Problem 46 - Chemistry The Science in Context

Question

Calculate the molarity of each ion in a \(0.025 M\) aqueous solution of: (a) \(\mathrm{KCl} ;\) (b) \(\mathrm{CuSO}_{4} ;\) (c) \(\mathrm{CaCl}_{2}\)

Step-by-Step Solution

Answer

Question: Calculate the molarity of each ion present in the given aqueous solutions of (a) \(KCl\), (b) \(CuSO_4\), and (c) \(CaCl_2\), each with an initial molarity of \(0.025 M\). Answer: (a) In the \(KCl\) solution, the molarity of \(\mathrm{K^+}\) and \(\mathrm{Cl^-}\) ions is \(0.025 M\). (b) In the \(CuSO_4\) solution, the molarity of \(\mathrm{Cu^{2+}}\) and \(\mathrm{SO_{4}^{2-}}\) ions is \(0.025 M\). (c) In the \(CaCl_2\) solution, the molarity of \(\mathrm{Ca^{2+}}\) ions is \(0.025 M\), and the molarity of \(\mathrm{Cl^-}\) ions is \(0.050 M\).

Step 1: (a) Calculation of molarity of ions in \(\mathrm{KCl}\) solution

For the \(\mathrm{KCl}\) aqueous solution, first, we need to identify the ions produced when \(\mathrm{KCl}\) dissociates in water: $$ \mathrm{KCl} \rightarrow \mathrm{K^+ + Cl^-} $$ Since for every one molecule of \(\mathrm{KCl}\), one \(\mathrm{K^+}\) ion and one \(\mathrm{Cl^-}\) ion are produced, the molarity of each ion is equal to the molarity of the original \(\mathrm{KCl}\) solution. Therefore, the molarity of \(\mathrm{K^+}\) and \(\mathrm{Cl^-}\) ions are both \(0.025 M\).

Step 2: (b) Calculation of molarity of ions in \(\mathrm{CuSO}_{4}\) solution

For the \(\mathrm{CuSO}_{4}\) aqueous solution, we need to identify the ions produced when \(\mathrm{CuSO}_{4}\) dissociates in water: $$ \mathrm{CuSO_{4}} \rightarrow \mathrm{Cu^{2+} + SO_{4}^{2-}} $$ Each molecule of \(\mathrm{CuSO_{4}}\) produces one \(\mathrm{Cu^{2+}}\) ion and one \(\mathrm{SO_{4}^{2-}}\) ion. Thus, the molarity of \(\mathrm{Cu^{2+}}\) and \(\mathrm{SO_{4}^{2-}}\) are both equal to the molarity of the original \(\mathrm{CuSO_{4}}\) solution, which is \(0.025 M\).

Step 3: (c) Calculation of molarity of ions in \(\mathrm{CaCl}_{2}\) solution

For the \(\mathrm{CaCl}_{2}\) aqueous solution, we need to identify the ions produced when \(\mathrm{CaCl}_{2}\) dissociates in water: $$ \mathrm{CaCl_{2}} \rightarrow \mathrm{Ca^{2+} + 2Cl^-} $$ Each molecule of \(\mathrm{CaCl}_{2}\) produces one \(\mathrm{Ca^{2+}}\) ion and two \(\mathrm{Cl^-}\) ions. Therefore, the molarity of \(\mathrm{Ca^{2+}}\) is equal to the molarity of the original \(\mathrm{CaCl_{2}}\) solution, which is \(0.025 M\). However, since two \(\mathrm{Cl^-}\) ions are produced for every one molecule of \(\mathrm{CaCl_{2}}\), the molarity of \(\mathrm{Cl^-}\) is twice the molarity of the original solution, which is \(0.050 M\).

Key Concepts

Understanding Aqueous Solutions

An aqueous solution is a solution where the solvent is water. This is quite common, as water is considered the "universal solvent" because of its ability to dissolve many substances. When a substance, known as a solute, dissolves in water, it breaks down into individual ions or molecules. These ions are dispersed evenly throughout the water.

This process turns each water molecule into a tiny magnet, attracting and holding onto ions. The resulting homogeneous mixture is what we call an aqueous solution. The concentration of these ions in the solution is measured in terms of molarity, which tells us how many moles of a given substance are present in one liter of solution.

For instance, when something like sodium chloride (table salt) dissolves in an aqueous solution, it separates into sodium and chloride ions.

The Process of Dissociation in Water

Dissociation is an essential concept when discussing aqueous solutions and ion concentration. When ionic compounds dissolve in water, they split into individual positive and negative ions. This process is known as dissociation.

For example, when potassium chloride ( KCl") dissolves in water, it dissociates into potassium ions ( K^+") and chloride ions ( Cl^-"). Similarly, calcium chloride ( CaCl_2") dissociates into one calcium ion ( Ca^{2+}") and two chloride ions ( Cl^-").

The degree of dissociation can vary from one compound to another and impacts how we calculate the molarity of each ion formed. This step is crucial for solving the problem of ion concentration, as it ensures that we accurately count each different type of ion released into the solution.

Calculating Ion Concentration

Ion concentration is a crucial concept to master when determining the properties of an aqueous solution. It involves calculating how many ions are present in a solution. The starting point is the molarity of the original solution, which tells us the number of moles of solute per liter of solution.

Once an ionic compound dissociates in water, each type of ion in the solution will have its own molarity based on the original substance's dissociation pattern. For example, if we have a solution with a molarity of 0.025 M KCl", the molarity of both the K^+" and Cl^-") ions will also be 0.025 M because each molecule of KCl" dissociates into one potassium ion and one chloride ion.

In cases where there are different ratios of ions, such as in CaCl_2", the number of resulting ions affects the ion concentration. Since CaCl_2" dissociates into one Ca^{2+}" ion and two Cl^-" ions, the resulting molarity for Cl^-") would be twice that of the original CaCl_2" solution.