Gold Standard Exam 3 Question: Molar Solubility

This forum made possible through the generous support of SDN members, donors, and sponsors. Thank you.

Necr0sis713

Full Member
7+ Year Member
Joined
Aug 1, 2014
Messages
650
Reaction score
242
So there's a question on their FL 3:

Ca(OH)2 has approximately the same Ksp as CaSO4. Which of them has the greater solubility in terms of mol L-1?

The answer ended up being Ca(OH)2. I don't get why CaSO4 doesn't have a higher molar solubility though.

Wouldn't the molar solubility of CaSO4 simply be the square root of Ksp, while for Ca(OH)2 it's the cubed root of 1/4?

So wouldn't CaSO4 "have the higher number" per say, so it has a higher molar solubility?

Members don't see this ad.
 
Wouldn't the molar solubility of CaSO4 simply be the square root of Ksp, while for Ca(OH)2 it's the cubed root of 1/4?

So wouldn't CaSO4 "have the higher number" per say, so it has a higher molar solubility?

Careful with your math there. Say Ksp = 10^-9. Square rt of that is 3*10^-5, while cube rt is 1*10^-3. The Cube rt is bigger than square root.
 
  • Like
Reactions: 1 user
Top