Question

A 10.0-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. $$ \mathrm{KI}(a q)+\mathrm{AgNO}_{3}(a q) \longrightarrow \mathrm{KNO}_{3}(a q)+\mathrm{AgI}(s) $$ If \(2.183 \mathrm{~g}\) of silver iodide was obtained, what was the molarity of the original KI solution?

Short answer

The molarity of the KI solution is 0.930 M.

Step-by-step solution

Calculate moles of AgI

First, determine the number of moles of silver iodide (AgI) that was produced. To do this, use the molar mass of AgI, which is approximately:\[ \text{Molar mass of AgI} = 107.87 \,\text{(Ag)} + 126.90 \, \text{(I)} = 234.77 \, \text{g/mol} \]Given mass of AgI is 2.183 g, the moles of AgI can be calculated using:\[ \text{Moles of AgI} = \frac{2.183 \, \text{g}}{234.77 \, \text{g/mol}} \approx 0.00930 \, \text{mol} \]

Determine moles of KI

According to the balanced chemical equation:\[ \mathrm{KI}(aq) + \mathrm{AgNO}_3(aq) \rightarrow \mathrm{KNO}_3(aq) + \mathrm{AgI}(s) \]Each mole of AgI produced corresponds to one mole of KI used. Thus, the moles of KI are equal to the moles of AgI:\[ \text{Moles of KI} = 0.00930 \, \text{mol} \]

Calculate molarity of KI solution

Molarity is defined as the number of moles of solute per liter of solution:\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{volume of solution in liters}} \]The given volume of KI solution is 10.0 mL, which is 0.0100 L. Thus, the molarity of the KI solution is:\[ \text{Molarity of KI} = \frac{0.00930 \, \text{mol}}{0.0100 \, \text{L}} = 0.930 \, \text{M} \]

Key concepts

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantitative relationships between the amounts of reactants and products in a chemical reaction. It allows us to use a balanced chemical equation to determine the proportions of each substance involved. In our example, the reaction between potassium iodide (KI) and silver nitrate (AgNO₃) is described by the equation: \[ \text{KI}(aq) + \text{AgNO}_3(aq) \rightarrow \text{KNO}_3(aq) + \text{AgI}(s) \] This equation shows us that one mole of KI reacts with one mole of AgNO₃ to produce one mole of AgI and one mole of KNO₃. Thus, the stoichiometric ratio is 1:1 for KI and AgI. Understanding this ratio is crucial because it helps us calculate the exact amount of reaction needed or the amount produced, which leads us to calculating moles, such as in this problem where the moles of AgI directly determine the moles of KI. Using stoichiometry is essential for calculating the concentrations and predictably conducting reactions in the lab or in industrial processes. It provides a bridge between the atomic scale and the macroscopic quantities we measure in experiments.

Moles and Molar Mass

Understanding moles and molar mass is fundamental in chemistry because they allow chemists to translate between the mass of a substance and the number of molecules or atoms it contains. A mole is a unit that symbolizes a very large number, specifically Avogadro's number, which is approximately \(6.022 \times 10^{23}\) entities (atoms, molecules, etc.). Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). In the given exercise, the molar mass of silver iodide (AgI) is crucial for converting its mass into moles. The molar mass of AgI is calculated by adding the atomic masses of silver (Ag) and iodine (I): - Ag: 107.87 g/mol - I: 126.90 g/mol - Total for AgI: 234.77 g/mol With the mass of AgI given as 2.183 g, the number of moles is calculated using the formula: \[ \text{Moles of AgI} = \frac{2.183 \text{ g}}{234.77 \text{ g/mol}} \approx 0.00930 \text{ mol} \] This calculation is essential for determining the amount of KI that reacted since it directly equates to the moles of KI present due to the 1:1 stoichiometric ratio.

Chemical Reactions

Chemical reactions describe processes where substances, called reactants, are transformed into different substances, called products. Each reaction follows the law of conservation of mass, meaning atoms are neither created nor destroyed; they are simply rearranged. Understanding chemical equations allows us to predict how much of each product will be formed or what amount of reactant is required. The reaction in this exercise is represented as: \[ \text{KI}(aq) + \text{AgNO}_3(aq) \rightarrow \text{KNO}_3(aq) + \text{AgI}(s) \] In this reaction, potassium iodide (KI) reacts with silver nitrate (AgNO₃) to form potassium nitrate (KNO₃) and silver iodide (AgI). The silver iodide forms a solid precipitate, visible as crystals. This type of reaction is known as a precipitation reaction, where two soluble salts react in solution to form an insoluble product. By understanding chemical reactions, we can determine not only the identity and quantities of our products but also calculate concentrations, such as the molarity of solutions. Molarity, determined as moles of solute per liter of solution, is used here to find the concentration of the original KI solution, which is done by relating moles of AgI to moles of KI and the volume of the solution.