Weak Acid and Base Equilibria

Sophie Anderson
7 min read
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Study Guide Overview
This study guide covers acids and bases, focusing on weak acids and bases. It reviews acid-base definitions (Arrhenius, Bronsted-Lowry), strong/weak acid/base identification, and the concept of equilibrium. It explains Ka and Kb expressions and calculations, including the use of ICE tables for determining pH. Finally, it provides practice questions and exam tips covering common question types, time management strategies, and high-priority topics for the AP Chemistry exam.
#AP Chemistry: Acids and Bases - The Night Before 🌙
Hey, future chemist! Let's get you prepped and confident for your AP Chemistry exam. We're diving into the world of acids and bases, focusing on those tricky weak ones. Buckle up, and let's make this stick!
#Acid-Base Basics: Strong vs. Weak
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Defining Acids and Bases
- Arrhenius Definition: Acids produce H⁺ ions in water, and bases produce OH⁻ ions.
- Bronsted-Lowry Definition: Acids are proton (H⁺) donors, and bases are proton acceptors.
- For our purposes, we'll primarily use the Arrhenius definition to keep things simple.
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Strong Acids and Bases
- Strong Acids: Completely dissociate in water (virtually 100% ionization).
- Memorize these 7: HCl, HBr, HI, HNO₃, H₂SO₄, HClO₃, HClO₄
- They break down into H⁺ and their conjugate base (e.g., HCl → H⁺ + Cl⁻).
- Strong Bases: Also completely dissociate in water.
- Alkali and alkaline earth metal hydroxides: NaOH, KOH, Ca(OH)₂, Sr(OH)₂, Ba(OH)₂
- They break down into metal cation and OH⁻ (e.g., NaOH → Na⁺ + OH⁻).
Conjugates of Strong Acids/Bases: Their conjugates (like Cl⁻ from HCl) are so weak they're practically neutral. They don't contribute to the pH.
Image: Visual representation of strong acids fully dissociating and weak acids partially dissociating.
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Weak Acids and Bases
- Weak Acids/Bases: Incompletely dissociate in water. They exist in equilibrium between the undissociated and dissociated forms.
- Weak Acid Example: Acetic acid (CH₃COOH) is a classic weak acid.
- Only some of it breaks down into H⁺ and CH₃COO⁻.
- Weak Base Example: Ammonia (NH₃) is the most common weak base.
- Reacts with water to form NH₄⁺ and OH⁻.
- Key takeaway: Any acid not in the list of 7 strong acids is weak.
Memory Aid: Think of strong acids and bases as 'all-in' - they fully commit to dissociation. Weak acids and bases are more 'hesitant' and only partially dissociate.
#Equilibrium: A Quick Refresher
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What is Equilibrium?
- Dynamic State: Forward and reverse reactions occur at the same rate. No net change in concentrations of reactants and products.
- Not Static: Reactions are still happening, but the overall concentrations remain constant.
Image: Graph showing reactant and product concentrations reaching a steady state at equilibrium.
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The Equilibrium Constant (K)
- Ratio of Products to Reactants: K = [Products]^coefficients / [Reactants]^coefficients
- Indicates Reaction Extent:
- Large K: Favors products
- Small K: Favors reactants
- ICE/RICE Tables: Used to find equilibrium concentrations. Remember to use initial concentrations, changes, and equilibrium concentrations.
#Weak Acid-Base Equilibria: Putting It All Together
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Weak Acid Dissociation (Ka)
- Ka Expression: HA ⇌ H⁺ + A⁻. Ka = [H⁺][A⁻] / [HA]
- Smaller Ka = Weaker Acid: Indicates less dissociation.
Memory Aid: Ka is like the 'acid strength score' - the smaller, the weaker!
Example Problem: Find the pH of a 2M solution of CH₃COOH (Ka = 1.8 × 10⁻⁵).
