Ba(OH)2 is a strong electrolyte. Determine the concentration of each of the individual ions in a 0.250 M Ba(OH)2 solution. Explain briefly.
The correct answer and explanation is :
To determine the concentration of each ion in a 0.250 M Ba(OH)₂ solution, it’s essential to recognize the dissociation process of Ba(OH)₂ in water.
Step 1: Write the Dissociation Equation
Barium hydroxide (Ba(OH)₂) is a strong electrolyte, meaning it dissociates completely in water. The dissociation equation for Ba(OH)₂ is as follows:
[ \text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^- ]
This equation shows that one formula unit of Ba(OH)₂ dissociates into one Ba²⁺ ion and two OH⁻ ions.
Step 2: Concentration of Ba²⁺ Ions
The concentration of Ba(OH)₂ is given as 0.250 M. Since one Ba(OH)₂ molecule dissociates into one Ba²⁺ ion, the concentration of Ba²⁺ ions will be equal to the concentration of Ba(OH)₂. Thus, the concentration of Ba²⁺ is:
[ [\text{Ba}^{2+}] = 0.250 \, \text{M} ]
Step 3: Concentration of OH⁻ Ions
From the dissociation equation, we can see that for every 1 mole of Ba(OH)₂ that dissociates, 2 moles of OH⁻ ions are produced. Therefore, the concentration of OH⁻ ions will be twice the concentration of Ba(OH)₂:
[ [\text{OH}^-] = 2 \times [\text{Ba(OH)}_2] = 2 \times 0.250 \, \text{M} = 0.500 \, \text{M} ]
Final Answer:
- The concentration of Ba²⁺ is 0.250 M.
- The concentration of OH⁻ is 0.500 M.
Explanation:
The strong electrolyte nature of Ba(OH)₂ means it dissociates completely in water, releasing ions in a 1:2 ratio of Ba²⁺ to OH⁻. The concentration of Ba²⁺ is directly equal to the initial concentration of Ba(OH)₂, while the concentration of OH⁻ is twice that value, because each formula unit of Ba(OH)₂ produces two OH⁻ ions.