Buffer Solutions and Henderson-Hasselbalch

🛡️ What Is a Buffer?

Part 1 of 7 — Resisting pH Change

What Makes a Buffer?

🔑 Why this matters: Buffers are essential in biochemistry (blood pH = 7.4), lab work, and industry — and they appear on every AP Chemistry exam.

What You'll Master in Part 1

• Defining what a buffer is and identifying buffer vs. non-buffer solutions
• Understanding why buffers need BOTH a weak acid AND its conjugate base
• Explaining qualitatively how buffers resist pH changes

📖 Buffer Definition

A buffer is a solution that resists changes in pH when small amounts of strong acid or strong base are added.

Composition

A buffer contains two key components:

Why Two Components?

• The weak acid neutralizes added base: $HA + OH^- \rightarrow A^- + H_2O$
• The conjugate base neutralizes added acid: $A^- + H^+ \rightarrow HA$

Neither component is consumed quickly because both are present in significant amounts — the pH changes only slightly!

What Does NOT Make a Buffer?

⚠️ Common AP Trap: Not every acid/base mixture is a buffer!

• Strong acid + strong base → complete reaction, no equilibrium
• Strong acid alone → no conjugate base reservoir
• Weak acid alone (no added conjugate base) → limited buffering
• A salt of a strong acid + strong base (e.g., $NaCl$)

🔧 How Buffers Maintain pH

Consider the acetic acid/acetate buffer ($CH_3COOH / CH_3COO^-$):

When Strong Acid ($H^+$) Is Added:

$CH_3COO^-(aq) + H^+(aq) \rightarrow CH_3COOH(aq)$

The conjugate base consumes the added $H^+$, converting it to weak acid. The pH barely changes because the ratio $[A^-]/[HA]$ changes only slightly.

When Strong Base ($OH^-$) Is Added:

$CH_3COOH(aq) + OH^-(aq) \rightarrow CH_3COO^-(aq) + H_2O(l)$

The weak acid consumes the added $OH^-$, converting it to conjugate base. Again, the ratio changes only slightly.

Key Insight

🔑 The buffer works because the added strong acid or base is completely consumed by reaction with one buffer component, and the $[A^-]/[HA]$ ratio changes only slightly if the buffer is concentrated enough.

Buffer Concept Check 🎯

🛡️ Common Buffer Systems

Biological Buffers

• Blood: Carbonic acid/bicarbonate system ($H_2CO_3/HCO_3^-$), maintained at pH 7.4
• Cells: Phosphate buffer system ($H_2PO_4^-/HPO_4^{2-}$), around pH 7.2
• Proteins: Amino acid side chains act as buffers

Buffer Identification 🔍

Buffer Component Identification 🧮

For each buffer, identify the missing component:

1) Buffer: $HNO_2$ / ___. What is the conjugate base? (Enter formula, e.g. NO2-)

2) Buffer: ___ / $NH_3$. What is the conjugate acid? (Enter formula, e.g. NH4+)

3) To make a phosphate buffer at pH ≈ 7.2, you mix $NaH_2PO_4$ with what? (Enter formula, e.g. Na2HPO4)

Exit Quiz — What Is a Buffer? ✅

⚔️ How Buffers Work — Neutralizing Added Acid or Base

Part 2 of 7 — Quantitative Buffer Calculations

The Two-Step Buffer Method

🔑 Why this matters: This two-step method is how every buffer problem on the AP exam is solved — master it and you can handle any buffer calculation.

What You'll Master in Part 2

• Setting up stoichiometry tables for buffer + strong acid/base
• Calculating new concentrations after neutralization
• Applying Henderson-Hasselbalch with the updated ratio

📋 The Two-Step Method

Step 1: Stoichiometry (Neutralization)

The added strong acid or base reacts completely with one buffer component:

Adding $H^+$: $A^- + H^+ \rightarrow HA$ (base component consumed)

Adding $OH^-$: $HA + OH^- \rightarrow A^- + H_2O$ (acid component consumed)

Calculate new moles of $HA$ and $A^-$ after reaction.

