what is the molarity of Cl^- in each solution? c. 0.100 M AlCl₃

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Step 1: Understand the dissociation of AlCl₃ in water. AlCl₃ dissociates into one Al³⁺ ion and three Cl^- ions. The balanced chemical equation is: AlCl₃ → Al³⁺ + 3Cl^-.

Step 2: Identify the initial concentration of AlCl₃. The problem states that the concentration of AlCl₃ is 0.100 M.

Step 3: Determine the stoichiometry of the dissociation. From the balanced equation, for every 1 mole of AlCl₃ that dissociates, 3 moles of Cl^- are produced.

Step 4: Calculate the concentration of Cl^- ions. Multiply the initial concentration of AlCl₃ by the stoichiometric coefficient of Cl^- from the balanced equation: 0.100 M AlCl₃ × 3 = concentration of Cl^-.

Step 5: Conclude with the molarity of Cl^- ions. The molarity of Cl^- in the solution is three times the molarity of AlCl₃ due to the 1:3 ratio in the dissociation equation.

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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 aluminum chloride (AlCl3), dissociate into their constituent ions when dissolved in water. For AlCl3, it dissociates into one aluminum ion (Al^3+) and three chloride ions (Cl^-). This dissociation is crucial for determining the concentration of specific ions in solution, as the total number of ions produced affects the overall molarity.

Stoichiometry of Dissociation

Stoichiometry involves the quantitative relationship between reactants and products in a chemical reaction. In the case of AlCl3, the stoichiometry indicates that one mole of AlCl3 produces three moles of Cl^-. Therefore, to find the molarity of Cl^- in a 0.100 M AlCl3 solution, one must multiply the molarity of AlCl3 by the stoichiometric coefficient of Cl^- from the dissociation equation.

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