*K*_{sp} – The Solubility Product Constant

*K*

_{sp}– The Solubility Product Constant

Remember, earlier when discussing precipitation reactions, we used tables with rules for solubility to classify compounds as water-soluble or insoluble. For example, **BaCl _{2}**, according to the rules, is

**soluble**, while

**BaSO**in water. However, these definitions are

_{4}is insoluble**not**to understand as

**black and white**. All the compounds are soluble in water or any other solvent, and the question is to what extent they are. To quantify how much a given compound dissolves in water, we use the equilibrium constant for its dissociation reaction.

For example, BaSO_{4} dissociates into Ba^{2+} and SO_{4}^{2-} ions, and because the salt is solid, this equation represents its dissolution:

BaSO_{4}(*s*) ⇆ Ba^{2+}(*aq*) + SO_{4}^{2-}(*aq*)

The equilibrium constant representing the dissolution of an ionic compound is called the **solubility product constant ( K_{sp})** where

*sp*stands for solubility product. For BaSO

_{4}, the expression of the solubility product would be:

*K*_{sp} = [Ba^{2+}][ SO_{4}^{2-}]

Remember, for a general reaction, the equilibrium constant **includes the stoichiometric coefficients** and is given by the following equation:

Therefore, the *K*_{sp} for CaF_{2} and Ag_{3}(PO_{4}) would be:

CaF_{2}(*s*) ⇆ Ca^{2+}(*aq*) + 2F^{–}(*aq*)

*K*_{sp} = [Ca^{2+}][F^{–}]^{2}

_{ }

Ag_{3}(PO_{4}) (*s*) ⇆ 3Ag^{+}(*aq*) + PO_{4}^{3-}(*aq*)

*K*_{sp} = [Ag^{+}]^{3}[PO_{4}^{3-}]

Notice that in all *K*_{sp} expressions, the salt was missing, and the reason for this is that, remember, **solids do not appear **in the equilibrium constant expressions for **heterogeneous equilibrium**.

The magnitude of *K*_{sp} is a measure of the solubility of a compound – the larger the value, the more soluble the compound is in water. However, we cannot compare the *K*_{sp} values of any two compounds to tell which is more soluble because of the concentrations raised to the power of coefficients. Therefore, use the *K*_{sp} values to compare the solubilities of compounds with the same cation-anion ratio such as AgBr and CdS, or CaF_{2} and PbCl_{2}.If the *K*_{sp} values vary in orders of magnitude, then, of course, we can pick the more soluble compounds.

The values of *K _{sp} *at 25 °C for many ionic solids can be found in an Appendix of a general chemistry textbook.

**Calculating ***K*_{sp} from Concentrations

*K*

_{sp}from Concentrations

One thing to keep in mind in calculations regarding *K*_{sp} and solubility is that they apply to **saturated solutions** because if we add for example, 1 g of a salt to gallon of water, the resulting concentrations may change if we add some more of the salt. The final and correct values of the concentrations are only achieved when the solution is saturated.

**Example**,

Calculate *K*_{sp} for silver chromate if the concentration of silver ions in its saturated solution is 1.3 x 10^{-4} *M.*

First, write the dissociation equation of silver chromate:

Ag_{2}(CrO_{4}) (*s*) ⇆ 2Ag^{+}(*aq*) + CrO_{4}^{2-}(*aq*)

Next, write the expression for the solubility product:

*K*_{sp} = [Ag^{+}]^{2}[ CrO_{4}^{2-}]

Now, the concentration of chromate ions is going to be twice smaller because, for every 2 moles of Ag^{+}, there is one mole of CrO_{4}^{2- }formed.

We can calculate this by the mole ratio method too:

\[M\,{\rm{(Cr}}{{\rm{O}}_{\rm{4}}}^{{\rm{2 – }}}{\rm{)}}\;{\rm{ = }}\,{\rm{1}}{\rm{.3}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 4}}}}\,\frac{{\cancel{{{\rm{mol}}\,{\rm{A}}{{\rm{g}}^{\rm{ + }}}}}}}{{\rm{L}}}\,{\rm{ \times }}\,\frac{{{\rm{1}}\,{\rm{mol}}\,{\rm{Cr}}{{\rm{O}}_{\rm{4}}}^{{\rm{2 – }}}\,}}{{{\rm{2}}\,\cancel{{{\rm{mol}}\,{\rm{A}}{{\rm{g}}^{\rm{ + }}}}}}}\; = \,6.5\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 5}}}}\,{\rm{mol/L}}\]

Therefore, the *K*_{sp} is:

*K*_{sp} = [Ag^{+}]^{2}[ CrO_{4}^{2-}] = (1.3 x 10^{-4})^{2 }x 6.5 x 10^{-5 }=^{ }1.1 x 10^{-12}

**Molar Solubility**

The terms **molar solubility **and** solubility product** may be confusing as they are similar but **not the same**. The solubility product is the *K*_{sp} which is the product of the ions raised to their coefficients.