Step 2: Equilibrium (Henderson-Hasselbalch)

Use the new amounts to find the new pH:

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

Since both species are in the same volume, you can use moles instead of concentrations:

$pH = pK_a + \log\frac{\text{mol } A^-}{\text{mol } HA}$

🧪 Worked Example: Adding Strong Acid

Problem: A buffer contains 0.15 mol $CH_3COOH$ and 0.15 mol $CH_3COO^-$ in 1.0 L. What is the pH after adding 0.020 mol $HCl$?

$K_a = 1.8 \times 10^{-5}$, $pK_a = 4.74$

Solution:

Before Addition

$pH = 4.74 + \log(0.15/0.15) = 4.74 + 0 = 4.74$

Step 1: Stoichiometry

$CH_3COO^- + H^+ \rightarrow CH_3COOH$

Step 2: Henderson-Hasselbalch

$pH = 4.74 + \log\frac{0.13}{0.17} = 4.74 + \log(0.765) = 4.74 + (-0.12) = 4.62$

The pH only dropped from 4.74 to 4.62 — a change of just 0.12 units!

Without the buffer, adding 0.020 mol $HCl$ to 1.0 L water would give pH = $-\log(0.020) = 1.70$ — a change of 5.3 pH units!

🔑 Key Comparison: Buffer changed pH by 0.12 units. Without the buffer, the same acid would change pH by 5.3 units. This is why buffers matter!

Buffer Calculation Concepts 🎯

🧪 Worked Example: Adding Strong Base

Problem: Same buffer: 0.15 mol $CH_3COOH$ and 0.15 mol $CH_3COO^-$ in 1.0 L. What is the pH after adding 0.030 mol $NaOH$?

Solution:

Step 1: Stoichiometry

$CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$

Step 2: Henderson-Hasselbalch

$pH = 4.74 + \log\frac{0.18}{0.12} = 4.74 + \log(1.50) = 4.74 + 0.18 = 4.92$

pH increased from 4.74 to 4.92 — only 0.18 units despite adding a strong base!

Buffer Calculation Drill 🧮

A buffer has 0.20 mol $HF$ and 0.20 mol $NaF$ in 1.0 L. ($pK_a = 3.17$)

1) What is the initial pH of this buffer? (2 decimal places)

2) After adding 0.050 mol $HCl$, how many moles of $F^-$ remain? (2 decimal places)

3) What is the pH after adding 0.050 mol $HCl$? (2 decimal places)

🛡️ When Is a Buffer Destroyed?

A buffer is destroyed when all of one component is consumed:

• Adding enough $H^+$ to consume all the $A^-$ → no more base component
• Adding enough $OH^-$ to consume all the $HA$ → no more acid component

⚠️ AP Exam Alert: After a buffer is destroyed, you must switch to a simple strong acid/base calculation — Henderson-Hasselbalch no longer applies!

Example

Buffer: 0.15 mol $HA$ + 0.15 mol $A^-$

• Adding 0.15 mol $HCl$ → all $A^-$ consumed → buffer destroyed
• Adding 0.20 mol $HCl$ → $A^-$ gone, excess $H^+$ remains → no longer a buffer

After destruction, treat as a simple strong acid or base problem!

Buffer Reaction Reasoning 🔍

Exit Quiz — How Buffers Work ✅

📐 The Henderson-Hasselbalch Equation

Part 3 of 7 — The Master Buffer Equation

The Henderson-Hasselbalch Equation

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

🔑 Why this matters: Henderson-Hasselbalch is the single most important equation for buffer calculations — and the ratio shortcut saves enormous time on the AP exam.

