Problem 48 - Chemistry and Chemical Reactivity

Question

For each solution, identify the ions that exist in aqueous solution, and specify the concentration of each ion. (a) \(0.12 \mathrm{M} \mathrm{BaCl}_{2}\) (b) \(0.0125 \mathrm{M} \mathrm{CuSO}_{4}\) (c) \(0.500 \mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\)

Step-by-Step Solution

Answer

(a) 0.12 M \(\mathrm{Ba}^{2+}\), 0.24 M \(\mathrm{Cl}^-\); (b) 0.0125 M each of \(\mathrm{Cu}^{2+}\) and \(\mathrm{SO}_4^{2-}\); (c) 1.00 M \(\mathrm{K}^+\), 0.500 M \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\).

Step 1: Dissociation Process

Understand the dissociation of the compound in water. When an ionic compound dissolves in water, it separates into its constituent ions.

Step 2: Analyze Part (a) - Dissociation of \(\mathrm{BaCl}_2\)

\(\mathrm{BaCl}_2\) dissociates in water as follows: \[\mathrm{BaCl}_2 \rightarrow \mathrm{Ba}^{2+} + 2\mathrm{Cl}^-\]This results in one \(\mathrm{Ba}^{2+}\) ion and two \(\mathrm{Cl}^-\) ions for every molecule of \(\mathrm{BaCl}_2\) that dissociates.

Step 3: Calculate Ion Concentration - Part (a)

Given \(0.12 \mathrm{M} \mathrm{BaCl}_2\):- The concentration of \(\mathrm{Ba}^{2+}\) ions is 0.12 M.- The concentration of \(\mathrm{Cl}^-\) ions is \(2 \times 0.12\, \mathrm{M} = 0.24\, \mathrm{M}\) due to the 2:1 ratio of \(\mathrm{Cl}^-\) to \(\mathrm{BaCl}_2\).

Step 4: Analyze Part (b) - Dissociation of \(\mathrm{CuSO}_4\)

\(\mathrm{CuSO}_4\) dissociates in water as follows:\[\mathrm{CuSO}_4 \rightarrow \mathrm{Cu}^{2+} + \mathrm{SO}_4^{2-}\]This results in one \(\mathrm{Cu}^{2+}\) ion and one \(\mathrm{SO}_4^{2-}\) ion for each \(\mathrm{CuSO}_4\) molecule.

Step 5: Calculate Ion Concentration - Part (b)

Given \(0.0125 \mathrm{M} \mathrm{CuSO}_4\):- The concentration of \(\mathrm{Cu}^{2+}\) ions is 0.0125 M.- The concentration of \(\mathrm{SO}_4^{2-}\) ions is also 0.0125 M.

Step 6: Analyze Part (c) - Dissociation of \(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\)

\(\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\) dissociates in water as follows:\[\mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7 \rightarrow 2\mathrm{K}^+ + \mathrm{Cr}_2\mathrm{O}_7^{2-}\]This provides two \(\mathrm{K}^+\) ions per formula unit.

Step 7: Calculate Ion Concentration - Part (c)

Given \(0.500 \mathrm{M} \mathrm{K}_2\mathrm{Cr}_2\mathrm{O}_7\):- The concentration of \(\mathrm{K}^+\) ions is \(2 \times 0.500\, \mathrm{M} = 1.00\, \mathrm{M}\) due to the 2:1 ratio.- The concentration of \(\mathrm{Cr}_2\mathrm{O}_7^{2-}\) ions is 0.500 M.

Key Concepts

Ionic Compounds

Ionic compounds are a type of chemical compound that consist of ions held together by ionic bonds. These compounds typically form when metals lose electrons to become positively charged ions, known as cations, and non-metals gain electrons to become negatively charged ions, known as anions. Ionic compounds are generally formed between elements with significantly different electronegativities, such as metals and non-metals.
- Examples of ionic compounds include sodium chloride ( NaCl ), barium chloride ( BaCl_2 ), copper(II) sulfate ( CuSO_4 ), and potassium dichromate ( K_2 Cr_2 O_7 ).
- In the context of solutions, ionic compounds dissolve in water to separate into their individual ions, a process known as dissociation.
This fundamental property of ionic compounds underlies their behavior in solution, allowing them to conduct electricity and to participate in chemical reactions.

Dissociation Process

Dissociation is a key concept in the study of solution chemistry, describing how ionic compounds dissolve in water to form ions.
When an ionic compound is placed in water, the polar molecules of the water surround the positive and negative ions, which overcomes the ionic bonds holding the compound together.
This causes the compound to dissociate into its constituent ions.
- For example, barium chloride (BaCl_2) dissociates in water to form one Ba^{2+} ion and two Cl^{-} ions
\[BaCl_2 \rightarrow Ba^{2+} + 2Cl^-\]Another example is copper(II) sulfate (CuSO_4), which dissociates into one Cu^{2+} ion and one SO_4^{2-} ion.
- This process is pivotal for understanding how the ions behave and interact in aqueous solutions, and is also essential for calculating ion concentrations in solutions.

Ion Concentration Calculation

Ion concentration calculation is an important aspect of solution chemistry, allowing us to understand the behavior of ions in a solution upon dissociation.
The concentration of an ion in solution after dissociation depends on the original concentration of the ionic compound and the stoichiometry of the dissociation process.
For instance:
- In a 0.12 M solution of barium chloride (BaCl_2), the concentration of Ba^{2+} ions is 0.12 M, and the concentration of Cl^{-} ions is doubled to 0.24 M due to the 1:2 stoichiometric ratio in the dissociation equation.
\[BaCl_2 \rightarrow Ba^{2+} + 2Cl^-\]- Similarly, in a 0.0125 M solution of copper(II) sulfate (CuSO_4), the concentration of Cu^{2+} ions and SO_4^{2-} ions remains 0.0125 M, as the dissociation produces these ions in a 1:1 ratio.
Accurately calculating ion concentrations is crucial for predicting the properties and reactivities of solutions.

Aqueous Solutions

Aqueous solutions are solutions where water is the solvent, and they play a central role in chemistry and biology.
In these solutions, water molecules surround and interact with dissolved substances, affecting their properties in various ways.
- When ionic compounds dissolve in water, they dissociate into ions, resulting in solutions that can conduct electricity due to the movement of these charged particles.
- Aqueous solutions are essential for many chemical reactions and biological processes; for example, most metabolic processes in living organisms occur in aqueous environments.
- The study of aqueous solutions involves understanding how compounds dissolve, how ions form and interact, and how these factors influence reactions.
This understanding is crucial for fields ranging from industrial chemistry to environmental science and pharmacology.