Question

Identify the ions that exist in each aqueous solution, and specify the concentration of each ion. (a) \(0.25 \mathrm{M}\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) (b) \(0.123 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\) (c) \(0.056 \mathrm{M} \mathrm{HNO}_{3}\)

Short answer

(a) 0.50 M NH4+ and 0.25 M SO4^2-; (b) 0.246 M Na+ and 0.123 M CO3^2-; (c) 0.056 M H+ and 0.056 M NO3^-.

Step-by-step solution

Identify the Ions in the First Solution

The first compound is \((NH_4)_2SO_4\), ammonium sulfate. In water, this compound dissociates into ammonium ions \((NH_4^+)\) and sulfate ions \((SO_4^{2-})\). Each formula unit of ammonium sulfate produces 2 ammonium ions and 1 sulfate ion.

Calculate Concentrations in the First Solution

Given the concentration of \((NH_4)_2SO_4\) is 0.25 M, the concentration of \(NH_4^+\) ions is twice that, since each \((NH_4)_2SO_4\) produces 2 \(NH_4^+\) ions. Thus, the concentration of \(NH_4^+\) is \(2 \times 0.25 \, \text{M} = 0.50 \, \text{M}\). The concentration of \(SO_4^{2-}\) remains 0.25 M, as one sulfate ion is produced per formula unit.

Identify the Ions in the Second Solution

The second compound is \(Na_2CO_3\), sodium carbonate. It dissociates into sodium ions \((Na^+)\) and carbonate ions \((CO_3^{2-})\). Each formula unit of sodium carbonate produces 2 sodium ions and 1 carbonate ion.

Calculate Concentrations in the Second Solution

Given the concentration of \(Na_2CO_3\) is 0.123 M, the concentration of \(Na^+\) ions is twice that because each \(Na_2CO_3\) produces 2 \(Na^+\) ions. Thus, the concentration of \(Na^+\) is \(2 \times 0.123 \, \text{M} = 0.246 \, \text{M}\). The concentration of \(CO_3^{2-}\) is 0.123 M.

Identify the Ions in the Third Solution

The third compound is \(HNO_3\), nitric acid. It dissociates into hydrogen ions \((H^+)\) and nitrate ions \((NO_3^-)\). Each formula unit of nitric acid produces 1 hydrogen ion and 1 nitrate ion.

Calculate Concentrations in the Third Solution

Given the concentration of \(HNO_3\) is 0.056 M, the concentrations of both \(H^+\) and \(NO_3^-\) ions are the same at 0.056 M, since each \(HNO_3\) produces 1 \(H^+\) ion and 1 \(NO_3^-\) ion.

Key concepts

Ammonium Sulfate Dissociation

Ammonium sulfate, represented by the chemical formula \((NH_4)_2SO_4\), is a classic example of an ionic compound that dissociates in water to form ions. When dissolved, each ammonium sulfate unit breaks apart into two ammonium ions \((NH_4^+)\) and one sulfate ion \((SO_4^{2-})\). This is key as it directly impacts the concentration of each ion in the solution.
Given a solution of 0.25 M ammonium sulfate, the dissociation leads to the generation of ions in specific concentrations:

• The concentration of ammonium ions \((NH_4^+)\) is twice the original because each unit contributes two \(NH_4^+\) ions. Therefore, the concentration is \(2 \times 0.25 \, \text{M} = 0.50 \, \text{M}\).
• The concentration of sulfate ions \((SO_4^{2-})\) remains the same as the initial concentration of the ammonium sulfate, at 0.25 M, because one sulfate ion is produced per formula unit.

Understanding this dissociation is fundamental when examining reactions in a solution, particularly in calculating ionic concentrations.

Sodium Carbonate Dissociation

Sodium carbonate, denoted as \(Na_2CO_3\), dissociates in water to form sodium ions \((Na^+)\) and carbonate ions \((CO_3^{2-})\). For each formula unit of sodium carbonate, two sodium ions and one carbonate ion are produced. This ratio is crucial as it helps in determining the ions' concentrations in solutions.
Consider a solution with a sodium carbonate concentration of 0.123 M:

• The resulting concentration of sodium ions \((Na^+)\) is doubled since each sodium carbonate unit provides two \(Na^+\) ions. Consequently, it's \(2 \times 0.123 \, \text{M} = 0.246 \, \text{M}\).
• The concentration of carbonate ions \((CO_3^{2-})\) mirrors the initial sodium carbonate concentration, maintaining at 0.123 M.

This understanding not only aids in chemical equilibrium situations but is also crucial when dealing with buffer solutions and precipitation reactions.

Nitric Acid Dissociation

Nitric acid, represented by \(HNO_3\), is a strong acid that completely dissociates in water into hydrogen ions \((H^+)\) and nitrate ions \((NO_3^-\)). Unlike the other compounds discussed, nitric acid dissociation results in a 1:1 ratio of ions.
For a solution with a nitric acid concentration of 0.056 M, each unit releases exactly one hydrogen ion and one nitrate ion:

• Both \([H^+]\) and \([NO_3^-]\) concentrations are equal to the initial nitric acid concentration, each at 0.056 M.

This information is crucial in various applications, such as titrations and pH calculations, where knowing the concentration of \(H^+\) is necessary for determining the acidity levels of a solution. Understanding the dissociation behavior of nitric acid ensures accurate chemical analysis and reaction predictions.