Problem 15 - General Chemistry: Principles and Modern Applications

Question

How many milligrams of \(\mathrm{MgI}_{2}\) must be added to \(250.0 \mathrm{mL}\) of \(0.0876 \mathrm{M} \mathrm{KI}\) to produce a solution with \(\left[\mathrm{I}^{-}\right]=0.1000 \mathrm{M} ?\)

Step-by-Step Solution

Answer

To achieve the solution with a desired I- concentration of 0.1 M, approximately 1.74 g or 1740 mg of MgI2 must be added to the 250 mL of 0.0876 M KI solution.

Step 1: Calculate initial moles of I- ions in KI

We can calculate the initial amount of I- ions in KI by using the formula for molarity and the volume. The formula for molarity is \[Molarity = \frac{Moles}{Volume}\]. Hence, moles are calculated by multiplying molarity (0.0876 M) with volume (0.250 L, converted from ml).

Step 2: Calculate final moles of I- ions required

To find the number of I- ions required in the final solution, multiply the final volume (0.250 L) with the desired molarity (0.1000 M). The difference between the final and initial moles will provide the moles of I- ions that should come from MgI2. Here, each MgI2 molecule dissociates to give 2 I- ions.

Step 3: Calculate the mass of MgI2 to be added

Once the moles of I- ions from MgI2 are known, these can be converted into mass by using the molar mass of MgI2 (278.11 g/mol). The mole-to-mass conversion is done by multiplying the moles with the molar mass. Convert this mass from grams to milligrams by multiplying by 1000.

Key Concepts

Ionic dissociation

Ionic dissociation is a crucial concept in chemistry, especially when dealing with ionic compounds. When an ionic compound like Magnesium Iodide (\(\mathrm{MgI}_2\)) is dissolved in water, it dissociates into its constituent ions. In this case, one molecule of \(\mathrm{MgI}_2\) dissociates into one \(\mathrm{Mg}^{2+}\) ion and two \(\mathrm{I}^{-}\) ions. Understanding this dissociation process helps us predict how many ions are contributed to the solution by each molecule of the compound. For example:

• \(\mathrm{MgI}_2 \rightarrow \mathrm{Mg}^{2+} + 2\mathrm{I}^{-}\)

Remember that the stoichiometry of this reaction shows us that every single molecule of \(\mathrm{MgI}_2\) adds two iodide ions to the solution. This doubling effect is vital for calculating changes in ion concentration in solutions, as seen in the exercise above. By accounting for each ion produced, you can accurately determine the changes in concentration needed for your desired chemical result.

Stoichiometry

Stoichiometry involves using the relationships between quantities of reactants and products in a chemical reaction to predict outcomes. In the context of ionic reactions, stoichiometry helps us understand the proportion of ions generated or consumed. For the exercise you are facing:

• You start by calculating the initial and desired concentrations of iodide ions (\(\mathrm{I}^{-}\)).
• Initially, from 250 mL of 0.0876 M \(\mathrm{KI}\), calculate the moles of \(\mathrm{I}^{-}\) using the formula:
  \[\text{moles} = \text{Molarity} \times \text{Volume (in liters)}\]
• Determine the desired final moles of \(\mathrm{I}^{-}\) in the 0.250 L solution at 0.1000 M concentration.
• The additional moles needed are the difference between the final and initial moles.

By acknowledging that \(\mathrm{MgI}_2\) contributes two iodide ions per formula unit, stoichiometry allows you to calculate how much \(\mathrm{MgI}_2\) is necessary. Stoichiometry ensures precise calculations in changing conditions or achieving specific solution concentrations.

Molar mass calculation

Molar mass calculation is an essential step in determining the amount of a substance needed in a reaction.The molar mass is the mass, typically expressed in grams per mole (g/mol), of one mole of a given substance. It is calculated by summing the atomic masses of all the atoms in the molecule.For Magnesium Iodide, \(\mathrm{MgI}_2\):

• Magnesium (Mg) has an atomic mass of approximately 24.31 g/mol.
• Iodine (I) has an atomic mass of approximately 126.90 g/mol, but there are two iodine atoms, so multiply by 2.

Therefore, the molar mass of \(\mathrm{MgI}_2\) is computed as:
\[\text{Molar Mass of } \mathrm{MgI}_2 = 24.31 + (126.90 \times 2) = 278.11 \text{ g/mol}\]To find out how much \(\mathrm{MgI}_2\) you need, convert the moles of \(\mathrm{I}^{-}\) required to mass using this molar mass, then multiply by 1000 to convert from grams to milligrams.This step is fundamental to transforming a molecular computation into a measurable gel or dry powder for practical laboratory or industrial uses.