#Chemical Reactions: Precipitation Reactions 🧪

This study guide focuses on precipitation reactions, a key type of chemical reaction you'll need to master for the AP Chemistry exam. Let's dive in!

#Types of Reactions: A Quick Overview

Before we focus on precipitation, let's briefly recap the three main types of reactions:

1. Acid-Base Reactions: Involve the transfer of protons (H⁺). Often, a strong acid reacts with a strong base to form a salt and water. Check out the next study guide for more details.

2. Oxidation-Reduction (Redox) Reactions: Involve the transfer of electrons. One substance is oxidized (loses electrons), and another is reduced (gains electrons). Combustion is a type of redox reaction. We'll cover these in depth later in this unit.

3. Precipitation Reactions: Involve the formation of an insoluble solid (precipitate) when two or more aqueous solutions are mixed. This is our focus for today!

Understanding the differences between these reaction types is crucial. Expect questions that require you to identify a reaction as acid-base, redox, or precipitation.

#Precipitation Reactions: The Details

Precipitation reactions occur when ions in aqueous solutions combine to form an insoluble or barely soluble solid called a precipitate. Think of it like mixing two clear liquids and suddenly seeing a solid form!

#Solubility Rules

While you don't need to memorize extensive solubility rules for the AP exam, familiarity helps. Usually, the question will tell you if a compound is soluble or not.

Table 4.1: Solubility of Common Ions in Water

#Net Ionic Equations

Here's a quick recap of how to write net ionic equations:

1. Identify Soluble and Insoluble Compounds: Use solubility rules or information given in the question.

2. Balance the Chemical Equation: Make sure the number of atoms of each element is the same on both sides.

3. Write the Complete Ionic Equation: Dissociate soluble ionic compounds into their respective ions.

4. Omit Spectator Ions: Remove ions that appear unchanged on both sides of the equation.

5. Write the Net Ionic Equation: Include the phases of matter (aq, s, l, g).

#Concentration of Ions in Precipitation Reactions

Let's tackle a challenging example that combines stoichiometry, limiting reactants, and ion concentrations. This is where things get interesting!

#Example Problem

Question: 20.0 mL of 0.100 M NaCl (aq) reacts with 30.0 mL of 0.0400 M Pb(C₂H₃O₂)₂ (aq).

Part a: What is the mass of the solid formed?

Part b: What are the concentrations of ions at the end of the reaction?

#Step-by-Step Solution

Step #1: Write and Balance the Equation

First, write the unbalanced equation:

NaCl (aq) + Pb(C₂H₃O₂)₂ (aq) → NaC₂H₃O₂ (aq) + PbCl₂ (s)

Then, balance it:

2 NaCl (aq) + Pb(C₂H₃O₂)₂ (aq) → 2 NaC₂H₃O₂ (aq) + PbCl₂ (s)

Step #2: Identify the Precipitate

Remember, sodium salts are always soluble. Therefore, PbCl₂ is the precipitate.

Step #3: Calculate Initial Moles

Use the given molarity and volume to calculate the moles of each reactant:

• Moles of NaCl: 0.100 M * 0.020 L = 0.00200 moles
• Moles of Pb(C₂H₃O₂)₂: 0.0400 M * 0.030 L = 0.00120 moles

Step #4: Determine the Limiting Reactant (LR)

To find the LR, use stoichiometry to convert both reactants to the product (PbCl₂):

• From NaCl: 0.00200 mol NaCl * (1 mol PbCl₂ / 2 mol NaCl) = 0.00100 mol PbCl₂
• From Pb(C₂H₃O₂)₂: 0.00120 mol Pb(C₂H₃O₂)₂ * (1 mol PbCl₂ / 1 mol Pb(C₂H₃O₂)₂) = 0.00120 mol PbCl₂

Since NaCl produces less PbCl₂, NaCl is the limiting reactant. The amount of product formed is based on the LR.

Step #5: Calculate Mass of Precipitate

Use the moles of PbCl₂ formed from the LR and its molar mass (278.2 g/mol) to find the mass:

0. 00100 mol PbCl₂ * 278.2 g/mol = 0.278 g of PbCl₂

Step #6: Calculate Final Ion Concentrations

• [Cl⁻]: Since Cl⁻ is part of the precipitate and NaCl is the LR, all Cl⁻ is used up. Therefore, [Cl⁻] = 0 M.

