# A saturated solution of lead(ll) nitrate was made at 50°C using 100 g of water, and this was then allowed to cool to 30 °C. What would happen as the solution cooled?

Nov 20, 2016

Possibly nothing.

#### Explanation:

You have described the formation of a saturated solution of $P b {\left(N {O}_{3}\right)}_{2}$. If the solution is saturated, then it contains the same amount of solute that would be in equilibrium with undissolved solute. This is an important definition that is poorly understood even at undergraduate level.

So, if the solution is saturated, the following equilibrium pertains:

$P b {\left(N {O}_{3}\right)}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s P {b}^{2 +} + 2 N {O}_{3}^{-}$

A temperature is usually specified, because a hot solution can generally hold more solute than a cold one. In your example, because we have a saturated solution, we might expect solid to precipitate out, as the temperature cools to $30$ ""^@C.

There is of course a catch. Saturated solutions are typically encountered, but supersaturated solutions are also possible. And a supersaturated solution is a solution that holds an amount of solute GREATER than that which would be in equilibrium with undissolved solute.

Conceivably, with your example, in a clean beaker, and a filtered solution, a supersaturated solution might be formed at lower temperature. Supersaturated solutions are, however, metastable phenomena. Once you introduce a seed crystal, or scracth the side of the vessel holding the solution, bulk crystallization occurs, and all the bulk solute might come crashing out. After this precipitation occurs, the solution has gone back to a saturated condition, because it demsonstrably holds an equilibrium quantity of solute.