What is the molarity of Br^- in each solution? b. 0.150 M CaBr₂

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Step 1: Understand the dissociation of CaBr₂ in water. Calcium bromide (CaBr₂) dissociates into one calcium ion (Ca²⁺) and two bromide ions (Br^-) in solution.

Step 2: Write the balanced chemical equation for the dissociation: CaBr₂ (aq) → Ca²⁺ (aq) + 2 Br^- (aq).

Step 3: Determine the relationship between the concentration of CaBr₂ and Br^-. From the balanced equation, for every 1 mole of CaBr₂ that dissociates, 2 moles of Br^- are produced.

Step 4: Calculate the molarity of Br^-. Since the initial concentration of CaBr₂ is 0.150 M, and it produces 2 moles of Br^- per mole of CaBr₂, multiply the molarity of CaBr₂ by 2 to find the molarity of Br^-.

Step 5: Conclude that the molarity of Br^- is twice the molarity of CaBr₂, which is 2 × 0.150 M.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Molarity

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is expressed in units of moles per liter (M). Understanding molarity is essential for calculating the concentration of ions in a solution, as it allows for the conversion between moles of solute and the volume of the solution.

Dissociation of Ionic Compounds

Ionic compounds, such as calcium bromide (CaBr2), dissociate into their constituent ions when dissolved in water. For CaBr2, it dissociates into one calcium ion (Ca²⁺) and two bromide ions (Br⁻). This dissociation is crucial for determining the concentration of specific ions in solution, as the number of ions produced depends on the stoichiometry of the compound.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced equations. In the context of solutions, stoichiometry helps in determining the molar ratios of ions produced from the dissociation of compounds. For example, knowing that 1 mole of CaBr2 produces 2 moles of Br⁻ allows for the calculation of the molarity of bromide ions in the solution.

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