Problem 70 - Chemistry The Central Science

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 \(0.25 \mathrm{M}\) KI solution, (b) \(100 \mathrm{~mL}\) of \(0.10 \mathrm{M}\) KI solution or \(200 \mathrm{~mL}\) of \(0.040 \mathrm{M} \mathrm{ZnI}_{2}\) solution, (c) \(3.2 \mathrm{M}\) HI solution or a solution made by dissolving \(145 \mathrm{~g}\) of NaI in water to make \(150 \mathrm{~mL}\) of solution.

Step-by-Step Solution

Answer

In summary, for each of the following pairs, the solution with a higher concentration of I⁻ ions is: (a) \(0.25\, \mathrm{M}\) KI solution (b) \(0.10\, \mathrm{M}\) KI solution (c) the solution made by dissolving \(145\, \mathrm{g}\) of NaI in water to make \(150\, \mathrm{mL}\) of solution.

Step 1: (a) Compare BaI₂ and KI concentrations

To find which solution has a higher I⁻ concentration, we must consider the balanced chemical equations for the dissolution of BaI₂ and KI in water. For BaI₂: \(\mathrm{BaI}_2 \rightarrow \mathrm{Ba^{2+}} + 2\mathrm{I^-}\) For KI: \(\mathrm{KI} \rightarrow \mathrm{K^+} + \mathrm{I^-}\) In a \(0.10 \mathrm{M}\) BaI₂ solution, one mole of BaI₂ produces 2 moles of I⁻ ions, so the I⁻ concentration will be: \(0.10 \, \mathrm{M} * 2 = 0.20 \, \mathrm{M}\) In a \(0.25\, \mathrm{M}\) KI solution, one mole of KI produces one mole of I⁻ ions, so the I⁻ concentration will be: \(0.25\, \mathrm{M} * 1 = 0.25\, \mathrm{M}\) Since \(0.25\, \mathrm{M} > 0.20\, \mathrm{M}\), the KI solution has a higher I⁻ concentration in this pair.

Step 2: (b) Compare KI and ZnI₂ Solutions

We'll again consider the balanced chemical equations for the dissolution of KI and ZnI₂ in water. For KI: \(\mathrm{KI} \rightarrow \mathrm{K^+} + \mathrm{I^-}\) For ZnI₂: \(\mathrm{ZnI}_2 \rightarrow \mathrm{Zn^{2+}} + 2\mathrm{I^-}\) First, we need to convert the volumes of these two solutions to liters. \(100\, \mathrm{mL} = 0.1\, \mathrm{L}\), and \(200\, \mathrm{mL} = 0.2\, \mathrm{L}\) Next, we'll calculate the moles of I⁻ ions in both solutions: For \(0.1\, \mathrm{L}\) of \(0.10\, \mathrm{M}\) KI solution: \(\mathrm{moles \, of \, I^-} = 0.1\, \mathrm{L} × 0.10\, \mathrm{M} = 0.01\, \mathrm{mol}\) For \(0.2\, \mathrm{L}\) of \(0.040\, \mathrm{M}\) ZnI₂ solution: \(\mathrm{moles \, of \, I^-} = 0.2\, \mathrm{L} × 0.040\, \mathrm{M} × 2 = 0.016\, \mathrm{mol}\) Now we can calculate the concentrations of I⁻ ions: For the KI solution: \(\mathrm{I^- \, concentration} = \frac{0.01\, \mathrm{mol}}{0.1\, \mathrm{L}} = 0.10\, \mathrm{M}\) For the ZnI₂ solution: \(\mathrm{I^- \, concentration} = \frac{0.016\, \mathrm{mol}}{0.2\, \mathrm{L}} = 0.08\, \mathrm{M}\) Since \(0.10\, \mathrm{M} > 0.08\, \mathrm{M}\), the KI solution has a higher I⁻ concentration in this pair.

Step 3: (c) Compare HI and NaI Solutions

We'll consider the balanced chemical equations for the dissolution of HI and NaI in water. For HI: \(\mathrm{HI} \rightarrow \mathrm{H^+} + \mathrm{I^-}\) For NaI: \(\mathrm{NaI} \rightarrow \mathrm{Na^+} + \mathrm{I^-}\) First, we need to find the number of moles of NaI given its mass: Given that the molar mass of NaI is approximately \(149\, \mathrm{g/mol}\): \(\mathrm{moles \, of \, NaI} = \frac{145\, \mathrm{g}}{149\, \mathrm{g/mol}} \approx 0.97\, \mathrm{mol}\) Next, calculate the concentration of I⁻ ions in both solutions: For the HI solution, the I⁻ concentration is \(3.2\, \mathrm{M}\) because HI completely dissociates in water. For the NaI solution, we need to find the concentration of I⁻ ions based on the provided mass and volume: \(\mathrm{I^- \, concentration} = \frac{0.97\, \mathrm{mol}}{0.150\, \mathrm{L}} \approx 6.47\, \mathrm{M}\) Since \(6.47\, \mathrm{M} > 3.2\, \mathrm{M}\), the NaI solution has a higher I⁻ concentration in this pair.

Key Concepts

Dissolution of Compounds

The process of dissolving a compound in a solvent, such as water, to produce ions is referred to as dissolution. When a compound dissolves, it separates into its constituent ions. For instance, when solid barium iodide (BaI₂) dissolves in water, it dissociates into one barium ion (Ba²⁺) and two iodide ions (I^-). Similarly, potassium iodide (KI) gives one potassium ion (K⁺) and one iodide ion (I^-) upon dissolution.

Dissolution is crucial in determining the concentration of a specific ion in solution. It's important for students to remember that not all compounds dissociate in the same ratio. For example, one mole of BaI₂ produces two moles of iodide ions, while one mole of KI produces only one mole of iodide. Understanding this concept allows for a better grasp of ionic concentrations in solution.

Chemical Equation Balancing

A balanced chemical equation ensures that the law of conservation of mass is upheld, indicating that the number of atoms of each element is the same on both sides of the equation. This is vital for calculating the changes in concentration when compounds dissolve in solution.

For instance, the dissolution of barium iodide (BaI₂) is represented by \( \mathrm{BaI}_2 \rightarrow \mathrm{Ba^{2+}} + 2\mathrm{I^-} \), which means that for each formula unit of BaI₂ that dissolves, there is one Ba²⁺ ion and two I^- ions produced. Balancing chemical equations gives a clear picture of these stoichiometric relationships and is essential for accurately calculating ion concentrations.

Molar Concentration Calculations

Molar concentration, often represented as molarity and expressed in moles per liter (M), is a measure of the concentration of a solute in a solution. To determine molar concentration, you need the number of moles of the solute and the volume of the solution in liters.

Calculating molar concentrations involves using the following formula: \( \text{Molarity} = \frac{\text{moles of solute}}{\text{liters of solution}} \). For example, to find the iodide ion concentration from a solid NaI dissolved in water, you first find the moles of NaI using the mass given and its molar mass. Then, you relate this to the volume of solution, which gives the molarity of the iodide ions. It is crucial to take into account that some compounds can produce more than one mole of an ion per mole of compound dissolved, affecting the final concentration.