# How do we describe the solution behaviour of PbCl_2(aq)?

Mar 21, 2016

Lead(II) chloride is very insoluble in aqueous solution. The filtrate is saturated with respect to $P {b}^{2 +}$.

#### Explanation:

A saturated solution is a solution in which the concentration of the solute is equal to THAT WHICH WOULD BE IN EQUILIBRIUM WITH UNDISSOLVED SOLUTE:

$P b C {l}_{2} \left(s\right) r i g h t \le f t h a r p \infty n s P b C {l}_{2} \left(a q\right)$, i.e.

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

The supernatant solution (the solution that lies atop undissolved solute) is said to be saturated with respect to lead chloride. Normally, a temperature is specified because a hot solution can usually hold more solute than a cold one.

AS for any equilibrium, we can express the reaction in terms of an equilibrium constant:

${K}_{s p} = \left[P {b}^{2 +}\right] {\left[C {l}^{-}\right]}^{2}$ $=$ $\text{A small number}$. The subscript $\text{sp}$ stands for solubility product. These are extensively tabulated for a range of sparingly soluble and insoluble salts; normally, laboratory temperature, $298$ $K$ is specified. So if $\left[C {l}^{-}\right]$ is artificially increased (by adding $C {l}^{-}$), the ion product, $\left[P {b}^{2 +}\right] {\left[C {l}^{-}\right]}^{2}$ will be momentarily GREATER than ${K}_{s p}$. In order to obey the ${K}_{s p}$ expression, lead chloride will precipitate in order to reduce $\left[P {b}^{2 +}\right]$ and $\left[C {l}^{-}\right]$.

I acknowledge that I have gone on a long time about nothing to the power of less. The important definition is the following:

A saturated solution is a solution in which the concentration of the solute is equal to THAT WHICH WOULD BE IN EQUILIBRIUM WITH UNDISSOLVED SOLUTE.

An examiner would be quite justified in rejecting a definition of saturated solution as a solution in which the solvent holds all the solute that it can .