate ion example 0.5 M HNO3, so there is two source of nitrate ions come from. Hence the equation of Ksp written as: Ksp = (s) (2s + 0.5) 2 So the solubility of product constant is related to the Gibb’s free energy: ln Ksp = -ΔG/RT this equation is valid for the very dilute solution since ΔG = ΔH – TΔS. Besided, it also can calculate as ln Ksp = (-ΔH/RT) + (ΔS/R). there also a way to calculated it by using the plotted graph ln Ksp versus 1/T so the general equation is : ln Ksp = (-ΔH/RT) + C.
Barium nitrate Ba(NO3)2
0.5 M nitric acid HNO3
50 mL clear glass volumetric pipette
10 mL clear glass volumetric pipette
100 mL, 250 mL, 500 mL
PROCEDURE Part A (i) 1. A clean, dry 250 mL beaker was weighed to ± 0.01 g. 2. 5 g of solid Ba(NO3)2 was placed in the beaker and was weighed to ±0.01 g. 3. 50.0 mL off deionized water was added with volumetric pipette. 4. Solution was stirred about 10 minutes. 5. The temperature of the solution was measured, and the saturated solution was decanted as much as possible into waste water container. 6. The beaker and the contents was heat on the hot plate. 7. The solution was evaporated to dryness about 5 minutes. The heating rate was reduced as evaporation near to completed. 8. The beaker and the contents were let to cool at room temperature, and then the beaker and the contents was weighed to ± 0.01 g. Part A (ii) 1. The procedure in Part A (i) was repeated except substitute the 50.0 mL deionised water with 50.0 mL of 0.5 M HNO3 and was evaporated the contents in the fume hood. Part B 1. Eight dry, clean 100 mL beakers was label and weighed to ± 0.0 g. 2. 40 g of solid Ba(NO3)2 was placed into 500 mL beaker and was weighed to ± 0.0 g. 3. 200 mL of deionized water was added with volumetric pipettes. 4. The solution with stirring was heat on hot plate until the temperature of the solution is about 70°C.
5. As soon as the solution was cooled to 45 °C, it was quickly decanted about 10 mL of the solution into one of the pre-weighed 100 mL beakers, the temperature was recorded. 6. The temperature of the solution was monitoring continuously. 7. When the temperature was cooled to 40 °C, another 10 mL of solution was decanted into a second pre- weighed 100 mL beaker. The step was repeated when the solutions cooled to 35, 30, 25, 20, 15, and 10°C. 8. The last temperature was attained by placed the solution into ice bath. Each of the solution in 100 mL beakers was allowed to come to room temperature and was weighed to ±0.01g. 9. Each of decant solution was heat on hot plate and was evaporated to dryness. 10. The heating rate was reduced as evaporation near completion to avoid splattering. 11. The beakers were allowed to cool at room temperature and each beaker and the remaining residue was weighed to ±0.01 g.
DATA AND RESULT Results for Part A: Temperature of the
Mass of the empty beaker
Mass of beaker + dry Ba(NO3)2
Results for Part B: Temperature of
Mass of the empty
Mass of beaker +
Mass of beaker +
the solution (0C)
saturated solution (g)
dry Ba(NO3)2 (g)
Process data: Part A:
Temperature of the solution (0C)
Mass of dry Ba(NO3)2 (g)
Temperature of the solution
Mass of H2O
Mass of dry Ba(NO3)2
CALCULATION 1. Calculate the Ksp and K’sp for Part A (i) and (ii), respectively. Part A (i): Molar solubility, M = initial mass Ba(NO3)2 – mass Ba(NO3)2 (0.050 L) (molar mass Ba(NO3)2) (5.0 g) – 0.33 g
(0.050 L) (261.37 g/mol) = 0.3573 Ksp = (M) (2M)2 = 4M3 = 4(0.3573) = 1.4292 Part A (ii): Molar solubility, M’ = initial mass Ba(NO3)2 – mass Ba(NO3)2 (0.050 L) (molar mass Ba(NO3)2) (5.0 g) – 2.22 g
(0.050 L) (261.37 g/mol) = 0.2127 K’sp = (M’) (2M + 0.5)2 = (0.2127) [(2)( 0.2127) + 0.5]2 = 0.1822
