To prepare 1.0 M of OH^- ions using a Ba(OH)_2 solution, what should be the molarity of the Ba(OH)_2 solution?

• A. 0.25 M
• B. 0.50 M
• C. 2.0 M
• D. 1.0 M

Answer: (B)

Verified step by step guidance

1. Identify the dissociation of barium hydroxide, Ba(OH)\_2, in water. It dissociates according to the equation: \(\text{Ba(OH)}_2 \rightarrow \text{Ba}^{2+} + 2\text{OH}^-\).
2. Recognize that each mole of Ba(OH)\_2 produces 2 moles of OH\(^-\) ions because there are two hydroxide ions per formula unit.
3. Set up the relationship between the molarity of Ba(OH)\_2 (let's call it \(M_{Ba(OH)_2}\)) and the molarity of OH\(^-\) ions: \([\text{OH}^-] = 2 \times M_{Ba(OH)_2}\).
4. Since the target concentration of OH\(^-\) ions is 1.0 M, write the equation: \$1.0 = 2 \times M_{Ba(OH)_2}$.
5. Solve for \(M_{Ba(OH)_2}\) by dividing both sides by 2: \(M_{Ba(OH)_2} = \frac{1.0}{2}\). This gives the molarity of Ba(OH)\_2 needed to prepare a 1.0 M OH\(^-\) solution.