CHEM 1112

The Solubility of Ba (NO3)2

 

Introduction

            The dissolution of barium nitrate, Ba(NO3)2, in water occurs according to the reaction

 

Ba(NO3)2(s) ® Ba2+(aq) + 2NO3-(aq)

 

(1)

 

 

 the solubility product constant is given by Ksp = (Ba2+)(NO3-)2.  If S is the molar solubility of barium nitrate in pure water, then (NO3-) = 2S.  For every one barium nitrate formula unit there are 2 nitrate ions.  The Ksp expression can be written as equation (1).

 

Ksp =  (S)(2S)2 = 4S3

(2)

 

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, in 0.5 M HNO3 solution (NO3-) = 2S’ + 0.5, where S’ is the solubility of Ba(NO3)2 in 0.5 M HNO3(aq).  The concentration of Ba2+  will be S’, and Ksp is given by equation (2).

 

Ksp = (S’)(2S’ + 0.5)2

(3)

 

In Part 1 of this experiment you will measure the solubility of barium nitrate at room temperature in pure water and in 0.5 M HNO3, and use your results to calculate Ksp.  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 enthalpy and entropy changes accompanying the dissolution of 1 mole of barium nitrate in water (reaction (1)). 

 

The solubility product constant is an equilibrium constant and is related to the Gibbs free energy change, DG, for the solution process by

 

ln Ksp =

(4)

where the units of DG are J/mol, T is in kelvins, and R is 8.3145 J/(K-mol).

Since DG = DH -TDS ( where DH and DS, are the enthalpy and entropy changes, respectively), equation (3) can be rewritten as

 

(5)

 

The units of DH and DS, are J/mol and J/(mol-K), respectively.  

Equation (4) shows that if Ksp is measured as a function of temperature, then a graph of ln Ksp versus 1/T should be straight line with a slope given by

 

Slope = -DH/R

(6)

 and an intercept given by

 

Intercept = DS/R

(7)

 

In Part 2 of the experiment, you will measure Ksp as a function of temperature, and use your results to obtain values of DH and  DS.  The values of Ksp will be calculated in terms of molal rather than molar concentration. 

 

The molal concentration, m, is given by

 

Then

Ksp = (mBa2+)(mNO3-)2 = 4(S’’)3

(8)

 

Where S’’ is the molal solubility of barium nitrate.

 

Procedure

Part 1a

 

Weigh a clean, dry 250 mL beaker to ±0.01 g.  Place about 5 g of solid Ba(NO3)2 in the beaker and weigh to ±0.01 g.  Add 50.0 mL of deionized water, measured with a graduated cylinder.  Stir for 10 min.  Measure the temperature of the solution, and then decant as much as possible of the saturated solution into a waste container.  Heat the beaker and contents on a hot plate.  Evaporate to dryness (about 5 min.) being careful to avoid spattering.  Let cool to room temperature, and then weigh the beaker and contents to ±0.01 g.  Save the dry solid for use in Part 2.

 

Part 1b

            Repeat the procedure in Part 1a, except substitute 50.0 mL of 0.5 M HNO3 for deionized water, and evaporate the contents in the hood.

 

Part 2

            Label and weigh to ±0.01 g, four dry, clean 100 mL beakers. To the solid saved from Part 1a in the 250 mL beaker, Add additional Ba(NO3)2 to bring the total mass to 20 g.  Add 100 mL of deionized water.  On a hot plate, heat the solution with stirring, until the temperature of the solution is about 70oC.  Stop heating.  As soon as the solution cools to 35oC, quickly decant about 10 mL of the solution into one of the pre-weighed 100 mL beakers, and record the temperature.  It is not necessary to know the exact volume of the solution decanted since the concentration will be expressed as molality and not molarity.  Be careful not to transfer any undissolved solid.  Continue monitoring the temperature of the solution.  When the temperature cools to 30oC, decant another 10 mL into a second pre-weighed beaker.  Repeat this when the solution cools to 25oC and then to 0oC.  The last temperature is attained by placing the solution in an ice bath.  Allow each of the solutions in the 100 mL beakers to come to room temperature, and then weigh to ±0.01 g.

 

            Heat each of the decanted solutions on a hot plate and evaporate to dryness.  To avoid splattering, reduce the heating rate as evaporation nears completion.  Allow the beakers to cool to room temperature, and then weigh each beaker and the remaining residue to ±0.01 g.

 

Calculations

Part 1a:

            The molar solubility of Ba(NO3)2 is equal to the moles of Ba(NO3)2 dissolved in the saturated solution divided by the volume of the solution, 0.050 L.

 

 

 

Once you have calculated S, use equation (2) to calculate Ksp.

 

Part 1b: 

            Calculate the molar solubility of Ba(NO3)2 using the same method as in Part 1a, and calculate Ksp using equation (3).

 

Part 2:

             A sample calculation for this part is shown in the example below.

Example: In the determination of the solubility of Ba(NO3)2 at 69oC, a sample of saturated solution was decanted into a pre-weighed beaker, and the following data were obtained.  Calculate the molal solubility and molal solubility product of Ba(NO3)2 at 69oC.

 

           

mass of beaker + sat. soln = 

111.08 g

mass of beaker + dry Ba(NO3)=

102.12 g

mass of water in sat. soln =

    8.96 g

 

 

mass of beaker + dry Ba(NO3)2 =

102.12 g

mass of empty beaker =

  99.62 g

mass of dry Ba(NO3)2 in sat. soln=

    2.50 g

 

 

The molal solubility, S’’ is

 

= 1.07 mol/kg

and Ksp is given by equation (8)

 

Ksp = 4(1.07)3 = 4.90

 

            Use your results to calculate the molal solubility and Ksp at four different temperatures.  Use Excel to calculate lnKsp and 1/T, and make a plot of lnKsp versus 1/T.  Use linear regression to obtain the slope and intercept of your graph.  From the slope and intercept, calculate DH and DS, respectively, for the solution process.

 

Conclusion

Answer the questions below in complete sentences.

 

1)     Report and compare the values of Ksp obtained in Parts 1a and 1b.  Are the two values the same?  Explain why or why not.

2)  Report the solubilities and the values of the Ksp’s obtained at the different temperatures in Part 2.  How does the solubility vary with temperature?

3)  What is the enthalpy change for the solution process?  Is this an exothermic or an endothermic processs?  Does your value, combined with your answer to question 2, agree with Le Chatelier’s Principle?  Please explain.

4)  What is the entropy change for the solution process?  Does its value agree with what you know about the relationship between entropy and disorder?  Please explain.