## What is K_{sp}?

K_{sp}, or the solubility product constant, is a unitless measure of how well a solid dissolves in an aqueous solution. To be precise, K_{sp} is an equilibrium constant, which tells how much a solid dissociates when the dissolution has reached equilibrium. Solubility is a specific case of acid or base dissociation, so solubility is calculated similarly to K_{a} or K_{b}.

The more soluble a salt is, the higher the K_{sp} value will be. This is because the salt will dissolve into more product ions, making the concentration of each product higher. According to the solubility constant expression, greater product = greater K_{sp}. A slightly soluble salt, as a result, will have a relatively low K_{sp} value.

## How to Calculate K_{sp}

As mentioned above, solubility is a specific case of acid or base dissociation, in which the original reactant is a solid. Since K_{eq} = [product][product]/[reactant][reactant], K_{a} = [product][product]/[acid].

In the case of acid dissociation, acid → H_{3}O^{+} + conj. base

Thus, K_{a} = [H_{3}O^{+}][conj. base]/[acid]

However, only aqueous substances are included in equilibrium expressions. Solids, liquids, and gases are not included. In the case of solubility, where the reactant is a solid, the equilibrium product should only include the aqueous products.

Thus, K_{sp} = [product][product]

## How to Calculate Molar Solubility

Using the K_{sp} equation, you can calculate solubility given the balanced dissolution equation and the K_{sp}.

In the dissolution equation: aA(s) → bB(aq) + cC(aq), where each of the lowercase letters symbolizes the coefficients of the corresponding substance, A is the solid reactant, and B and C are aqueous products.

The solubility constant expression for this reaction would be K_{sp} = [B]^{b}[C]^{c}. Since all products come from the same reactant, they will always be produced in proportion to each other. Thus, you can replace B and C with x and add the coefficients into the equation.

K_{sp} = [bx]^{b}[cx]^{c} = bc(x)^{b + c}

If the value of K_{sp} is given, the molar solubility of A can be found by solving for x.

For any questions, you will likely not need to memorize the K_{sp} or molar solubility values for any compounds – they should be given in the problem!