Question

What is a good analogy for a saturated solution?

Answer 1

As you know (or should know), `"saturation"` defines an equilibrium condition........which we will define shortly.........

Explanation:

A bus is licensed to carry 65 passengers, 45 seated, and 20 standing. As you know, we could probably fit more passengers on the bus than this, however, the licence is good is for 65 passengers only - this is the saturation point of relative comfort and ease of exit and entry off the bus (note that I pulled those figures out of the air; I am a chemist not a bus-driver - so hold onto your public transport tickets).

Saturation defines an equilibrium condition, i.e. there is (or would be) equilibrium between solute, and undissolved solute, viz. for a metal salt `MX`:

`MX_n(s) rightleftharpoons M^(n+) + nX^-`

And of course, we would normally specify a temperature, because a hot solution can generally hold more solute than a cold one. And we would define a `"solubility product"`, `K_"sp"` such that:

`K_"sp"=[M^(n+)][X^-]^n`

And given my analogy, `K_"sp"=65`; the bus is licensed to convey 65 punters. Of course, at peak hours, a lot of people would try to catch the bus, and the bus might hold MORE than 65 punters.....if the bus driver is not too bolshie. This represents a condition of `"supersaturation"`, i.e. (back to solutes):

`[M^(n+)][X^-]^n>K_"sp"`, i.e. there is greater concentration of solute than the equilibrium quantity, i.e. `"supersaturation"`, just as there were greater number of commuters than the given licence. Eventually punters will depart the bus so that the number of commuters is LESS than the licence, so that bus is `"unsaturated"`. Are you following me?

Anyway, I will have a think on this problem, and there might be better analogies available from other individuals.