Calculate the free energy in a saturated solution of Ca(OH)2 if the Ksp =6.5x10-6 at 25 degrees Celsius.
sorry to be a bother but I don't understand how to do this problem either.. can someone please explain this one to me as well?
I've been searching my chemistry book for hours and I can't find much on cell potential... I also tried watching youtube videos but I still can't figure it out... help would be appreciated.
dG = -RTlnKsp
Use 8.314 for R
To calculate the free energy in a saturated solution of Ca(OH)2, you can use the equation:
ΔG = -RT ln(Q)
where ΔG is the change in Gibbs free energy, R is the gas constant (8.314 J/mol·K), T is the temperature in Kelvin, and Q is the reaction quotient.
In this case, the reaction quotient (Q) can be calculated using the solubility product constant (Ksp):
Q = [Ca2+] [OH-]^2
Given that the Ksp of Ca(OH)2 is 6.5x10^-6, we know that:
Ksp = [Ca2+] [OH-]^2
Assuming the saturation is reached, we can assume that [Ca2+] and [OH-] are equal, let's call this value x.
Therefore, we can rewrite the Ksp equation as:
6.5x10^-6 = x (2x)^2
6.5x10^-6 = x^3 (since (2x)^2 = 4x^2)
Now, rearrange the equation and solve for x:
x^3 = 6.5x10^-6
x = (6.5x10^-6)^(1/3)
Calculate the value of x using a calculator, which is approximately 0.0143 M.
Substitute the values in the equation to calculate ΔG:
ΔG = -RT ln(Q)
ΔG = -8.314 J/mol·K * (273.15 + 25) K * ln(0.0143 M)
Calculate the natural logarithm of 0.0143 M using a calculator, and substitute the values to obtain the final value of ΔG in J/mol.