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Write the Equilibrium: CH₃COOH ⇌ H⁺ + CH₃COO⁻
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Set up ICE Table:
CH₃COOH H⁺ CH₃COO⁻ Initial (I) 2 0 0 Change (C) -x +x +x Equil (E) 2-x x x -
Ka Expression: 1.8 × 10⁻⁵ = x² / (2-x) ≈ x² / 2 (since x is small)
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Solve for x: x = [H⁺] ≈ 0.006
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Calculate pH: pH = -log(0.006) = 2.22
Image: ICE table setup for a weak acid dissociation.
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Weak Base Dissociation (Kb)
- Kb Expression: B + H₂O ⇌ BH⁺ + OH⁻. Kb = [BH⁺][OH⁻] / [B]
- Smaller Kb = Weaker Base: Indicates less dissociation.
Memory Aid: Kb is like the 'base strength score' - the smaller, the weaker!
Example Problem: Find the pH of 1M ammonia (NH₃). (Kb = 1.8 × 10⁻⁵).
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Write the Equilibrium: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻
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Set up ICE Table:
NH₃ NH₄⁺ OH⁻ Initial (I) 1 0 0 Change (C) -x +x +x Equil (E) 1-x x x -
Kb Expression: 1.8 × 10⁻⁵ = x² / (1-x) ≈ x² / 1
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Solve for x: x = [OH⁻] ≈ 0.0042
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Calculate pOH: pOH = -log(0.0042) = 2.38
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Calculate pH: pH = 14 - 2.38 = 11.62
Image: ICE table setup for a weak base dissociation.
Common Mistake Alert! Don't forget to convert pOH to pH when working with bases! pH + pOH = 14
#Final Exam Focus
- High-Priority Topics:
- Distinguishing between strong and weak acids/bases
- Setting up and solving equilibrium problems using ICE tables
- Calculating pH for both weak acids and weak bases
- Understanding the meaning of Ka and Kb
- Common Question Types:
- Multiple choice questions asking for pH calculations
- Free response questions involving equilibrium calculations and analysis of weak acid/base systems
- Time Management:
- Practice setting up ICE tables quickly.
- Make sure you are comfortable with log calculations.
- When possible, use the approximation (x is small) to simplify calculations.
- Strategies:
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Always write out the balanced equation first.
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Double-check whether you're solving for pH or pOH.
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Pay close attention to the given values (Ka, Kb, initial concentrations).
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Practice Question
#Practice Questions
#Multiple Choice Questions
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Which of the following is a strong acid? (A) CH₃COOH (B) HF (C) HNO₃ (D) H₂CO₃
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A 0.1 M solution of a weak base has a pH of 11. What is the pOH of the solution? (A) 1 (B) 3 (C) 7 (D) 11
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Which of the following statements is true regarding the dissociation of a weak acid in water? (A) The concentration of H+ ions is equal to the initial concentration of the acid. (B) The equilibrium favors the formation of the undissociated acid. (C) The equilibrium constant Ka is very large. (D) The acid completely dissociates into ions.
#Free Response Question
A 0.20 M solution of a weak acid, HA, has a pH of 3.00. (a) Write the equilibrium reaction for the dissociation of HA in water.
(b) Calculate the [H+].
(c) Calculate the Ka of the weak acid.
(d) If the concentration of the weak acid is increased, how will the pH of the solution change? Explain.
Scoring Breakdown:
(a) 1 point for correct equilibrium reaction: HA(aq) ⇌ H+(aq) + A-(aq)
(b) 1 point for correct [H+] calculation: [H+] = 10^-3 M = 0.001 M
(c) 3 points for correct Ka calculation: * 1 point for setting up the ICE table * 1 point for correct Ka expression: Ka = [H+][A-] / [HA] * 1 point for correct calculation: Ka = (0.001)(0.001) / (0.20 - 0.001) ≈ 5.0 x 10^-6
(d) 2 points for correct explanation: * 1 point for stating that the pH will decrease. * 1 point for explaining that increasing the concentration of the weak acid will increase the [H+] in the solution, thus lowering the pH.

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