# Question #f5130

Jan 29, 2015

Molar solubility represents the number of moles of solute that can be dissolved per liter of solution before the solution becomes saturated.

The solubility product constant, or ${K}_{s p}$, expresses the product of the molar concentrations of ions raised to the power of their respective stoichiometric coefficients from the equilibrium reaction.

Usually, you'll get either the value of ${K}_{s p}$ and have to determine the molar solubility of the dissociated ions, or vice versa. I'll demonstrate both cases. A general equation of a salt dissolved in aqueous solution is

${A}_{a} {B}_{b \left(s\right)} r i g h t \le f t h a r p \infty n s a {A}_{\left(a q\right)} + b {B}_{\left(a q\right)}$

The expression for this reaction's ${K}_{s p}$ is

${K}_{s p} = {\left[A\right]}^{a} \cdot {\left[B\right]}^{b}$

Let's say you know ${K}_{s p}$. You can determine the molar solubility of both ions by using the fact that both concentrations increase proportional to their respective stoichiometric coefficients.

For A, the increase in the concentration of the ion will be $a \cdot x$ (there are $\text{a}$ moles of $\text{A}$ dissociated in solution); for $\text{B}$, this increase will be $b \cdot x$. Therefore,

${K}_{s p} = {\left(a \cdot x\right)}^{a} \cdot {\left(b \cdot x\right)}^{b} = {a}^{a} {b}^{b} \cdot {x}^{\left(a + b\right)}$

Since $\text{a}$ and $\text{b}$ are derived from the equilibrium equation, solving for $x$ will produce the molar solubilities of both ions.

LIkewise, if you know the concentrations of the dissociated ions, ${K}_{s p}$ can then be determined by plugging these values in the equation

${K}_{s p} = {\left[A\right]}^{a} \cdot {\left[B\right]}^{b}$

Here are some links to actual examples:

http://socratic.org/questions/find-the-max-concentration-of-mg-2-ions-permissible-in-1l-0-01m-naoh-solution-if

http://socratic.org/questions/how-can-you-use-the-solubility-product-constant-to-calculate-the-solubility-of-a