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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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.

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01:13

Molarity

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.

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Ionic Compounds Naming

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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