Question

In each of the following pairs, indicate which has the higher concentration of I \({ }^{-}\)ion: (a) \(0.10 \mathrm{M} \mathrm{BaI}_{2}\) or \(0.25 \mathrm{M} \mathrm{KI}\) solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{KI}\) solution or \(200 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{ZnI}\) solution, (c) \(3.2 \mathrm{M}\) HI solution or a solution made by dissolving \(145 \mathrm{~g}\) of \(\mathrm{NaI}\) in water to make \(150 \mathrm{~mL}\) of solution.

Short answer

The solutions with higher I⁻ ion concentrations are: (a) \(0.25\,M\, KI\), (b) \(0.10\, M\, KI\), and (c) the solution made by dissolving \(145\,g\) of \(NaI\) in water to make \(150\,mL\) of solution.

Step-by-step solution

Convert given concentrations to moles of I⁻ per liter of solution

For each pair, we need to find the concentration of I⁻ ions in moles per liter. Remember to consider the stoichiometry of the compounds. ### Pair (a) ### (a) \(0.10\,M\, BaI_{2}\) and \(0.25\,M\, KI\). For \(BaI_2\), every 1 mole will dissociate into 2 moles of \(I^-\) ions: \[1\, BaI_2 \rightarrow 1\,Ba^{2+} + 2\,I^-\] For \(0.10\,M\, BaI_{2}\): \[0.10\,M\,I^-_{(from\,BaI_2)} = 0.10\,M\, BaI_2 \times 2 = 0.20\,M\] For \(KI\), every 1 mole will dissociate into 1 mole of \(I^-\) ions: \[1\, KI \rightarrow 1\,K^+ + 1\,I^-\] For \(0.25\,M\, KI\): \[0.25\,M\,I^-_{(from\,KI)} = 0.25\,M\, KI \times 1 = 0.25\,M\] ### Pair (b) ### (b) \(100\,mL\) of \(0.10\,M\, KI\) solution and \(200\,mL\) of \(0.040\, M\, ZnI_{2}\) solution. For \(KI\): \[0.10\,M\,I^{-}_{(from\,KI)} = 0.10\,M\, KI \times 1\] For \(ZnI_2\): \[1\, ZnI_{2} \rightarrow 1\, Zn^{2+} + 2\, I^-\] \[0.040\,M\,I^{-}_{(from\,ZnI_2)} = 0.040\,M\, ZnI_{2} \times 2\] ### Pair (c) ### (c) \(3.2\,M\) HI solution and a solution made by dissolving \(145\,g\) of NaI in water to make \(150\,mL\) of solution, For \(HI\): \[3.2\,M\,I^-_{(from\,HI)} = 3.2\,M\, HI \times 1\] For NaI, we need to calculate the molarity of NaI and then find I⁻ concentration: \[M_{NaI} = \frac{n_{NaI}}{V_{solution}}\] - We first calculate the number of moles of NaI (\(n_{NaI}\)) by dividing the mass (145 g) by its molar mass (149.89 g/mol): \[n_{NaI} = \frac{145\,g}{149.89\,g/mol} = 0.967\,mol\] - Next, we convert the volume from mL to L: \[V_{solution} = 150\,mL \times \frac{1\,L}{1000\,mL} = 0.150\,L\] - Now, we calculate the molarity of NaI: \[M_{NaI} = \frac{0.967\,mol}{0.150\,L} = 6.45\,M\] - Finally, we find the concentration of \(I^-\) ions: \[6.45\,M\,I^-_{(from\,NaI)} = 6.45\,M\, NaI \times 1\]

Compare I⁻ ion concentrations

Now, we will compare the concentrations of I⁻ ions in each pair to determine which solution has a higher concentration of I⁻ ions. ### Pair (a) ### - \(0.20\,M\,I^-_{(from\,BaI_2)}\) vs. \(0.25\,M\,I^-_{(from\,KI)}\) - \(0.25\,M\, KI\) solution has a higher concentration of I⁻ ions. ### Pair (b) ### - \(0.10\,M\,I^-_{(from\,KI)}\) vs. \(0.080\,M\,I^-_{(from\,ZnI_2)}\) - \(0.10\, M\, KI\) solution has a higher concentration of I⁻ ions. ### Pair (c) ### - \(3.2\,M\,I^-_{(from\,HI)}\) vs. \(6.45\,M\,I^-_{(from\,NaI)}\) - The solution made by dissolving \(145\,g\) of \(NaI\) in water to make \(150\,mL\) of solution has a higher concentration of I⁻ ions. In conclusion, the solutions with higher I⁻ ion concentrations are: \(0.25\,M\, KI\), \(0.10\, M\, KI\), and the solution made by dissolving \(145\,g\) of \(NaI\) in water to make \(150\,mL\) of solution.

Key concepts

Concentration

In chemistry, concentration refers to how much of a substance is present in a given volume of a solution. It's commonly expressed in terms of molarity, which is the number of moles of solute per liter of solution, denoted as M (mol/L). Concentration is crucial to understanding reactions and solutions because it provides insights into the potential reactivity and dynamic balance in chemical processes. For example:

• A solution with a higher concentration of iodine ions, I⁻, has more ions in the same volume, which can significantly affect the outcome of a chemical reaction.
• In the exercise, the concentration of I⁻ ions in various solutions is compared to determine which has the higher concentration from iodide sources such as KI, BaIₚ, HI, and NaI.

Understanding concentration helps in manipulating reactions and conducting experiments with precise control over reactant quantities.

Stoichiometry

Stoichiometry is the mathematical foundation of chemistry that deals with the quantitative relationships between reactants and products in a chemical reaction. It explains the proportions of elements and compounds necessary for a reaction and is crucial for predicting the yields of reactions. For dissociation reactions, stoichiometry tells us how many ions or molecules are produced from a compound. This is vital in solution chemistry when calculating concentrations.

Considering a Simple Dissociation Reaction

For instance, every mole of BaI₂ dissociates to yield two moles of I⁻ ions, while every mole of KI yields one mole of I⁻ ions. This relationship helps calculate the concentration of ions in the solution. For the exercise in question:

• The stoichiometry for BaI₂ -> Ba²⁺ + 2I⁻ helps determine that a 0.10 M solution of BaI₂ results in a 0.20 M concentration of I⁻ ions.
• Similarly, for KI -> K⁺ + I⁻, a 0.25 M solution results in a 0.25 M concentration of I⁻ ions.

Understanding stoichiometry is essential for balancing equations, understanding reactions, and performing precise calculations in solution chemistry.

Solution Chemistry

Solution chemistry revolves around the study of substances dissolved in solvents, mostly liquids. When a solute like salt, acid, or base is dissolved in a solvent, it forms a homogeneous mixture known as a solution.

Principles and Applications

- **Solubility and Dissociation**: These are the core principles determining how well solutes break into ions or interact with solvents. In the exercise, different compounds like BaI₂, KI, HI, and NaI are dissolved to find the resultant concentration of I⁻ ions. - **Calculating Molarity**: Knowing how to convert mass to moles and volume to liters is critical for finding molarity, which helps determine overall solute concentration in solutions. - **Reaction Dynamics**: The behavior of solutions in different chemical environments and at various concentrations can predict outcomes of reactions or improve industrial and laboratory procedures. Hence, solution chemistry is invaluable, allowing us to design experiments, understand natural processes, and create solutions with desired properties in fields from pharmacology to environmental science.