Solubility Product Constant – KSP

solution example

What is Ksp?

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 Ksp

As mentioned above, solubility is a specific case of acid or base dissociation, in which the original reactant is a solid. Given the generic formula for the equilibrium constant:

    \begin{gather*} K_{eq} = \frac{[Product][Product]}{[Reactant][Reactant]} \end{gather*}

In the case of acid dissociation,

     \begin{gather*} \text{Acid} \rightarrow H_{3}O^{+} + \text{Conj. Base} \end{gather*}


     \begin{gather*} K_{a} = \frac{[H_{3}O^{+}][\text{Conj. Base}]}{[\text{Acid}]} \end{gather*}

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.


     \begin{gather*} K_{sp} = [\text{Product}][\text{Product}] \end{gather*}

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:

     \begin{gather*} aA(\text{s}) \rightarrow bB(\text{aq}) + cC(\text{aq}) \end{gather*}

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

     \begin{gather*} K_{sp} = [B]^{b}[C]^{c} \end{gather*}

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.

     \begin{gather*} K_{sp} = [bx]^{b}[cx]^{c} = bc(x)^{b+c} \end{gather*}

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!

Solubility Product Constant Practice Problems

Problem 1

You dissolve  1.0\text{mol} of salt with a molecular formula of one cation and two cations ( XY_{2} ) into  1.0\text{L} . This salt dissolves according to the following reaction:

     \begin{gather*} XY_{2} \rightarrow X^{2+} + 2Y^{-} \end{gather*}

You measure the concentration of  X^{2+} to be  0.43\text{M} once the solution reaches equilibrium. Calculate the solubility product constant.

Problem 2

Under standard conditions in water, the salt  AgCl has the solubility product  K_{sp} = 1.6 \cdot 10^{-10} . You dissolve  1\text{mol} \: AgCl into  1.0\text{L} \: 0.5\text{M} \: HCl aqueous solution. Assume HCl has completely solubilized. Calculate the final equilibrium concentration of  Ag^{+} . (Hint: you will need to use the quadratic formula)

Solubility Product Constant Practice Problem Solutions

1: K_{sp} = 0.37

2:  [Ag^{+}] = 3.2 \cdot 10^{-10} \text{M}

Further Reading