For example, when we say the solubility product (*K*_{sp}) of CaF_{2} is 3.9 x 10^{-11}, it means that;

*K*_{sp} = [Ca^{2+}][ F^{–}]^{2 }= 3.9 x 10^{-11}

The **molar solubility **is the solubility in units of moles per liter (mol/L). Sometimes, the solubility is also given in grams per liter. Now, the molar solubility of CaF_{2 }can be calculated from the *K*_{sp} value by setting up an equation with an unknown. For this, we need the dissociation equation first:

CaF_{2}(*s*) ⇆ Ca^{2+}(*aq*) + 2F^{–}(*aq*)

If we assume that x mol/L of CaF_{2 }is dissolved in a saturated solution, the concentration of the Ca^{2+ }and F^{– }ions would be x mol/L and 2x mol/L respectively. This is based on their mole ratio of the chemical equation.

You can also set up an ICE table to determine these values:

x x 2x

CaF_{2}(*s*) ⇆ Ca^{2+}(*aq*) + 2F^{–}(*aq*)

[Ca^{2+}] |
[F^{–}] |
||

Initial | 0 | 0 | |

Change | +x | +2x | |

Equil | x | 2x |

So, the expression for *K*_{sp} can be written as:

*K*_{sp} = [Ca^{2+}][ F^{–}]^{2 }= (x)(2x)^{2} = 3.9 x 10^{-11}

Therefore,

(x)(2x)^{2} = 3.9 x 10^{-11}

4x^{3} = 3.9 x 10^{-11}

^{ }x = 2.1 x 10^{-4}

^{ }

Now, 2.1 x 10^{-4}^{ }mol/L is the concentration of dissolved CaF_{2 }because it is in 1:1 ratio with Ca^{2+}, and this is the **molar solubility of CaF _{2. }**So, in any (almost) saturated solution of CaF

_{2 }at 25

^{o}C, regardless of its volume the concentration of CaF

_{2}is 2.1 x 10

^{-4}

^{ }mol/L. There are factors such as the temperature, pH, and presence of other ions in the solution that affect the solubility, and we will discuss them in the next few articles.

**Calculating ***K*_{sp} from Solubility

*K*

_{sp}from Solubility

We can also calculate the *K*_{sp} from the molar solubility of the salt.

**For example, **

Calculate the *K*_{sp} value for bismuth sulfide (Bi_{2}S_{3}) if its solubility is 1.0 x 10^{-15} mol/L at 25 ^{o}C.

First, write the dissociation equation of Bi_{2}S_{3}:

Bi_{2}S_{3}(*s*) ⇆ 2Bi^{3+}(*aq*) + 3S^{2-}(*aq*)

The solubility product constant is:

*K*_{sp }= [Bi^{3+}]^{2}[S^{2-}]^{3}

The solubility value tells us how much of the salt dissociates to ions, so this is the concentration of the salt in the equation. Therefore, we can calculate the concentration of the ions based on their molar ratio with the salt.

\[M\,{\rm{(B}}{{\rm{i}}^{{\rm{3 + }}}}{\rm{)}}\;{\rm{ = }}\,{\rm{1}}{\rm{.0}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 15}}}}\,\frac{{\cancel{{{\rm{mol}}\,{\rm{B}}{{\rm{i}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}}}}}{{\rm{L}}}\,{\rm{ \times }}\,\frac{{{\rm{2}}\,{\rm{mol}}\,{\rm{B}}{{\rm{i}}^{{\rm{3 + }}}}\,}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{B}}{{\rm{i}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}}}}}\;{\rm{ = }}\,{\rm{2}}{\rm{.0}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 15}}}}\,{\rm{mol/L}}\]

\[M\,{\rm{(}}{{\rm{S}}^{{\rm{2 – }}}}{\rm{)}}\;{\rm{ = }}\,{\rm{1}}{\rm{.0}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 15}}}}\,\frac{{\cancel{{{\rm{mol}}\,{\rm{B}}{{\rm{i}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}}}}}{{\rm{L}}}\,{\rm{ \times }}\,\frac{{{\rm{3}}\,{\rm{mol}}\,{{\rm{S}}^{{\rm{2 – }}}}\,}}{{{\rm{1}}\,\cancel{{{\rm{mol}}\,{\rm{B}}{{\rm{i}}_{\rm{2}}}{{\rm{S}}_{\rm{3}}}}}}}\;{\rm{ = }}\,{\rm{3}}{\rm{.0}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{{\rm{ – 15}}}}\,{\rm{mol/L}}\]

Therefore,

*K*_{sp }= [Bi^{3+}]^{2}[S^{2-}]^{3 }= (2.0 x 10^{-15})^{2} (3.0 x 10^{-15})^{3} = 1.1 x 10^{-73}

**Check Also**

- Buffer Solutions
- The Henderson–Hasselbalch Equation
- The pH of a Buffer Solution
- Preparing a Buffer with a Specific pH
- The Common Ion Effect
- The pH and p
*K*_{a}Relationship - Strong Acid–Strong Base Titrations
- Titration of a Weak Acid by a Strong Base
- Titration of a Weak Base by a Strong Acid
- Titration of Polyprotic Acids
**Buffer Solutions Practice Problems**- The Effect of a Common Ion on Solubility
- The Effect of pH on Solubility
- Will a Precipitate Form?
*K*_{sp}and*Q* *K*_{sp}and Molar Solubility Practice Problems