#Acid-Base Reactions and Buffers: Your Ultimate Guide 🧪

Hey there, future AP Chem master! Let's dive into the exciting world of acid-base reactions and buffers. This is where things get really interesting, so buckle up! We'll make sure you're not just memorizing but truly understanding these concepts. Let's get started!

#Introduction to Buffers

#What are Buffers?

Buffers are like the superheroes of the chemistry world! They're mixtures of a weak acid (HA) and its conjugate base (A-) that resist drastic changes in pH. Think of them as the ultimate pH stabilizers. When you add acid or base to a buffer, it's like they shrug it off, keeping the pH relatively constant. How cool is that? 😎

Buffers are created through reactions between acids and bases, which we'll explore next.

Image from UOregon: Buffers in action!

#Reactions with Acids and Bases

#Strong Acid - Strong Base Reactions

These are the classic acid-base reactions you might have seen earlier. Think of reactions like: HCl + NaOH → NaCl + H2O

Let's break down a typical problem:

Example: What is the pH after adding 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl?

1. Write the reaction: HCl + NaOH <=> NaCl + H2O
2. Net ionic equation: H+ + OH- <=> H2O
3. Calculate millimoles (mmol):
   • NaOH: 10.0 mL * 0.100 M = 1 mmol OH-
   • HCl: 25.0 mL * 0.100 M = 2.5 mmol H+
4. Stoichiometry: The reaction consumes 1 mmol of both OH- and H+, leaving 1.5 mmol of H+

Image from Mister Chemistry: Stoichiometry in action!

5. Calculate [H+]: 1.5 mmol / 35 mL = 0.0429 M
6. Calculate pH: pH = -log(0.0429) = 1.37

#Weak Acid - Strong Base Reactions

These are similar to the previous reactions, but with a twist! The weak acid doesn't fully dissociate, changing the net ionic equation. For example: CH3COOH + NaOH <=> CH3COONa + H2O

Here comes the buffer! When you have a mixture of a weak acid and its conjugate base, you've got a buffer! To calculate the pH of a buffer, you can use the Henderson-Hasselbalch equation:

$$pH = pK_a + log \frac{[A^-]}{[HA]}$$

Image from MicrobeNotes: The Henderson-Hasselbalch Equation

Let's see this in action:

Example: Calculate the pH after adding 10.00 mL of 0.100 M NaOH to 25.0 mL of 0.100 M acetic acid (CH3COOH).

1. Calculate millimoles:
   • NaOH: 10.00 mL * 0.100 M = 1 mmol OH-
   • CH3COOH: 25.0 mL * 0.100 M = 2.5 mmol CH3COOH
2. Stoichiometry: 1 mmol of OH- reacts with 1 mmol of CH3COOH, leaving 1.5 mmol of CH3COOH and creating 1 mmol of CH3COO-

Image from Mister Chemistry: More Stoichiometry!

3. Use the Henderson-Hasselbalch equation:

$$pH = pK_a + log \frac{[CH_3COO^-]}{[CH_3COOH]}$$

$$pH = 4.74 + log \frac{1}{1.5} = 4.56$$

Image from Mister Chemistry: Applying the Henderson-Hasselbalch Equation

#Final Exam Focus

Alright, you've made it this far! Here's a quick rundown of the most important things to remember for exam day:

• Buffers are key: Understand their composition and how they resist pH changes.
• Net ionic equations: Always start with the net ionic equation.
• Stoichiometry: Use stoichiometry to find the amounts of reactants and products.
• Henderson-Hasselbalch equation: Know how and when to use it for buffer calculations.
• Regions in titrations: Be familiar with the different regions (acid only, buffer, equivalence, post-equivalence).

#Last-Minute Tips

• Time Management: Don't get bogged down on one question. Move on and come back if you have time.
• Common Pitfalls: Watch out for pH vs. pOH, and make sure you're using the correct equations.
• Challenging Questions: Break down complex problems into smaller, manageable steps.

#Practice Questions

Practice Question

Multiple Choice Questions

1. A buffer solution is prepared by mixing 50.0 mL of 0.10 M acetic acid (CH3COOH) and 50.0 mL of 0.10 M sodium acetate (CH3COONa). What is the approximate pH of this buffer solution? (The pKa of acetic acid is 4.74) (A) 2.87 (B) 3.74 (C) 4.74 (D) 5.74

2. Which of the following is the net ionic equation for the reaction of a weak acid, such as hydrofluoric acid (HF), with a strong base, such as sodium hydroxide (NaOH)? (A) H+ + OH- <=> H2O (B) HF + OH- <=> F- + H2O (C) Na+ + F- <=> NaF (D) HF <=> H+ + F-

Free Response Question

A 25.0 mL sample of 0.20 M hypochlorous acid (HClO), a weak acid, is titrated with 0.10 M sodium hydroxide (NaOH). The Ka of HClO is 3.0 x 10^-8. (a) Write the net ionic equation for the reaction between HClO and NaOH. (b) Calculate the initial pH of the 0.20 M HClO solution. (c) Calculate the pH at the half-equivalence point. (d) Calculate the volume of 0.10 M NaOH required to reach the equivalence point. (e) Calculate the pH at the equivalence point.

Answer Key and Point Breakdown for FRQ

(a) HClO + OH- <=> ClO- + H2O (1 point)

(b) Set up an ICE table for the dissociation of HClO:

Initial: [HClO] = 0.20, [H+] = 0, [ClO-] = 0 Change: [HClO] = -x, [H+] = +x, [ClO-] = +x Equilibrium: [HClO] = 0.20-x, [H+] = x, [ClO-] = x

Ka = [H+][ClO-]/[HClO] = 3.0 x 10^-8 = x^2 / (0.20 - x) Assume x is small, so 3.0 x 10^-8 ≈ x^2 / 0.20 x = 7.75 x 10^-5 = [H+] pH = -log(7.75 x 10^-5) = 4.11 (3 points)

(c) At the half-equivalence point, pH = pKa pKa = -log(3.0 x 10^-8) = 7.52 (1 point)

(d) Moles of HClO = 0.025 L * 0.20 M = 0.005 moles Moles of NaOH at equivalence = 0.005 moles Volume of NaOH = 0.005 moles / 0.10 M = 0.050 L = 50.0 mL (2 points)

(e) At the equivalence point, we have only ClO- [ClO-] = 0.005 moles / (0.025 L + 0.050 L) = 0.0667 M ClO- + H2O <=> HClO + OH- Kb = Kw/Ka = 1.0 x 10^-14 / 3.0 x 10^-8 = 3.33 x 10^-7 Kb = [HClO][OH-]/[ClO-] = 3.33 x 10^-7 = x^2 / 0.0667 x = 1.49 x 10^-4 = [OH-] pOH = -log(1.49 x 10^-4) = 3.83 pH = 14 - 3.83 = 10.17 (3 points)

You've got this! Remember, chemistry is all about understanding the concepts and applying them. You're well on your way to acing that exam! 🚀