You want to analyze a silver nitrate solution. What mass of NaCl is needed to precipitate Ag+ ions from 45.0 mL of 0.2500 M AgNO3 solution?

Name: Chemistry: The Central Science
Author: Brown
ISBN: 9780134414232

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Calculate the moles of AgNO_3 in the solution using the formula: \( \text{moles} = \text{molarity} \times \text{volume (L)} \). Convert 45.0 mL to liters by dividing by 1000.

Write the balanced chemical equation for the reaction: \( \text{AgNO}_3 (aq) + \text{NaCl} (aq) \rightarrow \text{AgCl} (s) + \text{NaNO}_3 (aq) \).

From the balanced equation, note that the mole ratio of AgNO_3 to NaCl is 1:1. Therefore, the moles of NaCl needed are equal to the moles of AgNO_3.

Calculate the mass of NaCl required using the formula: \( \text{mass} = \text{moles} \times \text{molar mass of NaCl} \). The molar mass of NaCl is approximately 58.44 g/mol.

Ensure all units are consistent and check calculations for accuracy.

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stoichiometry

Stoichiometry is the calculation of reactants and products in chemical reactions. It involves using balanced chemical equations to determine the proportions of substances involved. In this case, it helps to find the amount of NaCl needed to react with AgNO3 to form the precipitate AgCl.

Precipitation Reactions

Precipitation reactions occur when two soluble salts react to form an insoluble product, or precipitate. In this scenario, the reaction between NaCl and AgNO3 produces AgCl, which is insoluble in water. Understanding this concept is crucial for predicting the formation of the precipitate and calculating the required amounts of reactants.

Molarity

Molarity is a measure of concentration defined as the number of moles of solute per liter of solution. It is essential for determining how much of a substance is present in a given volume of solution. In this question, the molarity of the AgNO3 solution is used to calculate the moles of Ag+ ions available for precipitation with NaCl.

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