#Mastering Buffers: Your Ultimate Guide to the Henderson-Hasselbalch Equation

Hey AP Chem student! Let's dive into the Henderson-Hasselbalch equation, your secret weapon for tackling buffer problems. This guide will break it down, step-by-step, so you feel confident and ready for anything the exam throws at you. Let's get started!

#Introduction to Buffers and the Henderson-Hasselbalch Equation

Buffers are crucial in chemistry and biology, and the Henderson-Hasselbalch equation is your go-to tool for understanding them. Remember, buffers resist changes in pH and consist of a weak acid and its conjugate base (or a weak base and its conjugate acid). Let's get cozy with this equation:

#Breaking Down the Equation

• pH: The measure of acidity, calculated as -log[H+]. Often, this is what you're solving for.
• pKa: -log(Ka), which tells us about the acid's strength. Remember, a lower pKa means a stronger acid.
• log([A-]/[HA]): This is the log of the ratio of the conjugate base [A-] to the weak acid [HA]. This ratio is key to understanding buffer behavior. 💡

#Applying the Henderson-Hasselbalch Equation: Example Problems

Let's see how to use this equation in action. We'll tackle two common types of problems:

#Example Problem #1: Directly Stated Buffer

Problem: Find the pH of a buffer with 0.5M CH3COOH mixed with 0.25M CH3COONa (Ka = 1.8 * 10^-5).

Solution:

1. Calculate pKa: pKa = -log(1.8 * 10^-5) = 4.74

2. Apply the Henderson-Hasselbalch equation:

   pH = 4.74 + log(0.25/0.5) = 4.74 + log(0.5) = 4.74 - 0.30 = 4.44

#Example Problem #2: Using the Henderson-Hasselbalch During A Titration

Problem: Calculate the pH in the titration of 25.0 mL of 0.100M acetic acid with 0.100M NaOH after adding 15.0 mL of 0.100M NaOH.

Solution:

1. Net Ionic Equation: CH3COOH + OH- <=> CH3COO- + H2O

2. Stoichiometry:

   • Initial mmol CH3COOH: 25.0 mL * 0.100 M = 2.5 mmol
   • mmol OH- added: 15.0 mL * 0.100 M = 1.5 mmol
   • After reaction: 1.0 mmol CH3COOH, 0 mmol OH-, 1.5 mmol CH3COO-

3. Apply the Henderson-Hasselbalch equation:

   pH = 4.74 + log(1.5/1.0) = 4.74 + log(1.5) = 4.74 + 0.18 = 4.92

#Final Exam Focus

Okay, let's get down to the nitty-gritty for the exam. Here's what to focus on:

• Master the Equation: Know the Henderson-Hasselbalch equation inside and out. Understand what each component represents and how they relate to each other.
• Buffer Calculations: Be prepared to calculate the pH of buffers, both directly stated and those formed during titrations.
• Titration Curves: Understand how the Henderson-Hasselbalch equation applies to the buffer region of a titration curve. Recognize the half-equivalence point where pH = pKa.
• Conceptual Understanding: Don't just memorize formulas. Understand why buffers work and how they resist pH changes. 💡
• Time Management: Practice these problems under timed conditions to build speed and accuracy.

#Practice Questions

Let's test your knowledge with some practice questions!

Practice Question

#Multiple Choice Questions

1. A buffer solution is prepared by mixing 25 mL of 0.10 M weak acid HA with 25 mL of 0.10 M of its conjugate base NaA. The pH of the resulting buffer solution is 5.0. What is the pKa of the weak acid HA? (A) 2.0 (B) 3.0 (C) 5.0 (D) 7.0

2. Which of the following mixtures would create a buffer solution? (A) 100 mL of 0.1 M HCl and 50 mL of 0.1 M NaOH (B) 50 mL of 0.1 M HCl and 100 mL of 0.1 M NaOH (C) 100 mL of 0.1 M CH3COOH and 50 mL of 0.1 M NaOH (D) 50 mL of 0.1 M CH3COOH and 100 mL of 0.1 M NaOH

#Free Response Question

A 50.0 mL sample of 0.20 M hydrofluoric acid (HF) is titrated with 0.20 M sodium hydroxide (NaOH). The Ka of HF is 6.8 × 10⁻⁴.

(a) Write the balanced net ionic equation for the reaction between HF and NaOH.

(b) Calculate the pH of the solution before any NaOH is added.

(c) Calculate the pH of the solution after 25.0 mL of NaOH has been added.

(d) Calculate the volume of NaOH required to reach the equivalence point.

(e) Calculate the pH of the solution at the equivalence point.

#FRQ Scoring Breakdown

(a) (1 point)

• HF(aq) + OH⁻(aq) → F⁻(aq) + H₂O(l)

(b) (2 points)

• Set up an ICE table for the dissociation of HF:

  	HF	H+	F-
  Initial	0.20	0	0
  Change	-x	+x	+x
  Equil	0.20-x	x	x

• Ka = [H+][F-] / [HF] = x² / (0.20 - x) = 6.8 × 10⁻⁴

• Assume x is small: x² / 0.20 = 6.8 × 10⁻⁴

• x = [H+] = 0.0117 M

• pH = -log(0.0117) = 1.93

(c) (3 points)

• Initial moles of HF = 0.050 L * 0.20 M = 0.010 mol
• Moles of NaOH added = 0.025 L * 0.20 M = 0.005 mol
• Moles of HF remaining = 0.010 - 0.005 = 0.005 mol
• Moles of F⁻ formed = 0.005 mol
• pH = pKa + log([F⁻]/[HF])
• pKa = -log(6.8 × 10⁻⁴) = 3.17
• pH = 3.17 + log(0.005/0.005) = 3.17

(d) (1 point)

• At the equivalence point, moles of NaOH = moles of HF
• Volume of NaOH = (0.010 mol) / (0.20 M) = 0.050 L = 50.0 mL

(e) (3 points)

• At equivalence point, we have 0.010 mol of F⁻ in 100 mL of solution
• [F⁻] = 0.010 mol / 0.100 L = 0.10 M
• F⁻ + H₂O <=> HF + OH⁻
• Kb = Kw / Ka = 1.0 × 10⁻¹⁴ / 6.8 × 10⁻⁴ = 1.47 × 10⁻¹¹
• Kb = [HF][OH⁻] / [F⁻] = x² / (0.10 - x)
• Assume x is small: x² / 0.10 = 1.47 × 10⁻¹¹
• x = [OH⁻] = 1.21 × 10⁻⁶ M
• pOH = -log(1.21 × 10⁻⁶) = 5.92
• pH = 14 - 5.92 = 8.08

That's it! You've now got a solid grasp on the Henderson-Hasselbalch equation and how to apply it. Remember to practice, stay calm, and you'll do great on the AP Chemistry exam. You've got this! 💪