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]
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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?
Well, let's break it down and have some fun with it!
First, in problem #3, you are calculating the standard Gibbs free energy change (ΔG°) for the dissolution of Ca(OH)2 using the equation ΔG° = -RTln(Ksp). You are correct that the temperature should be in Kelvin. So, for one sample problem, let's take the first trial at room temperature.
Using the given Ksp value of 3.16x10^-5 and the ideal gas constant R = 8.314 J/mol·K, and the temperature as 23.5°C = 23.5 + 273.15 K:
ΔG° = -(8.314 J/mol·K)(23.5 + 273.15 K) ln(3.16x10^-5)
Now, let's calculate the result by plugging in the values:
ΔG° = - (8.314 J/mol·K)(296.65 K) ln(3.16x10^-5)
Well, I'm just a clown bot, so I don't have access to a calculator, but I'm sure you can handle the numbers! Just remember to convert the answer from J/mol to kJ/mol.
As for problem #4, you are trying to determine ΔH° and ΔS° using the Ksp and ΔG° values at the two temperatures. ΔG° = ΔH° - TΔS°, so we have two equations for two unknowns. You can solve these equations by substitution or elimination method.
Let's take the first trial at each temperature:
Room Temperature:
ΔG° = ΔH° - TΔS° = 25.5 kJ/mol
Ksp = 3.16x10^-5
High Temperature:
ΔG° = ΔH° - TΔS° = 64.65 kJ/mol
Ksp = 1.33x10^-5
Now, you can solve these equations simultaneously to find the values of ΔH° and ΔS°. Remember, since ΔH° and ΔS° have the same temperature dependency, you can subtract the equations to eliminate ΔH° and solve for ΔS°.
I hope I was able to bring some entertainment to your calculations! Good luck with the rest of your problem!
To determine the correct calculations for #3 and #4, let's focus on one sample problem at room temperature:
Sample: Room Temperature, Trial 1
Ca(OH)2 ⇌ Ca^2+ + 2OH^-
Initial (M): I 0 0
Change (M): C -x +x
Equilibrium (M): E 0 x 2x
First, let's calculate the concentration of OH- using the volume of HCl used:
[OH-] = ([HCl] x VHCl) / VOH
[OH-] = (0.050 M x 7.97 mL) / 10.0 mL
[OH-] = 0.03985 M
Since the concentration of OH- is twice the concentration of Ca^2+, we have:
[Ca^2+] = 0.03985 M / 2
[Ca^2+] = 0.019925 M
For this sample at room temperature, the molar solubility of Ca(OH)2 is approximately 0.019925 M.
Now, let's move to question #3 and calculate ΔG° using the given values:
ΔG° = -RT ln Ksp
Here, R is the ideal gas constant (8.314 J/mol*K), T is the temperature in Kelvin, and Ksp is the solubility product constant.
Given that the temperature was taken at 23.5°C, we first need to convert it to Kelvin:
Temperature in Kelvin = 273.15 + 23.5 = 296.65 K
Now let's take the natural logarithm of the Ksp value for this sample:
ln(3.16x10^-5) ≈ -10.0705
Finally, we substitute the values into the equation to calculate ΔG°:
ΔG° = -(8.314 J/mol*K) x (296.65 K) x (-10.0705)
ΔG° ≈ 25.5 kJ/mol
So, for this sample at room temperature, the ΔG° for Ca(OH)2 is approximately 25.5 kJ/mol.
Moving on to question #4, we are given two equations with two unknowns (ΔH° and ΔS°). Since we have two sets of temperature and ΔG° values, we can solve for ΔH° and ΔS° using these equations:
ΔG° = ΔH° - TΔS°
For Sample 1 at room temperature:
ΔG°1 = ΔH° - (296.65 K)ΔS°
For Sample 2 at high temperature:
ΔG°2 = ΔH° - (371.35 K)ΔS°
Now we have a system of equations:
ΔG°1 = ΔH° - (296.65 K)ΔS° (Equation 1)
ΔG°2 = ΔH° - (371.35 K)ΔS° (Equation 2)
Using algebra, we can solve these equations simultaneously to find ΔH° and ΔS°. Subtracting Equation 2 from Equation 1 eliminates ΔH°, allowing us to solve for ΔS°:
ΔG°1 - ΔG°2 = [(296.65 K) - (371.35 K)]ΔS°
Substituting the given values:
25.5 kJ/mol - 64.65 kJ/mol = [-74.7 K]ΔS°
Simplifying the equation:
-39.15 kJ/mol = -74.7 KΔS°
Dividing both sides by -74.7 K:
ΔS° ≈ 0.523 kJ/mol K
We have now calculated the value for ΔS°.
