Problem 80 - Chemistry The Central Science In Si Units

Question

You want to analyze a silver nitrate solution. What mass of \(\mathrm{NaCl}\) is needed to precipitate \(\mathrm{Ag}^{+}\) ions from \(45.0 \mathrm{~mL}\) of \(0.2500 \mathrm{M} \mathrm{AgNO}_{3}\) solution?

Step-by-Step Solution

Answer

The mass of NaCl needed is 0.6575 g.

Step 1: Determine moles of Ag+ ions

First, calculate the number of moles of Ag⁺ ions present in the solution. Use the formula: \( ext{moles} = ext{molarity} \times ext{volume} \). Convert 45.0 mL to liters, which is 0.045 L. Then, calculate the moles of Ag⁺: \( 0.2500 ext{ M} \times 0.045 ext{ L} = 0.01125 ext{ moles Ag}^{+} \).

Step 2: Use Stoichiometry to Find Moles of NaCl Needed

The balanced equation for the reaction is: \( \mathrm{AgNO}_{3} + \mathrm{NaCl} \rightarrow \mathrm{AgCl} + \mathrm{NaNO}_{3} \), showing a 1:1 molar ratio between \( \mathrm{AgNO}_{3} \) and \( \mathrm{NaCl} \). Therefore, you also need 0.01125 moles of NaCl to fully react with the Ag⁺ ions.

Step 3: Calculate Mass of NaCl Required

Use the molar mass of NaCl to find the required mass. The molar mass of NaCl is 58.44 g/mol. Multiply the moles of NaCl by its molar mass: \( 0.01125 ext{ moles} \times 58.44 \text{ g/mol} = 0.6575 \text{ g} \).

Key Concepts

Understanding Molarity

Molarity is a crucial concept in chemistry, particularly when working with solutions. It tells us how concentrated a solution is and is represented as moles per liter (M). Simply put, it measures how many moles of a substance (often called the solute) are dissolved in one liter of solution. This concept is vital for any calculations involving solution stoichiometry.To calculate molarity, use the formula:

• Molarity (M) = Moles of solute / Volume of solution in liters

In the exercise example, we had a 0.2500 M solution of silver nitrate (\(\text{AgNO}_3\)). This tells us there are 0.2500 moles of \(\text{AgNO}_3\) in one liter of solution. Knowing the molarity helps us understand how much of the solute is in a certain volume of the solution, which is essential for reacting our solution with another substance. When it comes to reactions, molarity helps predict how much reagent is needed or how much product will be formed, as it indicates the amount of chemical reactant available for a reaction.

Decoding Stoichiometry

Stoichiometry may sound like a complex term, but it's all about the math behind chemical reactions. It involves the quantitative relationships between the amounts of reactants and products in a chemical reaction, based on the balanced chemical equation. In the given exercise, the balanced equation is:

• \( \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \)

This equation shows a one-to-one relationship between \(\text{AgNO}_3\) and \(\text{NaCl}\). That means, for every mole of \(\text{AgNO}_3\) used, one mole of \(\text{NaCl}\) is needed.By using stoichiometry, we can determine exactly how much of each substance is required or produced in a reaction. In our exercise, we calculated the moles of \(\text{Ag}^+\) (which are in equal quantity to moles of \(\text{AgNO}_3\)) first, then used stoichiometry to find that the equivalent moles of \(\text{NaCl}\) needed are 0.01125 moles.This systematic approach ensures we have the correct amount of reactants to fully complete the reaction without any excess.

Precipitation Reaction Basics

Precipitation reactions are a specific type of chemical reaction where two solutions react and form an insoluble solid, known as a precipitate. These types of reactions are essential in various applications, including water treatment and analytical chemistry.In our example, when \(\text{AgNO}_3\) reacts with \(\text{NaCl}\), the product \(\text{AgCl}\) is a precipitate. It separates from the solution as a solid because it's not soluble in water. This happens because the attractive forces between \(\text{Ag}^+\) ions and \(\text{Cl}^-\) ions are stronger than those between ions and water molecules. Recognizing a precipitation reaction is key when predicting the outcome of mixing two solutions. Ensuring the complete precipitation of ions, like \(\text{Ag}^+\) in our solution, is important for accurate calculations and desired results in both lab and industrial settings.