2. Calculate the values of Ksp at 8 different temperatures.
Molal solubility, m =
[mass Ba(NO3)2] (1000 g/kg) [molar mass Ba(NO3)2] (mass H2O)
Ksp = 4m3
T (0C) 45
(1.51 g) (1000 g/kg) (261.37 g/mol) (10.43 g)
= 0.5539 40
(1.64 g) (1000 g/kg) (261.37 g/mol) (10.96 g)
= 0.5725 35
(1.63 g) (1000 g/kg) (261.37 g/mol) (10.80 g)
= 0.5774 30
(1.01 g) (1000 g/kg) (261.37 g/mol) (10.88 g)
= 0.3552 25
(1.28g) (1000 g/kg) (261.37 g/mol) (10.70g)
= 0.4577 20
(1.24 g) (1000 g/kg) (261.37 g/mol) (10.80 g)
= 0.4393 15
(0.87 g) (1000 g/kg) (261.37 g/mol) (10.67 g)
= 0.3120 10
(1.02 g) (1000 g/kg) (261.37 g/mol) (10.53 g)
Ksp Ksp = 4m3 = (0.5539)3 = 0.6798 Ksp = 4m3 = (0.5725)3 = 0.7506 Ksp = 4m3 = (0.5774)3 = 0.7700 Ksp = 4m3 = (0.3552)3 = 0.1793 Ksp = 4m3 = (0.4577)3 = 0.3835 Ksp = 4m3 = (0.4393)3 = 0.3391 Ksp = 4m3 = (0.3120)3 = 0.1215 Ksp = 4m3 = (0.3706)3 = 0.2036
3. Calculate and tabulate the values of In Ksp and 1/T.
3.14 x 10-3
3.19 x 10-3
3.25 x 10-3
3.30 x 10-3
3.36 x 10-3
3.41 x 10-3
3.47 x 10-3
3.53 x 10-3
4. Sketch a plot of In Ksp versus 1/T. from the slope and intercept, calculate ∆H and ∆S, respectively for the solution process.
10 y = -0.1012x - 9.5321 R² = 0.0309
-4 ln Ksp
Linear (ln Ksp) -8
From the graph; Slope
Intercept = -9.5321
∆H = - slope x R = - (-0.1012) (8.3145 J K-1 mol-1) = 0.8414 J mol-1
∆S= intercept x R = (-9.5321) ((8.3145 J K-1 mol-1) = -79.255 J mol-1 K-1
5. Calculate ∆G for the solubility of Ba(NO3)2 at 250C. ∆H and ∆S are constant with variation of temperature.
∆G = ∆H - T∆S = (0.8414 J mol-1) – (298 K) (-79.255 J mol-1 K-1) = 23 618.8 J mol-1 6. Calculate the ∆Gtheory from the standard data for the above solution process.
