Question

What is the total molar concentration of ions in each of the following solutions, assuming complete dissociation? (a) A \(0.750 \mathrm{M}\) solution of \(\mathrm{K}_{2} \mathrm{CO}_{3}\) (b) A \(0.355 \mathrm{M}\) solution of \(\mathrm{AlCl}_{3}\)

Short answer

(a) 2.250 M; (b) 1.420 M.

Step-by-step solution

Write the Dissociation Equation for Each Compound

First, write out the dissociation reaction for each compound when in solution. (a) \(\text{K}_2\text{CO}_3\) dissociates into 2 \(\text{K}^+\) ions and 1 \(\text{CO}_3^{2-}\) ion: \\[\text{K}_2\text{CO}_3 \rightarrow 2\text{K}^+ + \text{CO}_3^{2-}\](b) \(\text{AlCl}_3\) dissociates into 1 \(\text{Al}^{3+}\) ion and 3 \(\text{Cl}^-\) ions: \\[\text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^-\]

Determine the Ion Ratio from Dissociation

Identify how many ions of each type are produced from one molecule based on the dissociation reactions.(a) From \(\text{K}_2\text{CO}_3\), each molecule produces 3 ions in total: 2 \(\text{K}^+\) and 1 \(\text{CO}_3^{2-}\).(b) From \(\text{AlCl}_3\), each molecule produces 4 ions in total: 1 \(\text{Al}^{3+}\) and 3 \(\text{Cl}^-\).

Calculate Total Ion Concentration for Each Solution

Multiply the molarity of the solution by the total number of ions produced per formula unit to determine the total molar concentration of ions.(a) For \(0.750 \, \text{M} \, \text{K}_2\text{CO}_3\), total ions = \(0.750 \, \text{M} \times 3 = 2.250 \, \text{M}\).(b) For \(0.355 \, \text{M} \, \text{AlCl}_3\), total ions = \(0.355 \, \text{M} \times 4 = 1.420 \, \text{M}\).

Key concepts

Dissociation Equation

A dissociation equation is a crucial part of understanding chemical reactions in solution chemistry. When ionic compounds dissolve in water, they generally split into their constituent ions. This process is what we refer to as dissociation. For example, in the problem given:

• For \( \text{K}_2\text{CO}_3 \), the dissociation equation is written as: \[ \text{K}_2\text{CO}_3 \rightarrow 2\text{K}^+ + \text{CO}_3^{2-} \]
• For \( \text{AlCl}_3 \), the dissociation equation is: \[ \text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^- \]

With these equations, we identify the ions that are formed when each compound is dissolved. Recognizing these ions is essential for the next steps in solution chemistry, as it allows chemists to calculate concentrations and understand the type of ions in the solution.

Ion Ratio

The ion ratio is derived from the dissociation equation and indicates the number of ions each formula unit of a compound will form. Understanding the ion ratio is key to determining the total ionic concentration in a solution, which is often needed for stoichiometry calculations or predicting reaction outcomes. Let's break it down with our examples:

• From the dissociation of \( \text{K}_2\text{CO}_3 \), we get two \( \text{K}^+ \) ions and one \( \text{CO}_3^{2-} \) ion, totaling three ions per formula unit.
• From the dissociation of \( \text{AlCl}_3 \), we produce one \( \text{Al}^{3+} \) ion and three \( \text{Cl}^- \) ions, resulting in four ions per formula unit.

The ion ratio is a simple yet powerful concept that provides a direct pathway to calculating how concentrated a solution becomes in terms of its ionic constituents. It’s essential to note, however, that complete dissociation is assumed here, which is typical in solution chemistry calculations.

Solution Chemistry

Solution chemistry deals with how substances dissolve, react, and exist in a solvent like water. A major part of this branch of chemistry involves understanding concepts such as molar concentration and dissociation. In the context of molar concentration and ion formation, solution chemistry helps us calculate how much solute dissolves and what the total ionic contribution will be in a solution. To do this, we:

• Identify dissociation equations to know what ions are present.
• Apply the ion ratio to find out the number of ions per formula unit.
• Multiply the initial molar concentration by the number of ions produced, giving us the total molar concentration of ions.

For example, in a \(0.750 \text{M}\) solution of \(\text{K}_2\text{CO}_3\), assuming each formula unit splits into three ions, the total ion concentration becomes \(0.750 \text{M} \times 3 = 2.250 \text{M}\). Similarly, for \(\text{AlCl}_3\), with four ions per unit, we get \(0.355 \text{M} \times 4 = 1.420 \text{M}\). These calculations are fundamental in quantifying the ionic strength and reactivity of a solution in various chemical contexts.