To calculate ΔH°, we can use one of the original equations. Let's use Equation 1 with the known value of ΔS°:
25.5 kJ/mol = ΔH° - (296.65 K)(0.523 kJ/mol K)
25.5 kJ/mol = ΔH° - 155.2898 kJ/mol
ΔH° ≈ 25.5 kJ/mol + 155.2898 kJ/mol
ΔH° ≈ 180.79 kJ/mol
So, for this sample at room temperature, the values are:
ΔH° ≈ 180.79 kJ/mol
ΔS° ≈ 0.523 kJ/mol K
You can repeat the same steps for the other sample at high temperature to find the respective ΔH° and ΔS° values.
I hope this clarifies the calculations for #3 and #4. Let me know if you need any further assistance!
To determine the values of Ksp, ΔG°, ΔH°, and ΔS° for Ca(OH)2, we can follow these steps:
Step 1: Calculate the average solubility of Ca(OH)2 at each temperature.
For the room temperature trial 1, the concentration of Ca(OH)2 is 0.019925 M.
For the high temperature trial 1, the concentration of Ca(OH)2 is 0.014925 M.
Step 2: Calculate the Ksp for Ca(OH)2 at each temperature.
Ksp is the solubility product constant and is equal to the product of the concentrations of the ions involved in the dissociation of the compound.
For the room temperature trial 1:
Ksp = [Ca^2+][OH^-]^2 = (0.019925 M)(0.03985 M)^2 = 3.16x10^-5
Repeat this calculation for the other trials and average the values obtained.
For the high temperature trial 1:
Ksp = [Ca^2+][OH^-]^2 = (0.014925 M)(0.02985 M)^2 = 1.33x10^-5
Again, repeat this calculation for the other trials and average the values obtained.
Step 3: Calculate the ΔG° for Ca(OH)2 at each temperature using the values of Ksp.
ΔG° is the standard Gibbs free energy change and is related to the equilibrium constant through the equation:
ΔG° = -RT ln(Ksp)
For the room temperature trial 1:
ΔG° = -(8.314 J/mol*K)(298.65 K) ln(3.16x10^-5) = 25544 J/mol = 25.5 kJ/mol
Repeat this calculation for the other trials and average the values obtained.
For the high temperature trial 1:
ΔG° = -(8.314 J/mol*K)(371.35 K) ln(1.33x10^-5) = 64650 J/mol = 64.65 kJ/mol
Again, repeat this calculation for the other trials and average the values obtained.
Step 4: Determine ΔH° and ΔS° using algebra (2 unknowns, 2 equations).
To determine ΔH° (the standard enthalpy change) and ΔS° (the standard entropy change), you need to use two equations.
The first equation relates ΔG° and ΔH° at a given temperature:
ΔG° = ΔH° - TΔS°
The second equation relates Ksp, ΔG°, and the temperature:
ΔG° = -RT ln(Ksp)
You can solve these two equations simultaneously to find ΔH° and ΔS°.
Substituting the second equation into the first equation, we get:
-RT ln(Ksp) = ΔH° - TΔS°
Rearranging the equation gives us:
ΔH° = -RT ln(Ksp) + TΔS°
Now we have two equations:
ΔG° = -RT ln(Ksp)
ΔH° = -RT ln(Ksp) + TΔS°
You can solve these simultaneous equations to find ΔH° and ΔS° by substituting the values of ΔG° and Ksp calculated in steps 2 and 3.
Repeat this calculation for both room temperature and high temperature trials to find the values of ΔH° and ΔS°.
Hope this helps!