Lecture 31

Exam 2 Preparation

General Information

• Calculator, periodic table, and clicker are required.
• Session ID will be needed.
• Quiz 5 is due tonight before 11:55 PM.
• Continue reading Chapter 6.
• Chapter 6 homework is due Monday, April 21st.
• Exam 2 is on Tuesday, April 15th at 7 PM.
• The exam will be held in the same rooms as the last exam, covering chapters 4.2-6.8.
• It is recommended to start Chapter 6 homework for practice before the exam.

Exam 3 Reviews

1. Sunday 11 AM - 1 PM at Mossman 102
2. Monday 7 PM - 9 PM at Min Kao 622

Precipitation Question

• Problem: Separate Ag+1, Ba+2, and Fe+3 in a water solution using NaCl, Na2SO4, and NaOH. Determine the order of precipitation.
• Correct Order: AgCl, BaSO4, Fe(OH)3
• Solubility Rules:
  • Cl-: AgCl is insoluble, others are soluble.
  • SO42-: BaSO4 is insoluble, others have varying solubility.
  • OH-: Fe(OH)3 is insoluble, Ba(OH)2 slightly soluble, others have varying solubility.

Amphoteric Molecules

• Question: Can the same molecule act as an acid or a base?
  • Answer: YES
• Example Reactions:
  • $NH<em>4^+(aq) + H</em>2O(l) \rightleftharpoons NH<em>3(aq) + H</em>3O^+(aq)$
  • $NH<em>3(aq) + H</em>2O(l) \rightleftharpoons NH_4^+(aq) + OH^−(aq)$
• Amphoteric: A substance that can act as both an acid and a base.
• Determination: Whether an amphoteric substance acts as an acid or a base depends on the chemical it reacts with.

Glucose in Water

• Question: What happens when glucose (C6H12O6) is put in water?
  • Answer: Neither an acid nor a base is formed.
• Explanation: Covalent compounds generally do not dissociate in H2O, thus they cannot produce or accept H+1 ions or produce OH-1 ions.
• Exception: Organic acids (e.g., acetic acid) dissociate slightly in water and act as weak acids.

KOH in Water

• Question: How does KOH behave when dissolved in water?
  • Answer: It's a base that forms K+1 and OH–1 ions.

Al(OH)3 in Water

• Question: What are the major species present when the weak base Al(OH)3 is mixed with H2O?
  • Answer: Al(OH)3 and H2O

Identifying Acids and Bases

• Acids:
  • Generally have “H” listed first in the formula or have “acid” in the name.
  • Examples: HCl, HClO4, H2SO4, HCH3CO2 (acetic acid), HC6H7O6 (ascorbic acid).
• Bases:
  • Composed of a metal ion or NH4+1 with OH-1, O-2, or NH2-1.
  • Examples: NaOH, K2O, LiNH2, Mg(OH)2, CaO.
  • Important base to know: NH3

Acid-Base Neutralization Reactions

• Definition: Processes in which an acid reacts with a base to produce water and an ionic compound (salt).
• Example: $HCl(aq) + NaOH(aq) \rightarrow H_2O(l) + NaCl(aq)$
• Ionic Equation: $H^{+1}(aq) + Cl^{-1}(aq) + Na^{+1}(aq) + OH^{-1}(aq) \rightarrow H_2O(l) + Na^{+1}(aq) + Cl^{-1}(aq)$
• Net Ionic Equation: $H^{+1}(aq) + OH^{-1}(aq) \rightarrow H_2O(l)$

Neutralization Equations

• Reaction of strong acid (sulfuric acid) with strong base (potassium hydroxide).
• Molecular equation: $2 KOH(aq) + H<em>2SO</em>4(aq) \rightarrow 2 H<em>2O(l) + K</em>2SO_4(aq)$
• Complete ionic equation: $2 K^+(aq) + 2 OH^-(aq) + 2 H^+(aq) + SO<em>4^{-2}(aq) \rightarrow 2 H</em>2O(l) + 2 K^+(aq) + SO_4^{-2}(aq)$
• Net ionic equation: $OH^-(aq) + H^+(aq) \rightarrow H_2O(l)$

Weak Acid and Strong Base Reaction

• Reaction between weak acid (H3PO4) and strong base (NaOH).
• Molecular equation: $H<em>3PO</em>4(s) + 3 NaOH(aq) \rightarrow Na<em>3PO</em>4(aq) + 3 H_2O(l)$
• Complete ionic equation: $H<em>3PO</em>4(s) + 3 Na^{+1}(aq) + 3 OH^{-1}(aq) \rightarrow 3 Na^{+1}(aq) + PO<em>4^{-3}(aq) + 3 H</em>2O(l)$
• Net ionic equation: $H<em>3PO</em>4(s) + 3 OH^{-1}(aq) \rightarrow PO<em>4^{-3}(aq) + 3 H</em>2O(l)$

Weak Base and Strong Acid Reaction

• Reaction of weak base Mg(OH)2 with strong acid HNO3.
• Molecular Equation: $Mg(OH)<em>2(s) + 2 HNO</em>3(aq) \rightarrow Mg(NO<em>3)</em>2(aq) + 2 H_2O(l)$
• Complete Ionic Equation: $Mg(OH)<em>2(s) + 2 H^{+1}(aq) + 2 NO</em>3^{-1}(aq) \rightarrow Mg^{+2}(aq) + 2 NO<em>3^{-1}(aq) + 2 H</em>2O(l)$
• Net Ionic Equation: $Mg(OH)<em>2(s) + 2 H^{+1}(aq) \rightarrow Mg^{+2}(aq) + 2 H</em>2O(l)$

Titration

• Titration: Procedure in which a substance of known concentration (standard solution) is reacted with a substance of unknown concentration to determine the unknown’s concentration.
• An indicator may be added to mark when the reaction is complete.

Titration Calculation

• Steps:
  1. Use the volume or molarity of substance A to find moles of substance A.
  2. Use stoichiometric coefficients to convert moles of A to moles of B.
  3. Use the moles of substance B to find the volume or molarity of substance B.
     $M = \frac{mol}{L}$

Example Titration Problem

• Problem: Reacting 35.17 mL of 0.5065 M sodium hydroxide solution with 50.00 mL acetylsalicylic acid solution. Find the original concentration of the acid.
• Balanced equation: $HC<em>9H</em>7O<em>4 + NaOH \rightarrow NaC</em>9H<em>7O</em>4 + H_2O$
• Calculations:
  • Moles of $NaOH = 0.5065 \frac{mol NaOH}{L} * 0.03517 L = 0.01781 mol NaOH$
  • Moles of $HC<em>9H</em>7O<em>4 = 0.01781 mol NaOH * \frac{1 mol HC</em>9H<em>7O</em>4}{1 mol NaOH} = 0.01781 mol HC<em>9H</em>7O_4$
  • Molarity of $HC<em>9H</em>7O<em>4 = \frac{0.01781 mol HC</em>9H<em>7O</em>4}{0.05000 L} = 0.3562 M HC<em>9H</em>7O_4$
• Answer: 0.3562 M acetylsalicylic acid