In the previous post, we talked about the solubility and solubility product constant (*K*_{sp}) of ionic compounds with low solubility.

For example, the dissolution equation and the *K*_{sp} for CaF_{2} are:

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

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

By assigning x mol/L as the concentration of CaF_{2 }dissolved in a saturated solution, we were able to determine the molar solubility of CaF_{2 }from the *K*_{sp}.

Setting up an ICE table helps determine the concentrations correctly.

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) = 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}

^{ }

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. }**

**The Effect of a Common Ion on Solubility**

Let’s now assume that the solution of CaF_{2} contains Ca(NO_{3})_{2 }with a concentration of 0.20 *M. *What is the molar solubility of CaF_{2} in this solution?

Before doing the calculations, we can predict the solubility will go down because Ca(NO_{3})_{2 }is a strong electrolyte and completely dissociates into Ca^{2+} and NO_{3}^{–} ions. Importantly, the Ca^{2+ }is also formed when CaF_{2} dissolves in water, so it is a common ion, and as we discussed earlier it pushes the equilibrium to the opposite direction according to the Le Châtelier’s principle.

And now, let’s calculate the molar solubility of CaF_{2}. It is going to be the same procedure, except the initial concentration of Ca^{2+} ions is not zero since the dissociation of Ca(NO_{3})_{2 }produces an equivalent amount of the cation.

0.20 0.20 0.40

Ca(NO_{3})_{2 }(*aq*) ⇆ Ca^{2+}(*aq*) + 2NO_{3}^{–}(*aq*)

x x 2x

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

[Ca^{2+}] |
[F^{–}] |
||

Initial | 0.20 | 0 | |

Change | +x | +2x | |

Equil | 0.20 + x | 2x |

So, the expression for *K*_{sp} is:

*K*_{sp} = [Ca^{2+}][ F^{–}]^{2 }= (0.20 + x)(2x)^{2} = 3.9 x 10^{-11}

Now, because the ionization of CaF_{2} is negligible compared to the one of Ca(NO_{3})_{2, }we assume that 0.20 + x ≈ 0.20, and the simplified equation will be:

(0.20)(2x)^{2} = 3.9 x 10^{-11}

0.80x^{2} = 3.9 x 10^{-11}

Therefore,

x = 7.0 x 10^{-6}

^{ }^{ }

*The x* is very small compared to 0.20, and therefore, the approximation was valid, and 7.0 x 10^{-6 }*M* is the molar solubility of CaF_{2 }in the presence of 0.20 *M* Ca(NO_{3})_{2. }As expected it is lower than the solubility of CaF_{2 }in pure water (2.1 x 10^{-4}).

Let’s do **another example,** where the solution contains 0.15 *M* NaF.

In this case, the common ion is F- and therefore, its initial concentration is going to be equal to 0.15 *M* since NaF is a strong electrolyte and completely dissociates in aqueous solutions.

0.15 0.15 0.15

NaF(*aq*) ⇆ Na^{+}(*aq*) + F^{–}(*aq*)

x x 2x

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

[Ca^{2+}] |
[F^{–}] |
||

Initial | 0 | 0.15 | |

Change | +x | +2x | |

Equil | x | 0.15 + 2x |

So, the expression for *K*_{sp} is:

*K*_{sp} = [Ca^{2+}][ F^{–}]^{2 }= (x)(0.15 + 2x)^{2} = 3.9 x 10^{-11}

Now, because the ionization of CaF_{2} is negligible compared to the one of NaF_{, }we assume that 0.15 + x ≈ 0.15, and the simplified equation will be:

(x)(0.15)^{2} = 3.9 x 10^{-11}

0.0225x = 3.9 x 10^{-11}

Therefore,

x = 1.7 x 10^{-9}

*The x* is very small compared to 0.15, and therefore, the approximation was valid, and 1.7 x 10^{-9 }*M* is the molar solubility of CaF_{2 }in the presence of 0.15 *M* NaF_{. }As expected it is lower than the solubility of CaF_{2 }in pure water (2.1 x 10^{-4}).

So, to summarize, remember that in general, the solubility of a slightly soluble ionic compound decreases with the presence of a common ion in the solution.

**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***K*_{sp}and Molar Solubility- The Effect of pH on Solubility
- Will a Precipitate Form?
*K*_{sp}and*Q* *K*_{sp}and Molar Solubility Practice Problems