Molar Solubility — Overview & Calculation

We've been studying equilibrium! Recently, we began examining a unique equilibrium expression. It's the solubility product constant, $K_{sp}$. The solubility product is the relationship between a solid compound and its ions. But remember, we always exclude pure solids from equilibrium expressions. So, the solubility product only uses ion concentrations.

With the solubility product, we can determine the molar solubility. The molar solubility is the maximum moles of dissolved solute per one liter of solvent. Notice that the unit for molar solubility is moles per liter. That's molarity! So what's the difference between solubility and concentration? The solubility is the greatest concentration! We cannot surpass the solubility. Beyond it, the solution reaches saturation. Do you remember when we learned about precipitates? Once saturated, any extra compound undergoes a precipitation reaction.

Calculating Molar Solubility

So, how do we calculate the molar solubility? Thankfully, we only need the solubility product constant. It's the concentration of the products at equilibrium. Typically, we would use an ICE table. But, for solubility problems, we have a shortcut. Let's make an ICE table for a generic dissociation reaction, $AB \rightleftharpoons A^{+} + B^{-}$

$$\begin{array}{|r|c|c|c|} \hline \text{}&AB&A^{+}&B^{-}\\ \hline \text{Initial}&-&0&0\\ \hline \text{Change}&-&+x&+x\\ \hline \text{Equilibrium}&-&x&x\\ \hline \end{array}$$

We see that the equilibrium concentrations of $A^{+}$ and $B^{-}$ both equal x. So, we could have just skipped to $K_{sp}=[A^{+}][B^{-}]=[x][x]=x^{2}$.

What happens if the mole ratio isn't 1:1? For example, $AB_{y} \rightleftharpoons A^{Y+} + Y B^{-}$. Let's make another ICE table.

$$\begin{array}{|r|c|c|c|} \hline \text{}&AB&A^{Y+}&B^{-}\\ \hline \text{Initial}&-&0&0\\ \hline \text{Change}&-&+x&+Y(x)\\ \hline \text{Equilibrium}&-&x&Y(x)\\ \hline \end{array}$$

For a problem like this, $K_{sp}=[A^{Y+}][B^{-}]^{Y}$. From our ICE table, $K_{sp}=[x][Yx]^{Y}=Yx^{Y+1}$. We see the concentrations are just related to the mole ratios. So, we could have skipped the ICE table. But, if it's helpful to make one, you should. Let's look at a couple of examples.

Example Problems

Problem 1:

First, let's look at silver bromide. Often, the halogen compounds are very soluble! But, silver, a transition metal, is one of the exceptions. Silver bromide's $K_{sp}$ is $5.35\times10^{-13}$. Let's calculate the molar solubility.

First, let's write our balanced chemical equation.

$AgBr \rightleftharpoons Ag^{+} + Br^{-}$

Here we see that our mole ratio is 1:1. So, our solubility product constant is just the concentrations of silver and bromide. So, we can set both concentrations equal to "x."

$K_{sp}=[Ag^{+}][Br^{-}]=[x][x]=5.35\times10^{-13}\rightarrow x=7.31\times10^{-7}$

Problem 2:

Now let's look at calcium hydroxide. Do you remember our qualitative solubility rules? They taught us that alkaline-earth hydroxides have low solubilities. So, we would expect that the molar solubility is minimal. The $K_{sp}$ of calcium hydroxide is $5.02\times10^{-6}$. Notice that the $K_{sp}$ is much larger than the previous example. So, even if the solubility is low, we expect it to be larger.

First, let's write our balanced chemical equation.

$Ca(OH)_{2} \rightleftharpoons Ca^{2+} + 2 OH^{-}$

Here we see that our mole ratio is 1:2. Our solubility product constant needs to account for the mole ratio.

$K_{sp}=[Ca^{2+}][OH^{-}]^{2}=5.02\times10^{-6}$

Now we need to consider the concentration of the products. Each mole of calcium hydroxide yields one mole of calcium and two moles of hydroxide. Let's make our calcium concentration "x." What is the hydroxide ion concentration? It's "2x." Now, we can plug this into our solubility product expression.

$5.02\times10^{-6}=[Ca^{2+}][OH^{-}]^{2}=[x][2x]^{2}\rightarrow 4x^{3}=5.02\times10^{-6}\rightarrow x=0.011M$

(Video) Ksp - Molar Solubility, Ice Tables, & Common Ion Effect

by The Organic Chemistry Tutor

In this video, The Organic Chemistry Tutor explains molar solubility. You'll see nine different example problems that relate the molar solubility with the solubility product constant. After watching this video, you'll know how to solve any molar solubility problem.