What You'll Master in Part 3

• Deriving Henderson-Hasselbalch from the $K_a$ expression
• Using the equation to find pH, p$K_a$, or concentration ratios
• Applying the basic buffer version: pOH = p$K_b$ + log([$BH^{+}$]/[B])

📌 Derivation

Starting from the $K_a$ expression:

$K_a = \frac{[H^+][A^-]}{[HA]}$

Solve for $[H^+]$:

$[H^+] = K_a \cdot \frac{[HA]}{[A^-]}$

Take $-\log$ of both sides:

$-\log[H^+] = -\log K_a - \log\frac{[HA]}{[A^-]}$

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

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

Key Features

• When $[A^-] = [HA]$: $pH = pK_a + \log(1) = pK_a$
• When $[A^-] > [HA]$: $pH > pK_a$ (more basic)
• When $[A^-] < [HA]$: $pH < pK_a$ (more acidic)
• The log term adjusts pH relative to $pK_a$

🔑 Key Insight: The $pK_a$ is the "anchor" of a buffer’s pH. The log ratio just shifts the pH up or down from that anchor.

📌 Using the Henderson-Hasselbalch Equation

Example 1: Find pH

Problem: A buffer contains 0.30 M $CH_3COOH$ and 0.50 M $CH_3COO^-$. ($pK_a = 4.74$)

Solution:

$pH = 4.74 + \log\frac{0.50}{0.30} = 4.74 + \log(1.67) = 4.74 + 0.22 = 4.96$

Example 2: Find Ratio

Problem: What ratio of $[A^-]/[HA]$ gives pH = 5.00 for a buffer with $pK_a = 4.74$?

Solution:

$5.00 = 4.74 + \log\frac{[A^-]}{[HA]}$

$\log\frac{[A^-]}{[HA]} = 0.26$

$\frac{[A^-]}{[HA]} = 10^{0.26} = 1.82$

So you need about 1.8 times more conjugate base than acid.

Example 3: Equal Concentrations

When $[A^-] = [HA]$:

$pH = pK_a + \log(1) = pK_a$

This is a critically important result: the pH of a buffer with equal concentrations equals the $pK_a$!

🔑 AP Must-Know: When $[A^-] = [HA]$, $pH = pK_a$. This is the half-equivalence point during a titration and the point of maximum buffer capacity.

Henderson-Hasselbalch Concept Check 🎯

Henderson-Hasselbalch Drill 🧮

1) Buffer: 0.40 M $HF$ and 0.60 M $NaF$ ($pK_a = 3.17$). Find the pH. (2 decimal places)

2) What pH does a buffer with $pK_a = 9.25$ and $[NH_3]/[NH_4^+] = 0.50$ have? (2 decimal places)

3) For a buffer with $pK_a = 4.74$ at pH = 5.50, what is $[A^-]/[HA]$? (1 decimal place)

🛡️ Henderson-Hasselbalch for Basic Buffers

For a buffer made from a weak base ($NH_3$) and its conjugate acid ($NH_4^+$):

Method 1: Use $pK_a$ of the conjugate acid

$pH = pK_a(NH_4^+) + \log\frac{[NH_3]}{[NH_4^+]}$

where $pK_a(NH_4^+) = 14 - pK_b(NH_3) = 14 - 4.74 = 9.25$

Method 2: Use $pOH$ form

$pOH = pK_b + \log\frac{[BH^+]}{[B]}$

Then $pH = 14 - pOH$.

Both methods give the same answer. Method 1 is usually preferred for consistency.

💡 Tip: On the AP exam, always convert to $pK_a$ and use Method 1. This avoids sign errors with the $pOH$ form.

Henderson-Hasselbalch Reasoning 🔍

Exit Quiz — Henderson-Hasselbalch ✅

🎯 Buffer Capacity and Effective Range

Part 4 of 7 — How Much Can a Buffer Handle?

Buffer Capacity and Range

Effective range: p$K_a$ ± 1

🔑 Why this matters: The AP exam tests whether you know the limits of a buffer — not just how to calculate pH, but when the buffer fails.

What You'll Master in Part 4

• Defining buffer capacity and the factors that affect it
• Understanding why the effective range is p$K_a$ ± 1
• Predicting when a buffer has been overwhelmed

🛡️ Buffer Capacity

Buffer capacity = the amount of strong acid or base that can be added before the pH changes significantly (usually > 1 unit).