• [Na⁺]: All Na⁺ ions remain in solution as spectator ions. Use the moles of NaCl and the total volume (20.0 mL + 30.0 mL = 50.0 mL = 0.050 L):

  (0.00200 mol Na⁺) / 0.050 L = 0.0400 M Na⁺

• [C₂H₃O₂⁻]: All C₂H₃O₂⁻ ions remain in solution as spectator ions. Use the moles of Pb(C₂H₃O₂)₂ and the total volume:

(0.00120 mol * 2 C₂H₃O₂⁻) / 0.050 L = 0.0480 M C₂H₃O₂⁻

• [Pb²⁺]: To find the excess Pb²⁺, first find the moles of Pb²⁺ that reacted using the LR (NaCl). Since it's a 2:1 ratio, 0.00100 moles of Pb²⁺ reacted. Subtract this from the initial moles of Pb²⁺ (0.00120) to find the unreacted moles: 0.00120 - 0.00100 = 0.00020 moles. Then, divide by the total volume:

0.00020 mol / 0.050 L = 0.0040 M Pb²⁺

#Final Answers

Part a: 0.278 g of PbCl₂

Part b:

• [Cl⁻] = 0 M

• [Na⁺] = 0.0400 M

• [C₂H₃O₂⁻] = 0.0480 M

• [Pb²⁺] = 0.0040 M

#Final Exam Focus

• High-Priority Topics:
  • Solubility rules and predicting precipitates.
  • Writing balanced net ionic equations.
  • Stoichiometry calculations involving limiting reactants.
  • Calculating ion concentrations in precipitation reactions.
• Common Question Types:
  • Multiple-choice questions testing solubility rules and identifying precipitates.
  • Free-response questions requiring you to write net ionic equations and perform stoichiometric calculations.
  • Questions that combine precipitation with other concepts, such as acid-base reactions or titrations.
• Last-Minute Tips:

  • Time Management: Practice solving problems quickly and efficiently. Allocate your time wisely during the exam.

  • Common Pitfalls: Be careful with units, always balance equations, and double-check your calculations.

  • Strategies: Break down complex problems into smaller steps. If you get stuck, move on and come back later.

Practice Question

#Multiple Choice Questions

1. Which of the following compounds is insoluble in water? (A) NaCl (B) KNO₃ (C) AgCl (D) Na₂CO₃

2. When solutions of Pb(NO₃)₂ and KI are mixed, a precipitate forms. Which of the following is the correct net ionic equation? (A) Pb²⁺(aq) + 2 NO₃⁻(aq) → Pb(NO₃)₂(s) (B) K⁺(aq) + NO₃⁻(aq) → KNO₃(s) (C) Pb²⁺(aq) + 2 I⁻(aq) → PbI₂(s) (D) Pb²⁺(aq) + 2 I⁻(aq) + 2 K⁺(aq) + 2 NO₃⁻(aq) → PbI₂(s) + 2KNO₃(aq)

3. If 100 mL of 0.2 M AgNO₃ is mixed with 100 mL of 0.1 M NaCl, what is the concentration of silver ions (Ag⁺) after the reaction goes to completion? (A) 0.05 M (B) 0.1 M (C) 0.15 M (D) 0 M

#Free Response Question

A 50.0 mL sample of 0.200 M BaCl₂ (aq) is mixed with a 50.0 mL sample of 0.300 M Na₂SO₄ (aq). A precipitate of BaSO₄ forms. Assume that the volumes are additive.

(a) Write the balanced net ionic equation for the reaction.

(b) Identify the limiting reactant.

(c) Calculate the mass of BaSO₄ (molar mass = 233.4 g/mol) formed.

(d) Calculate the concentrations of all ions remaining in the solution after the reaction is complete.

Scoring Breakdown:

(a) 1 point for the correct net ionic equation: Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)

(b) 2 points for identifying the limiting reactant: * 1 point for calculating moles of Ba²⁺: (0.050 L)(0.200 M) = 0.010 mol * 1 point for calculating moles of SO₄²⁻: (0.050 L)(0.300 M) = 0.015 mol * Ba²⁺ is the limiting reactant.

(c) 2 points for the correct mass of BaSO₄: * 1 point for using the correct mole ratio: 0.010 mol BaSO₄ * 1 point for calculating the mass: (0.010 mol)(233.4 g/mol) = 2.334 g

(d) 4 points for the correct concentrations of all ions: * 1 point for [Ba²⁺] = 0 M (limiting reactant) * 1 point for [SO₄²⁻]: (0.015 mol - 0.010 mol) / 0.100 L = 0.050 M * 2 points for [Na⁺] and [Cl⁻]: * [Na⁺] = (0.050 L)(0.300 M) * 2 / 0.100 L = 0.300 M * [Cl⁻] = (0.050 L)(0.200 M) * 2 / 0.100 L = 0.200 M

Good luck with your exam! You've got this! 💪