What happens to the red blood cell in CaCl₂ solution? Here's the problem: One red blood cell is placed in a hypertonic solution of NaCl, another is placed in a solution of CaCl2 equimolar with the NaCl solution. What would you expect to happen and why? My reasoning is that the red blood cell will shrink due to water loss by osmosis when placed in hypertonic NaCl. It will shrink even further due to the larger osmosis gradient due to there being such low concentration of Ca2+ normally in the cell. However, I'm not sure if this is right so could someone please explain the correct answer?

The reason why the cell would shrink more in CaCl2 solution is because it has a higher van't Hoff factor i.e. total number of dissociated ionic species per solute molecule (it is 2 for NaCl whereas it is 3 for CaCl2).

( _Nonionic solutes do not dissociate and will therefore have a van't Hoff factor of 1_ )

Osmotic pressure (and other colligative properties) are proportional to van't Hoff factor. Therefore osmotic pressure in CaCl2 solution will be 3/2 times of that in an equimolar NaCl solution.

Note that the ionic chemical potential is not what drives osmosis; it is the differential concentration of water (or any other solvent) that drives it.