Factors That Affect Capacity

1. Concentration: More concentrated buffers have greater capacity

   • 1.0 M buffer > 0.10 M buffer > 0.010 M buffer
   • More moles of $HA$ and $A^-$ = more acid and base can be neutralized

2. Ratio of components: Buffers work best when $[A^-] \approx [HA]$

   • Equal concentrations = maximum capacity in both directions
   • If $[A^-] \gg [HA]$: good capacity for added acid, poor for added base
   • If $[HA] \gg [A^-]$: good capacity for added base, poor for added acid

Maximum Capacity

A buffer can neutralize added acid equal to the moles of $A^-$ present, and added base equal to the moles of $HA$ present. Beyond that, the buffer is destroyed.

🔑 Key Rule: Max acid neutralized = mol $A^-$. Max base neutralized = mol $HA$.

📊 Effective Buffer Range

A buffer is effective when the $[A^-]/[HA]$ ratio stays between 0.1 and 10:

$\frac{[A^-]}{[HA]} = 0.1 \rightarrow pH = pK_a + \log(0.1) = pK_a - 1$

$\frac{[A^-]}{[HA]} = 10 \rightarrow pH = pK_a + \log(10) = pK_a + 1$

The Rule

$\boxed{\text{Effective buffer range: } pK_a \pm 1}$

Examples

Why $pK_a \pm 1$?

Outside this range, one component is less than 10% of the other. There's not enough of it to provide meaningful buffering.

⚠️ AP Trap: If a problem gives you a buffer with $[A^-]/[HA] > 10$ or $< 0.1$, the buffer is outside its effective range and won’t resist pH changes well.

Buffer Capacity & Range Check 🎯

Buffer Capacity Calculations 🧮

A buffer contains 0.40 mol $CH_3COOH$ and 0.60 mol $CH_3COO^-$ in 2.0 L. ($pK_a = 4.74$)

1) What is the maximum moles of $HCl$ this buffer can absorb? (2 decimal places)

2) What is the maximum moles of $NaOH$ this buffer can absorb? (2 decimal places)

3) What is the pH after adding 0.30 mol $NaOH$? (2 decimal places)

📊 Effect of Dilution on Buffers

Adding water (dilution) to a buffer:

• Does NOT change pH (both $[HA]$ and $[A^-]$ decrease by the same factor, so the ratio stays the same)
• DOES decrease capacity (fewer moles of each component per liter)

🔑 Key Distinction: Dilution preserves pH but weakens the buffer. This is a common AP free-response question.

Example

Buffer: 0.50 M $HA$ / 0.50 M $A^-$ in 1.0 L

$pH = pK_a + \log(0.50/0.50) = pK_a$

Dilute to 2.0 L: 0.25 M $HA$ / 0.25 M $A^-$

$pH = pK_a + \log(0.25/0.25) = pK_a$ (same!)

But now there's half the buffering capacity.

Buffer Range & Capacity Reasoning 🔍

Exit Quiz — Buffer Capacity & Range ✅

🧪 Preparing Buffers

Part 5 of 7 — Choosing the Right Acid and Designing a Buffer

Buffer Preparation Strategy

🔑 Why this matters: Lab-based AP questions ask you to design a buffer at a specific pH — you need to know how to choose the acid and calculate amounts.

What You'll Master in Part 5

• Selecting a weak acid with p$K_a$ close to the target pH
• Calculating the required conjugate base-to-acid ratio
• Understanding alternative preparation methods (partial neutralization)

🧪 Step 1: Choose the Right Weak Acid

Rule: Choose a weak acid whose $pK_a$ is as close as possible to the desired pH.

🔑 AP Strategy: Always pick the acid with $pK_a$ nearest your target pH. This gives the ratio closest to 1:1 and the strongest buffer.

Why?

• The buffer is most effective at $pH = pK_a$ (equal concentrations of $HA$ and $A^-$)
• The buffer works in the range $pK_a \pm 1$
• Closer $pK_a$ to target pH → closer to 1:1 ratio → better capacity

Common Buffer Acids and Their $pK_a$ Values

🔢 Step 2: Calculate the Required Ratio

From the Henderson-Hasselbalch equation:

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

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

$\boxed{\frac{[A^-]}{[HA]} = 10^{(pH - pK_a)}}$

Worked Example

Problem: Prepare 1.0 L of a pH 5.00 buffer using acetic acid ($pK_a = 4.74$) with a total buffer concentration of 0.20 M.

