Question

(a) Determine the chloride ion concentration in each of the following solutions: \(0.150 \mathrm{M} \mathrm{BaCl}_{2}, 0.566 \mathrm{M} \mathrm{NaCl}\), \(1.202 \mathrm{M} \mathrm{AlCl}_{3}\) (b) What is the concentration of a \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) solution that is \(2.55 \mathrm{M}\) in nitrate ion?

Short answer

(a) Chloride ion concentrations: 0.300 M for BaCl2, 0.566 M for NaCl, 3.606 M for AlCl3. (b) The Sr(NO3)2 concentration is 1.275 M.

Step-by-step solution

Identify the Dissociation of Compounds

First, identify how each compound dissociates in water.- \( \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \)- \( \text{NaCl} \rightarrow \text{Na}^+ + \text{Cl}^- \)- \( \text{AlCl}_3 \rightarrow \text{Al}^{3+} + 3\text{Cl}^- \)- \( \text{Sr(NO}_3)_2 \rightarrow \text{Sr}^{2+} + 2\text{NO}_3^- \)

Calculate Chloride Ion Concentration for BaCl2

For \(0.150 \text{ M BaCl}_2\), each formula unit produces 2 chloride ions.Thus, the concentration of chloride ions \( [\text{Cl}^-] \) = \( 2 \times 0.150 \text{ M} = 0.300 \text{ M} \).

Calculate Chloride Ion Concentration for NaCl

For \(0.566 \text{ M NaCl} \), each formula unit produces 1 chloride ion.Thus, the concentration of chloride ions \( [\text{Cl}^-] \) = \(1 \times 0.566 \text{ M} = 0.566 \text{ M} \).

Calculate Chloride Ion Concentration for AlCl3

For \(1.202 \text{ M AlCl}_3\), each formula unit produces 3 chloride ions.Thus, the concentration of chloride ions \( [\text{Cl}^-] \) = \( 3 \times 1.202 \text{ M} = 3.606 \text{ M} \).

Determine the Strontium Nitrate Solution's Molarity

Given that \( \text{Sr(NO}_3)_2 \) gives 2 nitrate ions per formula unit, if the nitrate ion concentration is \(2.55 \text{ M}\), then the original concentration of \( \text{Sr(NO}_3)_2\) is \( \frac{2.55 \text{ M}}{2} = 1.275 \text{ M} \).

Key concepts

Chloride Ion Concentration

Chloride ion concentration in solutions often depends on the dissociation of compounds that contain the chloride ion. Let's explore a few examples to get a better understanding.
For a solution of BaCl eous 2](https://www.khanacademy.org/science/chemistry/atomic-structure-and-properties/properties-of-covalent-bonds/a/for example, when BaCl2 dissociates, it produces two chloride ions. This means we multiply the molarity of BaCl aired 2 by 2 to find the chloride concentration, resulting in 0.300 M. It's important to note how many chloride ions each compound releases upon dissociation.

• NaCl simply releases one chloride ion per unit, so its chloride concentration remains 0.566 M.
• In contrast, AlCl3 releases three chloride ions, resulting in a chloride concentration of 3.606 M.

Understanding these basics allows us to predict chloride ion concentrations in any solution given the compound and its molarity.

Compounds Dissociation

Compounds dissociate into ions when they dissolve in water. This process is essential for calculating ion concentrations in a solution.
Consider the dissociation equations:

• BaCl eous 2 dissociates into one Ba^{2+} ion and two Cl^{-} ions. Thus, in solution, it provides twice the amount of chloride ions compared to its concentration.
• NaCl separates into one Na^{+} ion and one Cl^{-} ion, so the amount of chloride ions is equal to its concentration.
• AlCl3 dissolves into one Al^{3+} ion and three Cl^{-} ions. Here, three times the molarity of AlCl3 results in the cloride ion concentration.
• Sr(NO3)2 provides one Sr^{2+} ion and two NO3^{-} ions upon dissolution.

Once you recognize these patterns of dissociation, calculating concentrations becomes straightforward and predictable.

Molarity Calculation

Molarity is a measure of the concentration of solute in a solution. It's calculated as moles of solute divided by liters of solution. This concept is invaluable when working with solutions in chemistry.
To calculate the molarity of compounds like Sr(NO3)2, we first need to know how many ions each formula unit contributes. For Sr(NO3)2, the formula unit yields two NO3^- ions. If we're given the nitrate concentration, we can determine the overall molarity of Sr(NO3)2 by dividing the nitrate ion concentration by two (since two nitrate ions come from one Sr(NO3)2).
Using the given nitrate concentration of 2.55 M, divide by 2 to find the original Sr(NO3)2 molarity, resulting in 1.275 M. This method is applicable across various compounds by modifying it based on the number and type of ions dissociated.

Nitrate Ion Concentration

Nitrate ion concentration helps us understand the amount of nitrate available in a solution following dissociation. For compounds like Sr(NO3)2, which dissociate into Sr^{2+} and two NO3^{-} ions, this is particularly important.
Given that each unit of Sr(NO3)2 dissociates to give two nitrate ions, a solution with 2.55 M nitrate ions originates from a 1.275 M concentration of Sr(NO3)2. This step involves identifying how many ions are contributed by each molecule of the compound.
Understanding nitrate ion concentration also helps us analyze chemical reactions and processes where nitrates play a crucial role, such as in fertilizer formulations or environmental chemistry. By mastering this concept, you can better assess and predict the behavior of solutions involving nitrates.