Q19.63P - Chemistry: Molecular Nature Of Matter And Change

Question

Why does  affect the solubility of ${\mathbf{\text{BaF}}}_{\mathbf{\text{2}}}$but not of ${\mathbf{\text{BaCl}}}_{\mathbf{\text{2}}}$? 

Step-by-Step Solution

Answer

When pH is reduced, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$ ions in the solution increases, which increases the reaction between ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ and F}}^{\text{ - }}$. As per the principle of LeChatelier, there is a higher solubility of ${\text{BaF}}_{\text{2}}$.

Because hydronium ions have no influence on the concentration of ${\text{Cl}}^{\text{ - }}$, any change in pH value has no effect on ${\text{BaCl}}_{\text{2}}$’s solubility.

Step 1: Concept Introduction

The solubility of ionic compounds containing basic anions increases as the pH of the solution decreases. There is no effect on the solubility of ionic compounds that generally contain anions having basicity that is negligible when there is a change in pH.

Step 2: Effect of pH on the solubility of BaF 2

Let us construct the dissolution equation for ${\text{BaF}}_{\text{2}}$ as follows:

${\text{BaF}}_{\text{2}}{\text{ > > Ba}}^{\text{2 + }}{\text{ + 2F}}^{\text{ - }}$

In the solution, there are ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$ ions that will react with ${\text{F}}^{\text{ - }}$ ions to form HF:

${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ + F}}^{\text{ - }}{\text{ > > HF + H}}_{\text{2}}\text{O}$

 

Because HF is a weak acid, it will partially dissolve into ${\text{F}}^{\text{ - }}$ and hydronium ions, which will then react to produce HF once more. When  is reduced, the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$ ions in the solution increases, which increases the reaction between ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}{\text{ and F}}^{\text{ - }}$.

According to the Le Chatelier principle, this results in a higher solubility of ${\text{BaF}}_{\text{2}}$ (when the concentration of a product is decreased, the equilibrium is moved to the right).

Step 3: Effect of pH on the solubility of BaCl 2

We can now construct the dissolving equation for ${\text{BaCl}}_{\text{2}}$ as follows: 

 ${\text{BaCl}}_{\text{2}}{\text{ > > Ba}}^{\text{2 + }}{\text{ + 2Cl}}^{\text{ - }}$

We also have ${\text{H}}_{\text{3}}{\text{O}}^{\text{ + }}$ ions in the solution, but they have no effect on solubility since HCL is a strong acid that will entirely dissociate into ${\text{Cl}}^{\text{ - }}{\text{ and H}}_{\text{3}}{\text{O}}^{\text{ + }}$  which will not react again to form HCL: ${\text{HCl + H}}_{\text{2}}{\text{O > > Cl}}^{\text{ - }}{\text{ + H}}_{\text{3}}{\text{O}}^{\text{ + }}$.