Solution:

Step 1: Find the ratio

$\frac{[A^-]}{[HA]} = 10^{(5.00 - 4.74)} = 10^{0.26} = 1.82$

Step 2: Set up equations

Let $[HA] = x$ and $[A^-] = 1.82x$

$x + 1.82x = 0.20 \text{ M}$

$2.82x = 0.20$

$x = 0.071 \text{ M} = [HA]$

$[A^-] = 1.82(0.071) = 0.129 \text{ M}$

Step 3: Calculate moles for 1.0 L

• $CH_3COOH$: 0.071 mol
• $NaCH_3COO$: 0.129 mol

Buffer Preparation Concepts 🎯

📌 Alternative Preparation Methods

Method 2: Partial Neutralization

Instead of mixing weak acid + salt, you can add strong base to excess weak acid:

Example: To make an acetate buffer at pH 4.74:

Start with 0.20 mol $CH_3COOH$, then add 0.10 mol $NaOH$:

$CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$

• $CH_3COOH$ remaining: $0.20 - 0.10 = 0.10$ mol
• $CH_3COO^-$ formed: $0.10$ mol
• Ratio = 1:1 → $pH = pK_a = 4.74$ ✓

Method 3: Partial Neutralization of Base

Start with weak base + add strong acid:

Start with 0.20 mol $NH_3$, add 0.10 mol $HCl$:

$NH_3 + H^+ \rightarrow NH_4^+$

• $NH_3$ remaining: 0.10 mol
• $NH_4^+$ formed: 0.10 mol
• Buffer at $pH = pK_a(NH_4^+) = 9.25$

💡 Tip: Partial neutralization is how buffers are often made in lab. You don’t need to buy both the acid and its salt—just start with excess weak acid (or base) and neutralize part of it.

Buffer Preparation Calculations 🧮

1) What ratio $[A^-]/[HA]$ is needed for a buffer at pH 4.00 using acetic acid ($pK_a = 4.74$)? (2 decimal places)

2) You have 0.30 mol $CH_3COOH$. How many moles of $NaOH$ should you add to make a buffer at pH = 4.74? (2 decimal places)

3) To prepare 500 mL of pH 9.55 buffer from $NH_4Cl$ and $NH_3$ ($pK_a = 9.25$), with total concentration 0.40 M, how many moles of $NH_3$ are needed? (3 decimal places)

Buffer Design Reasoning 🔍

Exit Quiz — Preparing Buffers ✅

🛠️ Problem-Solving Workshop

Part 6 of 7 — Buffer Solutions & Henderson-Hasselbalch

Problem Types in This Workshop

🔑 Why this matters: AP free-response problems often chain 3-4 buffer calculations together — this workshop builds the stamina and accuracy you need.

What You'll Master in Part 6

• Solving multi-addition buffer problems step by step
• Distinguishing buffer solutions from non-buffer mixtures
• Calculating buffer capacity and identifying when a buffer is overwhelmed

🛡️ Problem 1: Buffer After Multiple Additions

Problem: A 1.0 L buffer contains 0.30 mol $HCOOH$ (formic acid, $pK_a = 3.75$) and 0.30 mol $HCOONa$ (sodium formate). Find the pH after adding 0.10 mol $NaOH$, then 0.05 mol $HCl$.

Solution:

Initial pH: $pH = 3.75 + \log(0.30/0.30) = 3.75$

Add 0.10 mol $NaOH$:

$HCOOH + OH^- \rightarrow HCOO^- + H_2O$

After: $HCOOH = 0.20$ mol, $HCOO^- = 0.40$ mol

$pH = 3.75 + \log(0.40/0.20) = 3.75 + 0.30 = 4.05$

Then add 0.05 mol $HCl$:

$HCOO^- + H^+ \rightarrow HCOOH$

After: $HCOO^- = 0.35$ mol, $HCOOH = 0.25$ mol

$pH = 3.75 + \log(0.35/0.25) = 3.75 + 0.15 = 3.90$

Your Turn: Sequential Additions 🧮

A 1.0 L buffer has 0.25 mol $CH_3COOH$ and 0.25 mol $CH_3COO^-$ ($pK_a = 4.74$).