In Ksp = ∆H + ∆S RT =
R 0.8414 J mol-1
(8.3145 J K-1 mol-1) (298 K) = -0.2656
∆Gtheory = -In Ksp (RT) = - (-0.2656) (8.3145 J K-1 mol-1) (298 K) = 658.083 J mol-1
-79.255 J mol-1 K-1 298 K
DISCUSSION In this experiment, there are two parts which are Part A (i and ii) and Part B. Based on Calculation 1 above, the value of Ksp for Part A (i) and (ii) are different that are 1.4292 and 0.1822. Both of these values should be same since Ksp does not affected by the same temperature of the solutions. However, the values of Ksp for both solutions are not equal even their temperatures are same (28.80C). This condition occurred due to the errors either personal or instrument errors. One of the sources of personal errors is lacking of skills while handling the apparatus. Others, dry Ba(NO3)2 and beaker may not be cooled enough before they are weighed. Calculation 2 above shows the values of Ksp at 8 different temperatures for Part B. The table shows Ksp changes as temperature changes. The slope and intercept can be obtained from a plot of In Ksp versus 1/T as shown on from the graph plotted. ∆H and ∆S can be calculated by using these equations: ∆H = -(slope)(R)
∆S = (intercept)(R)
The value of ∆H is 0.8414 J mol-1. It is a positive value and it shows that this reaction is endothermic. The heat of the surrounding is absorbed by the system. However, it does not agree with Le Chatelier’s Principle. This principle states that, decreasing temperature of the system will favor the exothermic reaction whereby the system released heat to the surrounding thus produce negative value of ∆H. The entropy change for the solution process, ∆S is in negative value that is – 79.255 J mol-1 K-1. It means the reaction in this experiment is less disorder or more ordered. ∆Gexperiment that calculated in Calculation 5 is 23 618.8 J mol-1 is different with ∆Gtheory that calculated in Calculation 6 above; 658.083 J mol-1. The positive values of both ∆G show that the process that occurred in this experiment is non-spontaneous since ∆H is more than 0 and ∆S is less than zero. There might be some errors in this experiment. There are splattering occur while heating the decanted solution thus it effect the mass of the Ba(NO3)2 which might reduce the real mass of it. The beaker and its content may not cooled enough after they are heated on the hot plate. The heat that come from the beaker or content may affect the actual mass of Ba(NO3)2 too. This will give the incorrect values for Ksp then it affected the values of ∆H, ∆S and ∆G.
QUESTION 1. What is the value of Ksp for Ag2SO4 if 5.4g is soluble in 1 L of water? Molar solubility, M =
mass Ag2SO4 (1 L) (molar mass Ag2SO4)
5.4 g (1L)(311.87 g/mol)
= 0.0173 3
Ksp = 4M
= 4(0.0173)3 = 2.076 X 10-5 2. The molar solubility of Pb(IO3)2 at 26 °C is 4.0 X 10-5. Determine the Ksp of lead (II) iodate Ksp = 4M3 = 4(4.0 X 10-5)3 = 2.56X10-13
3. Balance the chemical equations and predict the spontaneity of a reaction for the following 4 cases. Explain your answer. Case 1 and case 4 CH4 + 2O2
CO2 + 2H2O
CO2 + 2H2O
CH4 + 2O2
: Spontaneous process which is it is a reversible process in case 1 there is increase in randomness or increase in disorder. Hence, the case 4 it is nonspontaneous because it less disorder. Case 2 and case 3 H2O (s)
: The process change the states form solid to liquid make particles more randomness and it is spontaneous process. H2O (g)
: it is non-spontaneous process where it become less disorder.
CONCLUSION As the conclusion the value for ksp obtained in the part I and part (ii) is not same. From the result we get its not agree with the Le Chatelier’s Principle.
REFERENCES 1. Standard Thermodynamic Functions of Reaction, (page 141- 164).Levin, Ira N. (1988). Physical Chemistry. Third edition. Singapore: McGraw – Hill.(ISBN 0-07100563-3). 2. Reaction Condition and the Equilibrium state: Le Chatelier’s Principle, (page 573580).Martin S. Silberberg. (2010). Principles Of General Chemistry. Second edition. New York: McGraw- Hill. (ISBN 978- 0- 07- 017263- 0). 3. Entropy, (page328- 350).Yunus A.Cengel, Michael A.Boles (2011). Thermodynamics an engineering approach. Seventh edition.New York: McGraw- Hill. (ISBN 978- 007131111-3). 4. Energy and Equilibrium, (page 123- 162).D.Eisenberg, D. Crothers (1979). Physical Chemistry with applications to the life Sciences. Philippines: Benjamin. (ISBN 08053-2402-x).