Problem 145 - General Chemistry

Question

A \(10.0-\mathrm{mL}\) sample of potassium iodide solution was analyzed by adding an excess of silver nitrate solution to produce silver iodide crystals, which were filtered from the solution. \(\mathrm{KI}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{KNO}_{3}(a q)+\operatorname{AgI}(s)\) If \(2.183 \mathrm{~g}\) of silver iodide was obtained, what was the molarity of the original KI solution?

Step-by-Step Solution

Answer

The molarity of the KI solution is 0.930 M.

Step 1: Write the reaction equation

The balanced chemical equation for the reaction between potassium iodide (KI) and silver nitrate (AgNO₃) is given as: \[ \text{KI} (aq) + \text{AgNO}_3 (aq) \rightarrow \text{KNO}_3 (aq) + \text{AgI} (s) \]. This equation shows that 1 mole of KI reacts with 1 mole of AgNO₃ to produce 1 mole of AgI.

Step 2: Calculate moles of AgI formed

To determine the number of moles of silver iodide (AgI) formed, use its molar mass. The molar mass of AgI is approximately 234.77 g/mol. The mass of AgI formed is 2.183 g. Calculate the moles using the formula: \[ \text{moles of } \text{AgI} = \frac{\text{mass of AgI}}{\text{molar mass of AgI}} = \frac{2.183 \text{ g}}{234.77 \text{ g/mol}} \approx 0.00930 \text{ moles.} \]

Step 3: Relate moles of AgI to moles of KI

The balanced chemical equation from Step 1 shows a 1:1 mole ratio between KI and AgI. This means the moles of AgI produced is equal to the moles of KI that reacted. Therefore, the moles of KI in the original solution are 0.00930 moles.

Step 4: Calculate the molarity of the KI solution

Molarity is defined as moles of solute per liter of solution. The original volume of the KI solution is 10.0 mL, which needs to be converted to liters: \[ 10.0 \text{ mL} = 0.0100 \text{ L} \]. Now, calculate the molarity (M) using the moles of KI from Step 3 and the volume of the solution: \[ \text{Molarity (M)} = \frac{\text{moles of KI}}{\text{volume in liters}} = \frac{0.00930}{0.0100} = 0.930 \text{ M}. \]

Key Concepts

Stoichiometry

Stoichiometry is like a recipe for chemical reactions. It tells us how much of each ingredient is needed to make a certain product. Think of it as the math behind chemistry that ensures we have the right proportions of reactants. In the context of the given exercise, stoichiometry involves calculating how many moles of potassium iodide (KI) are needed to produce a given amount of silver iodide (AgI). To achieve this, we rely on the balanced chemical equation that represents the reaction. The equation informs us that 1 mole of KI reacts with 1 mole of silver nitrate (AgNO₃) to give 1 mole of AgI. This 1:1:1 ratio is crucial because it allows us to translate between moles of reactants and products accurately.

Key points to remember about stoichiometry include:

• Understanding the mole concept to count atoms, molecules, or ions.
• Using balanced equations to find the relationship between reactants and products.
• Applying conversion factors derived from these relationships to perform calculations.

This process essentially makes stoichiometry indispensable for solving many types of chemistry problems.

Chemical Reactions

Chemical reactions are processes where substances, known as reactants, are transformed into different substances called products. In our exercise, we witness a classic chemical reaction where potassium iodide (KI) reacts with silver nitrate (AgNO₃) to produce silver iodide (AgI) and potassium nitrate (KNO₃). This specific reaction is a single displacement reaction, where the ions switch places to form new compounds.

Key characteristics of chemical reactions include:

• Formation of new substances with different properties.
• Changes in energy, often observable as heat, light, or gas production.
• Conservation of mass, meaning the total mass of reactants equals the total mass of products.

Understanding chemical reactions is essential because they explain how substances interact and transform, providing insights into their properties and uses.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react in solution to form one or more insoluble products, known as precipitates. In the given exercise, mixing solutions of potassium iodide (KI) and silver nitrate (AgNO₃) results in the formation of silver iodide (AgI), a precipitate that appears as a solid. This happens because AgI has low solubility in water.

Several important ideas surrounding precipitation reactions are:

• They often result in the removal of ions from the solution, forming a solid mass that can be filtered out.
• These reactions are useful for identifying the presence of certain ions in a solution.
• The formation of a precipitate can indicate the completion of a reaction.

These reactions are prevalent in various fields, including analytical chemistry and environmental testing, as they help detect and quantify specific components in a mixture.

Solution Chemistry

Solution chemistry deals with understanding how different substances dissolve in a solvent to form solutions. It's a vast field that covers various concepts related to solute concentration, solubility, and the nature of solutions. In our problem, we are primarily concerned with the molarity of the potassium iodide (KI) solution, which is an expression of how concentrated the solution is.

Some core aspects of solution chemistry include:

• Molarity, which indicates the number of moles of solute per liter of solution.
• Solubility, determining how well a solute dissolves in a solvent.
• Interactions between solute and solvent that define solution properties.

Understanding solution chemistry enables us to prepare solutions with precise concentrations, predict reaction outcomes, and control the conditions necessary for chemical reactions."}. This knowledge is essential in both academic settings and real-world applications such as pharmaceuticals and industrial processes.