Question

Data

Note: Salt Chosen is PbCl2 and initial temperature given is 37.78

Solution 1

1. Make a saturated solution of your salt at the initial temperature.

a. Add 10g of the salt to a beaker.

b. Add 200 mL of water to the beaker

c. Change the temperature to your assigned temperature by right clicking on the beaker, choose Thermal Properties, check the Insulated from Surroundings box, and enter the desired temperature.

d. Right click on the beaker with the solution and choose “Remove Solid”. In reality we would have filtered off the solid

2. Record the concentration of the ions in the saturated solution and the temperature of the solution.

Solution 2

3. Take 10mL of the solution 1 and add it to another beaker or flask.

4. Add 1mL (0.5mL if your salt is Ag2SO4) of water to this solution

5. Return the solution to your assigned temperature – if no solid is present you should adjust temperature down until you determine within 0.01 oC the temperature at which it is saturated If solid is present you will have to increase temperature to make it completely saturated. You accomplish this by adjusting the temperature until you find the 0.01 degree at which you go from no solid present to solid present as shown on the left-hand window. Watch the tutorial if you are not sure how to accomplish this.

6. Record the concentration of ions in the saturated solution and the temperature of this solution.

Solution 3

7. Take 10mL of the solution 2 and add it to another beaker or flask.

8. Add 1mL (0.5mL if your salt is Ag2SO4) of water to this solution

9. Determine the temperature within 0.01oC at which the solution is saturated just like in procedure number 5.

10. Record the concentration of ions in the saturated solution and the temperature of this solution.

Solution 4

11. Take 10mL of the solution 3 and add it to another beaker or flask.

12. Add 1mL (0.5mL if your salt is Ag2SO4) of water to this solution

13. Determine the temperature within 0.01oC at which the solution is saturated.

14. Record the concentration of ions in the saturated solution and the temperature of this solution.

Solution 5

15. Take 10mL of the solution 4 and add it to another beaker or flask.

16. Add 1mL (0.5mL if your salt is Ag2SO4) of water to this solution

17. Determine the temperature within 0.01oC at which the solution is saturated.

18. Record the concentration of ions in the saturated solution and the temperature of this solution.

Experiment Results:

Solution	Temperature (°C)	Cation Concentration (M)	Anion Concentration (M)
1	37.78	0.0101477	0.0202954
2	41.53	0.00922516	0.0184503
3	45.36	0.00838651	0.0167730
4	49.29	0.00762410	0.0152482
5	53.32	0.00693100	0.0138620

Questions:

1)  For each of the five temperatures, calculate the Ksp of the salt.

2) Make an excel plot of lnKsp versus (1/T). Insert a linear trendline and have the equation displayed on the chart. Ensure that your graph has all of the hallmarks of a good graph. Include the graph here or at the end of the report.

3) a)From your plot, calculate the value of ΔHo for the dissolving reaction. Include units with your answer.

b) From your plot, calculate the value of ΔSo for the dissolving reaction. Include units with your answer.

c) Calculate the value of ΔG at 25 ^oC using the thermodynamic values you obtained from the previous two questions.

Answer 1

For salt PbCl₂ (s)

Pb2+ is the cation and Cl- is the anion.

PbCl2 Pb2+ (aq) + 2C1- (aq)

and

K sp = [P62+] x [C1-12

Q1. See the table below, calculated values using excel spreadsheet

Note that T in K

In Ksp T(OC) T (in K) (1/T) in K^-1 [Pb2+] [CI-] [CI-]^2 Ksp 37.78 310.78 0.00321771 0.010148 0.0202954 0.000412 4.17987E-06 -12.3852 41.53 314.53 0.00317935 0.009225 0.0184503 0.00034 3.14037E-06 -12.6712 45.36 318.36 0.0031411 0.008387 0.016773 0.000281 2.35941E-06 -12.9571 49.29 322.29 0.0031028 0.007624 0.0152482 0.000233 1.77266E-06 -13.243 53.32 326.32 0.00306448 0.006931 0.013862 0.000192 1.33183E-06 -13.529

Q2. excel plot of lnKsp versus (1/T) :

In Ksp vs (1/T) -12.2 y = 7465.x - 36.40 -12200305 0.0031 0.00315 0.0032 0.00325 R = 1 -12.6 -12.8 In Ksp -13 -In Ksp vs (1/T) -Linear (In Ksp vs (1/T)) -13.2 -13.4 -13.6 1/T

Equation of linear trendline :

Y = 7465 X - 36.40

For Q3.

Von't hoff equation relates change of equilibrium K vs T as:

InK ΔΗ RT + ΔS R

The plot of ln K vs (1/T) gives a straight line with

slope =

ΔΗ" R

and

intercept =

ASO R

so, from the equation of line: Y = 7465 X - 36.40

slope = 7465 K

Hence, $\Delta H^{\Theta }$ = -(7465 K) * (8.314 J/mol.K) = -62064.01 J/mol = -62.6 kJ .mol          ...Answer 3 a)

and intercept = -36.40

$\Delta S^{\Theta }$ = -36.40 * (8.314 J/mol.K) = -302.63 J/mol.K            ...Answer 3 b)

Note that 1 kJ = 1000 J and $\Delta S^{\Theta }$ = -0.3102 kJ/mol.K           ; this unit useful for Q 5.

3 c) . value of ΔG at 25 ^oC = (25 + 273) K = 298 K

Using , $\Delta G^{\Theta } =\Delta H^{\Theta } -T\Delta S^{\Theta }$

at T = 298 K and from Q3 and 4 above,

$\Delta G^{\Theta } =(-62.6 kJ/mol) -(298 K)(-0.30263kJ/mol)$

             = + 27.58 kJ/mol             ....Answer 3 c)

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