What is the total molar concentration of ions in each of the following solutions, assuming complete dissociation? (b) A 0.355 M solution of AlCl3

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Identify the dissociation equation for AlCl_3 in water: AlCl_3 \rightarrow Al^{3+} + 3Cl^{-}.

Determine the number of ions produced per formula unit of AlCl_3: 1 Al^{3+} ion and 3 Cl^{-} ions, totaling 4 ions.

Recognize that the initial concentration of AlCl_3 is 0.355 M.

Calculate the concentration of each ion: Al^{3+} is 0.355 M and Cl^{-} is 3 \times 0.355 M.

Add the concentrations of all ions to find the total molar concentration of ions: 0.355 M (Al^{3+}) + 3 \times 0.355 M (Cl^{-}).

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Key Concepts

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

Molarity

Molarity (M) is a measure of concentration defined as the number of moles of solute per liter of solution. It is commonly used in chemistry to express the concentration of a solution. In this case, a 0.355 M solution of AlCl3 indicates that there are 0.355 moles of aluminum chloride dissolved in one liter of solution.

Dissociation of Ionic Compounds

Ionic compounds, such as 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 complete dissociation is crucial for calculating the total molar concentration of ions in the solution.

Total Ion Concentration

The total ion concentration in a solution is the sum of the concentrations of all individual ions produced from the dissociation of the solute. For a 0.355 M solution of AlCl3, the total ion concentration would be calculated by adding the concentration of Al^3+ (0.355 M) and the concentration of Cl^- ions (3 × 0.355 M), resulting in a total concentration of 1.420 M.

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