# How can you use the solubility product constant to calculate the solubility of a sparingly soluble ionic compound?

Jan 8, 2015

The solubility product constant, "K"_("sp"), represents the product of the molar concentrations of its dissolved ions raised to the power of their respective stoichiometric coefficients determined from the equilibrium reaction.

Since you're dealing with a sparingly soluble ionic compound, ions are released when the compound is placed in water. This is what will get you from ${K}_{s p}$ to the compound's solubility.

Let's say you have a generic ionic compound placed in water:

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

For this reaction, the solubility product constant is equal to

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

A useful tool to have in your arsenal is the ICE table (more here: http://en.wikipedia.org/wiki/RICE_chart)

.....${A}_{\left(s\right)} r i g h t \le f t h a r p \infty n s b {B}_{\left(a q\right)} + c {C}_{\left(a q\right)}$
I:....$-$...........0................0
C:..$-$.........(+bx)...........(+cx)
E:..$-$..........bx...............cx

The concentration of the solid is presumed to be constant. Initially, the concentrations of its ions are zero, since the solid was not yet placed in water. When equilibrium is reached, the concentrations of the ions have increased proportional to their stoichiometric coefficients, b and c. So,

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

The only unknown in this equation is x, since the balanced chemical equation will produce the values for b and c. Solving for x will get you the solubility of the original ionic compound.

A quick example:

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

For this reaction, ${K}_{s p}$ is $1.6 \cdot {10}^{- 5}$. Calculate the solubility of $P b C {l}_{2}$ in $\text{g/L}$.

The values for b and c are 1 and 2, respectively, so you get

${K}_{s p} = 1 \cdot {2}^{2} \cdot {x}^{\left(1 + 2\right)} = 4 {x}^{3} = 1.6 \cdot {10}^{- 5}$

This equation produces the value of $x = 1.59 \cdot {10}^{- 2}$ $\text{mol/L}$, which represents $P b C {l}_{2}$'s molar solubility.

To transform this into $\text{g/L}$, multiply it by $P b C {l}_{2}$'s molar mass, $277.2$ $\text{g/mol}$

$1.59 \cdot {10}^{- 2}$ $\text{mol/L} \cdot 277.2$ $\text{g/mol} = 4.41$ $\text{g/L}$

Here's a link to a similar problem:

http://socratic.org/questions/what-s-the-solubility-in-grams-per-liter-of-laf3-in-pure-water