Posted by **Airin** on Tuesday, March 18, 2014 at 3:30am.

Determination of Ksp, ΔG°, ΔH° and ΔS° for Ca(OH)2.

*** Need help on problem #3 & #4 ***

CALCULATIONS:

1. Calculate the average solubility of calcium hydroxide, Ca(OH)2, at each temperature.

2. Calculate the Ksp for Ca(OH)2 at each temperature.

3. Calculate the ΔG ̊ for Ca(OH)2 at each temperature using the values of Ksp.

4. Determine ΔH ̊ and ΔS ̊ using algebra (2 unknowns, 2 equations).

[Note: I am assuming I am to use the values of Ksp and ΔG ̊ at the two temperatures]

-----------------

During our titrations,this is what we got for our results:

Concentration of standardized HCl (M): 0.050 M

@ Room Temperature: 23.5°C

Vol. Ca(OH)2 (mL): Trial 1: 10.0 mL, Trial 2: 10.0 mL, Trial 3: 10.0 mL

Vol. HCl (mL): Trial 1: 7.97 mL, Trial 2: 7.89 mL, Trial 3: 7.79 mL

@ High Temperature:98.2°C

Vol. Ca(OH)2 (mL): Trial 1: 10.0 mL, Trial 2: 10.0 mL, Trial 3: 10.0 mL

Vol. HCl (mL): Trial 1: 5.97 mL, Trial 2: 5.98 mL, Trial 3: 5.60 mL

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Calculations:

1) Sample: Room Temperature: Trial 1:

__Ca(OH)2 __ <---> Ca^2+ + 2OH-

__I____ some __________ o ______ o __

__C____ -x ___________ +x _____ +2x __

__E____ less ______0.019925____ 0.03985

[OH-] = ([HCl] x VHCl)/VOH

= (0.050 M x 7.97 mL)/(10.0 mL)

= 0.03985 M

Ca(OH)2 is 0.03985 M / 2

= 0.019925 M Ca^2+

Thus, after plugging everything in there, I got:

For room temperature: 0.019925M, 0.019725M, 0.019475M; (Avg. is 0.0197083M)

For high temperature, the molar solubility of Ca(OH)2 are: 0.014925M, 0.01495M, 0.014M; (Avg is 0.014625M)

2) Ksp = [Ca^2+][OH^-]^2

Room Temp: 3.16x10^-5, 3.07x10^-5, 2.95x10^-5

High Temp: 1.33x10^-5, 1.34x10^-5, 1.10x10^-5

3) ΔG ̊ = -RT ln Ksp [*Note: my professor said this was at 0°C?)

Sample: Room Temp. Trial 1:

-(8.314 J/mol*K)(273+23.50K)* ln(3.16x10^-5)

= 25, 544 J/mol x (1 kJ/1,000 J)

= 25.5 kJ/mol

So...

Room Temp: 25.5 kJ/mol, 25.6 kJ/mol, 25.71 kJ/mol

High Temp: 64.65 kJ/mol, 34.63 kJ/mol, 35.24 kJ/mol

The problem is...I am not sure if I am doing #3 correctly....because of the kelvin and Ksp part. Can someone please show and explain to me for one sample problem? and how would you exactly start #4?

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