Problem 70 - Chemistry The Central Science

Question

In each of the following pairs, indicate which has the higher concentration of \(\mathrm{I}^{-}\) ion: (a) 0.10 \(\mathrm{M}\) 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 \(\mathrm{MZnI}_{2}\) solution, \((\mathbf{c}) 3.2 \mathrm{M}\) HI solution or a solution made by dissolving 145 g of Nal in water to make 150 \(\mathrm{mL}\) of solution.

Step-by-Step Solution

Answer

(a) 0.25 M KI solution has a higher concentration of I⁻ (0.25 M) compared to 0.10 M BaI₂ solution (0.20 M). (b) 200 mL of 0.040 M ZnI₂ solution has a higher concentration of I⁻ (0.016 mol) compared to 100 mL of 0.10 M KI solution (0.01 mol). (c) 6.45 M NaI solution has a higher concentration of I⁻ compared to the 3.2 M HI solution.

Step 1: (a) Comparing 0.10 M BaI2 and 0.25 M KI solutions

BaI2 (barium iodide) dissociates into Ba²⁺ and 2I⁻ ions in the solution. Therefore, for every 1 mole of BaI2 dissolved, two moles of I⁻ are released. On the other hand, KI (potassium iodide) dissociates into K⁺ and I⁻ ions in the solution. Hence, for every 1 mole of KI dissolved, one mole of I⁻ is released. For BaI2: \( 0.10\, M \times 2 = 0.20\, M\) For KI: \( 0.25\, M \times 1 = 0.25\, M\) Thus, 0.25 M KI solution has the higher concentration of I⁻.

Step 2: (b) Comparing 100 mL of 0.10 M KI and 200 mL of 0.040 M ZnI₂ solutions

As mentioned earlier, 1 mole of KI releases 1 mole of I⁻, while 1 mole of ZnI₂ (zinc iodide) releases 2 moles of I⁻ upon dissociation. For KI, iodide concentration in 100 mL solution (0.10 L) is: \(0.10\,M \times 0.10 L = 0.01 mol\) For ZnI₂, iodide concentration in 200 mL solution (0.20 L) is: \(2 \times 0.040 M \times 0.20 L = 0.016 mol\) Since 0.016 mol of I⁻ in 200 mL ZnI₂ > 0.01 mol of I⁻ in 100 mL KI, 200 mL of 0.040 M ZnI₂ solution has a higher concentration of I⁻.

Step 3: (c) Comparing 3.2 M HI and 145 g NaI in 150 mL solutions

1 mole of HI (hydroiodic acid) dissociates into 1 mole of I⁻ upon dissociation. Molar concentration of I⁻ in HI: \(3.2\,M HI \times 1 = 3.2\,M\ I^{-}\) For NaI (sodium iodide), we first find the number of moles in 145 g of NaI. Molar mass of NaI = 22.99 (Na) + 126.90 (I) = 149.89 g/mol Number of moles of NaI: \(\frac{145 g}{149.89 \frac{g}{mol}} = 0.967\, mol\) For every 1 mole of NaI dissolved, 1 mole of I⁻ is released. Molar concentration of I⁻ in NaI: \(\frac{0.967\, mol}{0.150 L} = 6.45\,M\) Thus, the 6.45 M NaI solution has a higher concentration of I⁻ compared to the 3.2 M HI solution.

Key Concepts

Dissociation of Ionic Compounds

Understanding how ionic compounds dissociate in water is crucial for grasping the concentration of ions in a solution. When an ionic compound such as KI dissolves in water, it separates into its constituent ions, in this case, potassium (K⁺) and iodide (I⁻). The ratio of these ions depends on the compound's chemical formula. For example, BaI₂ splits into one barium ion (Ba²⁺) and two iodide ions (2I⁻), releasing twice the number of iodide ions into the solution per formula unit compared to KI, which only releases one iodide ion.

Dissociation is a reversible process where the solid form and dissolved ions are in dynamic equilibrium in a saturated solution. However, this removal of ions from the equilibrium by reactions or dilutions drives the reaction toward complete dissociation, making it essentially irreversible in dilute solutions. This knowledge helps to anticipate the concentration of specific ions in a solution, a vital concept in understanding chemical reactions in a solution.

Molar Concentration

Molar concentration—also known as molarity—is the number of moles of a solute per liter of solution. It is expressed in units of moles per liter (M). Calculating the molar concentration is straightforward when you have a pure compound: simply divide the amount of substance in moles by the volume of the solution in liters. For ionic compounds that dissociate, each ion type must be considered separately. When determining which solution has a higher concentration of a specific ion, you need to factor in not only the molarity of the ionic compound but also the dissociation pattern.

For example, as highlighted in the given solution for BaI₂, you multiply the molarity of the compound by the number of the specific ion it releases upon dissociation. The product yields the concentration of that ion in the solution. Recognizing these relationships helps in comparing different solutions to determine the one with a higher concentration of a particular ion.

Stoichiometry of Dissolution

The stoichiometry of dissolution is a part of chemical stoichiometry that focuses on the proportions in which substances dissolve to form solutions. The stoichiometry of an ionic compound provides the ratio of cations to anions released upon dissolution. For example, with ZnI₂, the ratio is 1:2 (one Zn²⁺ to two I⁻ ions). This stoichiometry is essential for calculating the concentration of ions after dissolution.

In practical terms, if you dissolve 1 mole of ZnI₂, you obtain 2 moles of iodide ions. So, for a solution with a known volume, you multiply the molarity of ZnI₂ by 2 to find the molarity of I⁻ in that solution. Similarly, the calculation for the molar concentration of I⁻ from NaI involves using the molar mass of NaI and the weight of NaI to find the moles of solute, which then allows for the calculation of molarity when you divide by the solution volume. This highlights why understanding stoichiometry is pivotal for determining concentrations in chemistry.