Some solutes dissolve more readily than others in a solvent such as water, and chemists have defined a quantity called the solubility product (K_{sp}) to quantify this. It's the product of the concentrations of ions in solution when the solution has reached equilibrium, and no more of the solid will dissolve. Although K_{sp} is not the same thing as the solubility of the dissolving solid, it's related, and you can easily derive solubility from K_{sp}. To do this, you need to know the dissociation equation for the solid, which tells you how many ions the solid produces when it dissolves.

## How are Ksp and Solubility Related?

Ionic compounds are those that dissolve in water. They break into positive and negative ions, and the solubility of the original solid is the amount of the solid that will dissolve. It's expressed in moles/liter or molarity.

The solubility product K_{sp} , on the other hand, is a ratio of the products of the concentrations of the ions to that of the original solid when the solution reaches equilibrium. If a solid AB splits into A^{+} and B^{-} ions in solution, the equation is AB <=> A^{+} + B^{-} and the solubility product is Ksp = [A^{+}][B^{-}]/{AB]. The undissolved solid AB gets a concentration of 1, so the equation for the solubility product becomes K_{sp} = [A+][B-]

In general, the solubility product for a compound A_{x}B_{y} that dissolves according to the equation A_{x}B_{y} <=> xA^{+} + yB^{-} is K_{sp} = [A+]^{x}[B-]^{y}

So it's important to know the dissociation equation before you can calculate K_{sp}. The solubility product doesn't have any units associated with it, but when converting to solubility, you use units of molarity.

## Procedure for Converting from Ksp to Solubility

When you have the solubility product for an ionic compound. you can calculate the solubility of the compound as long as you know the dissociation equation. The general procedure is this:

## Write the equilibrium equation and that for Ksp

## Assign a Variable

## Solve for x

For the general equation A_{m}B_{n} <=> mA^{+} + nB^{-}, the expression for Ksp is

K_{sp} = [A+]^{m}[B-]^{n}

Let the amount of solute that dissolves be x. Each mole of solute dissolves into the number of component ions indicated by the subscripts in the chemical formula. This puts a coefficient in front of the x and raises x multiplied by that coefficient to the same power. The equation for K_{sp} becomes:

K_{sp} = (nx)^{n} • (mx)^{m}

The variable x tells you how many moles of solute will dissolve, which is its solubility.

## Sample Calculations

**1. Barium sulfate has a solubility product (K _{sp}) of 1.07 x 10^{-10}. What is its solubility?**

The dissociation equation for barium sulphate is BaSO_{4}(s) <=> Ba^{2+} + SO_{4}^{2-}

K_{sp} = [Ba^{2+}][SO_{4}^{2-}]

One mole of solute produces one mole of barium ions and one mole of sulphate ions. Letting the concentration of barium sulphate that dissolves be x, you get: K_{sp} = x^{2}, so x = square root (K_{sp}).

Solubility = square root (1.07 x 10-^{10}) = 1.03 x 10^{-5} M

**1. The Ksp of tin hydroxide is 5.45 x 10 ^{-27}. What is its solubility?**

The dissociation equation is: Sn(OH)_{2}(s) <=> Sn^{2+} + 2OH¯

K_{sp} is [Sn^{2+}] [OH¯]^{2}

Assigning the molar solubility of Sn(OH)_{2} the variable x, you can see that [Sn^{2+}] = x and [OH¯] = 2x. In other words, each mole of solute produces two moles of OH^{-} ions for every mole of Sn^{2+} ions. The equation for Ksp becomes:

K_{sp} = **5.45 x 10**** ^{-27} =** (x) (2x)

^{2}= 4x

^{3}

Solve for x to find the solubility to be 1.11 x 10¯^{9} M.