Dissociation of Barium Hydroxide and Calculation of H3O and OH- Concentrations

In this article, we will explore the dissociation process of barium hydroxide (Ba(OH)2) and how to calculate the concentrations of hydroxide ions (OH-) and hydronium ions (H3O ) in a 0.050 M solution of Ba(OH)2.

Introduction to Barium Hydroxide Dissociation

Ba(OH)2 is a strong base that dissociates completely in water. To understand its dissociation, we start with the balanced equation for the dissociation of Ba(OH)2.

Dissociation Equation:

Ba(OH)2 → Ba2 2OH-

Calculation of OH- Concentration

Based on the balanced equation and the molar ratio between Ba(OH)2 and OH-, we can calculate the concentration of OH-. Given a 0.050 M solution of Ba(OH)2, the calculation is as follows:

Concentration of OH-:

Concentration of OH- 0.050 M × 2 0.1 M

Calculation of H3O Concentration

To find the concentration of H3O ions, we use the relationship between the concentrations of H3O and OH- ions in water. This relationship is given by the ion product constant for water, Kw:

Kw [H3O ][OH-] 1.0 x 10-14

Using this equation, we can solve for the concentration of H3O :

[H3O ] 1.0 x 10-14 / [OH-] 1.0 x 10-14 / 0.1 M 1.0 x 10-13 M

Alternative Calculation Method

Alternatively, the calculation can be done using a simpler approach:

[OH-] 2 × 0.050 M 0.100 M 10-1 because each mole of Ba(OH)2 dissociates into 2 moles of OH- ions.

Therefore:

[H3O ] 10-14 / 10-1 10-13 M

Conclusion

In summary, when a 0.050 M solution of barium hydroxide is present, the concentration of OH- ions is 0.1 M and the concentration of H3O ions is 1.0 x 10-13 M. This relationship highlights the strong nature of Ba(OH)2 as a base and demonstrates the intricate balance of ion concentrations in aqueous solutions.

Keywords

dissociation, Ba(OH)2, concentration calculation, H3O , OH-