# Solubility Product Constant (Ksp)

The solubility product constant, or* K _{sp},* is the equilibrium constant of a solute dissolving in a solvent to form a solution. A solute is soluble if more than 1 g dissolves in 100 mL of water. The solubility product constant is generally used for a sparingly soluble solute that does not entirely dissolve. It represents the extent to which the solute can go into the solution. The more soluble a solute, the higher the

*K*value

_{sp}^{[1-4]}.

## How to Find Solubility Product Constant

When a soluble ionic compound is added to water, it disassociates into its constituent ions and goes into the solution. There is an equilibrium between the undissociated solid compound and the ionic solution. The general dissolution of a substance M_{x}N_{y} is given below ^{[1-4]}.

M_{x}N_{y} (s) ⇋ xM^{y+} (aq.) + yN^{x-} (aq.)

The following equation gives the solubility product constant.

K_{sp} = [M^{y+}]^{x} [N^{x-}]^{y}

To solve for *K _{sp}*, one must raise the concentrations of the products to their stoichiometric coefficients and multiply them. The solid reactant M

_{x}N

_{y}is not included in the expression. The reason is that its concentration does not change during the reaction. Hence,

*K*denotes the maximum extent that a solid can dissolve in a solution.

_{sp}Since it is the product of the ion concentrations present in a saturated solution of an ionic compound, *K _{sp}* is called the solubility product constant.

## Example

Consider the dissolution of calcium phosphate (Ca_{3}(PO_{4})_{2})

Ca_{3}(PO_{4})_{2} (s) ⇋ 3 Ca^{2+} (aq.) + 2 PO_{4}^{3-} (aq.)

The solubility product constant for the dissolution of (Ca_{3}(PO_{4})_{2} is

K_{sp} = [Ca^{2+}]^{3} [PO_{4}^{3-}]^{2}

The value of *K _{sp}* at 25 ˚C for a pH of 7.0 is 2.07 x 10

^{-33}. It means that the concentrations of Ca

^{2+}and PO

_{4}

^{3-}ions in equilibrium with Ca

_{3}(PO

_{4})

_{2}are extremely low.

The *K _{sp} *values for some common salts at 25 ˚C are listed in the table below.

Solid | Formula | K_{sp} |
---|---|---|

Silver bromide | AgBr | 5.35 x 10^{-13} |

Calcium carbonate | CaCO_{3} | 3.36 x 10^{-9} |

Lead carbonate | PbCO_{3} | 7.40 x 10^{-14} |

Silver chloride | AgCl | 1.77 x 10^{-10} |

Lead chloride | PbCl_{2} | 1.70 x 10^{-5} |

Calcium chromate | CaCrO_{4} | 7.40 x 10^{-4} |

Barium fluoride | BaF_{2} | 1.84 x 10^{-7} |

Calcium hydroxide | Ca(OH)_{2} | 5.02 x 10^{-6} |

Lead iodide | PbI_{2} | 9.80 x 10^{-9} |

Silver phosphate | Ag_{3}PO_{4} | 8.89 x 10^{-17} |

Barium sulfate | BaSO_{4} | 1.08 x 10^{-10} |

Silver sulfate | Ag_{2}SO_{4} | 1.20 x 10^{-5} |

Cadmium sulfide | CdS | 8.00 x 10^{-27} |

Lead sulfide | PbS | 8.00 x 10^{-28} |

## Problems and Solutions

**Problem 1**: Calcium fluoride (CaF_{2}) is slightly soluble in water and dissolves as follows:

CaF_{2} (s) ⇋ Ca^{2+} (aq.) + 2 F^{–} (aq.)

The concentration of Ca^{2+} in a saturated solution of CaF_{2} is 1.5 × 10^{–4 }M. What is the solubility product constant of CaF_{2}?

**Solution**

Given

[Ca^{2+}] = 1.5 × 10^{–4 }M

The number of moles of fluoride (F^{–}) is twice that of calcium. Therefore,

[F^{–}] = 2 x 1.5 × 10^{–4 }M = 3 x 10^{-4} M

The following equation gives the solubility product constant (Ksp):

K_{sp} = [Ca^{2+}] [F^{–}]^{2}

=> K_{sp} = (1.5 × 10^{–4 }) (3 x 10^{-4})^{2}

=> K_{sp} = 1.35 × 10^{-11}

We do not include the units for *K _{sp}*.

### How to Calculate Molar Solubility from K_{sp}

To understand how to calculate molar solubility from *K _{sp}*, let us take an example of silver sulfate (Ag

_{2}SO

_{4}).

**Problem 2**. Ag_{2}SO_{4} has a *K _{sp}* value of 1.20 x 10

^{-5}at 25 ˚C. The dissociation of Ag

_{2}SO

_{4}is given by

Ag_{2}SO_{4} (s) ⇋ 2 Ag^{+} (aq.) + SO_{4}^{2-} (aq.)

Calculate the molar solubilities of Ag^{+} and SO_{4}^{2-}.

**Solution.**

Given

K_{sp} = 1.20 x 10^{-5}

The dissociation equation is given by

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

Let* x* be the number of moles of SO_{4}^{2-}. Then, the number of moles of Ag^{+} is *2x*. Let us plug in the value of *K _{sp} *to get an algebraic expression.

1.20 x 10^{-5} = (2x)^{2} (x)

=> 1.20 x 10^{-5} = 4x^{3}

=> x^{3} = (1.2 x 10^{-5})/4

=> x = 0.014

Therefore,

[SO_{4}^{2-}] = 0.014 M and [Ag^{+}] = 2 x 0.014 = 0.028 M

## Factors Affecting Solubility Product Constant

### 1. Temperature

Most solutes become soluble with an increase in temperature. An example is coffee dissolving more quickly in hot water than in cold water. Temperature affects the solubility of solids and gases, but not liquids.

### 2. Pressure

Pressure affects the solubility of gases in liquids. According to Henry’s law, the solubility of gases is directly proportional to the partial pressure. The solubility increases if the partial pressure increases and vice versa.

### 3. Molecular Size

Solutes having smaller molecular sizes are more soluble than ones with larger sizes. So, those molecules can dissolve faster than larger molecules.