The free energy change for dissolution of CaSO4 at 25C is 28.774kJ/mole. What is the solubility of this compound.
i got ksp
lnK = deltaG/RT
lnK= 28774/(-8.31*298)
Ksp = 2.39e-12
i don't know waht to do after this..
CaSO4 ==> Ca^+2 + SO4^-2
Ksp = (Ca^+2)(SO4^-2)
S = solubility CaSO4; then
S = (Ca^+2)
S = (SO4^-2)
Substitute and solve for S.
thanks for the help but im still not getting the right answer
im getting 1.54e-6
After calculating the value of Ksp, you can use it to find the solubility of CaSO4. The solubility of a compound represents the maximum amount of that compound that can dissolve in a solvent under certain conditions, typically expressed in moles per liter (mol/L) or grams per liter (g/L).
In the case of CaSO4, the dissolution process can be represented by the following equation:
CaSO4 (s) ⇌ Ca2+ (aq) + SO42- (aq)
The solubility product constant (Ksp) is defined as the product of the concentration of the dissolved ions raised to the power of their stoichiometric coefficients, in the equilibrium equation. For CaSO4, the Ksp expression is:
Ksp = [Ca2+] * [SO42-]
Since the compound CaSO4 dissociates into one calcium ion (Ca2+) and one sulfate ion (SO42-), the concentration of these ions will be equal to the solubility of CaSO4.
Using the Ksp value you calculated (2.39e-12), you can set up the equilibrium expression as:
2.39e-12 = [Ca2+] * [SO42-]
Since the concentration of calcium ion (Ca2+) will be equal to the concentration of sulfate ion (SO42-) due to the 1:1 stoichiometry, you can substitute [Ca2+] with [SO42-]:
2.39e-12 = [SO42-] * [SO42-]
Simplifying the equation:
2.39e-12 = [SO42-]^2
To solve for [SO42-], take the square root of both sides:
[SO42-] = sqrt(2.39e-12)
Calculating the square root:
[SO42-] = 4.89e-6 mol/L (or 4.89e-3 g/L)
Therefore, the solubility of CaSO4 is approximately 4.89e-6 mol/L or 4.89e-3 g/L. This indicates that at 25°C, under the given conditions, up to 4.89e-6 moles or 4.89e-3 grams of CaSO4 can dissolve per liter of solution.