Barium Nitrate is a not very soluble nitrate. Calculate the Ksp (at 25 degrees) from the following ΔG's:
Ba2+ = -561kj/mole
NO3 1- = - 109kj/mole
Ba(NO3)2 = - 797 kj/mole
I disagree that Ba(NO3)2 isn't very soluble. The solubility is about 10g/100 mL H2O at 25 C. It doesn't have a Ksp in the sense of it being a slightly soluble salt.
dGrxn = (n*dGf products) - (n*dGf reactants)
Then dG = -RTlnK
I am working on the same problem. I did the first step DrBob gave and I ended up with dGrxn = -127 kj/mol. When I plugged it into the second equation I got -127000/(-298)(8.314) = 51.2599. Then if I take the e of this I get 1.8x10^22. I feel like this is way off, since the options are:
a. 3.5 x 10^-3
b. 1.05 x 10^-5
c. 7.0 x 10^-4
d. 1.4 x 10^-6
To calculate the solubility product constant (Ksp) for Barium Nitrate (Ba(NO3)2) at 25 degrees Celsius, we can use the relation between ΔG and Ksp:
ΔG = -RT ln(Ksp)
Where:
- ΔG represents the Gibbs free energy change
- R is the gas constant (8.314 J/(mol·K))
- T is the temperature in kelvin
- Ksp is the solubility product constant
First, we need to convert the given ΔG's from kilojoules (kJ) to joules (J), since the gas constant is given in joules/mol·K.
ΔG(Ba2+) = -561 kJ/mol = -561,000 J/mol
ΔG(NO3-) = -109 kJ/mol = -109,000 J/mol
ΔG(Ba(NO3)2) = -797 kJ/mol = -797,000 J/mol
Next, we need to convert the temperature from degrees Celsius to Kelvin.
T = 25 degrees Celsius + 273.15 = 298.15 K
Now we can calculate ln(Ksp) using the given ΔG values:
ln(Ksp) = (ΔG(Ba2+) + 2ΔG(NO3-)) / (-RT)
ln(Ksp) = (-(561,000 J/mol) + 2(-109,000 J/mol)) / (-8.314 J/(mol·K) * 298.15 K)
Simplifying the expression:
ln(Ksp) = (-561,000 J/mol - 218,000 J/mol) / (-8.314 J/(mol·K) * 298.15 K)
ln(Ksp) = (-779,000 J/mol) / (-2,472.86 J/(mol·K))
ln(Ksp) ≈ 314.68
Finally, we can calculate Ksp by taking the exponential of ln(Ksp):
Ksp = e^(ln(Ksp))
Ksp = e^(314.68)
Ksp ≈ 1.267 x 10^136
Therefore, the solubility product constant (Ksp) for Barium Nitrate (Ba(NO3)2) at 25 degrees Celsius is approximately 1.267 x 10^136.