First, 0.08 mol $NaOH$ is added. Then, 0.05 mol $HCl$ is added.

1) After adding $NaOH$, how many moles of $CH_3COO^-$ are present? (2 decimal places)

2) After adding $NaOH$, what is the pH? (2 decimal places)

3) After then adding $HCl$, what is the final pH? (2 decimal places)

🛡️ Problem 2: Buffer or Not?

Determine whether each mixture forms a buffer:

A) 50 mL of 0.20 M $HF$ + 50 mL of 0.10 M $NaOH$

$HF + OH^- \rightarrow F^- + H_2O$

mol $HF$ = 0.010, mol $OH^-$ = 0.005

After: $HF = 0.005$ mol, $F^- = 0.005$ mol, $OH^-$ = 0 ✅ Buffer!

B) 50 mL of 0.20 M $HF$ + 50 mL of 0.20 M $NaOH$

mol $HF$ = 0.010, mol $OH^-$ = 0.010

After: $HF = 0$, $F^- = 0.010$, $OH^-$ = 0 ❌ Not a buffer (only $F^-$ remains)

C) 50 mL of 0.20 M $HCl$ + 50 mL of 0.10 M $NaCl$

$HCl$ is a strong acid. ❌ Not a buffer (no weak acid/base pair)

⚠️ AP Trap: A mixture of a strong acid + its salt (like $HCl$ + $NaCl$) is NEVER a buffer. You need a WEAK acid/base pair.

Buffer Identification 🎯

Problem 3: Buffer Design 🧮

Design a 500 mL phosphate buffer at pH 7.40 ($pK_a = 7.21$, total phosphate = 0.20 M).

1) What is the required $[HPO_4^{2-}]/[H_2PO_4^-]$ ratio? (2 decimal places)

2) What concentration of $H_2PO_4^-$ is needed? (Enter in M, 3 decimal places)

3) How many moles of $Na_2HPO_4$ are needed for 500 mL? (3 decimal places)

Workshop Synthesis 🔍

Exit Quiz — Problem-Solving Workshop ✅

🎓 Synthesis & AP Review

Part 7 of 7 — Buffer Solutions & Henderson-Hasselbalch

Buffer Mastery Checklist

🔑 Why this matters: Buffers connect to titrations, equilibrium, and biochemistry — expect cross-topic AP questions that use buffers as the foundation.

What You'll Master in Part 7

• Tackling AP-style questions that integrate buffers with titrations and equilibrium
• Writing clear free-response explanations of buffer mechanisms
• Avoiding the most common AP exam mistakes in buffer problems

📋 Complete Summary

Buffer Essentials

⚠️ Remember: Once a buffer is destroyed, Henderson-Hasselbalch no longer applies. Switch to a simple acid/base or salt calculation.

Problem-Solving Strategy

🔑 5-Step Buffer Method for the AP Exam:

1. Identify if a buffer exists (weak acid + conjugate base?)
2. Stoichiometry first — react any added strong acid/base completely
3. Check if buffer survives (both components still present?)
4. Henderson-Hasselbalch to find new pH using remaining moles
5. If destroyed — treat as simple acid, base, or salt problem

AP-Style Questions — Set 1 🎯

AP Calculation Practice 🧮

1) A formate buffer ($pK_a = 3.75$) has pH = 4.05. What is the $[HCOO^-]/[HCOOH]$ ratio? (1 decimal place)

2) 0.020 mol $NaOH$ is added to 500 mL of a buffer with 0.15 M $CH_3COOH$ and 0.15 M $CH_3COO^-$ ($pK_a = 4.74$). What is the new pH? (2 decimal places)

3) What is the effective buffer range for ammonium/ammonia ($pK_a = 9.25$)? Enter the lower limit of the range. (2 decimal places)

AP-Style Questions — Set 2 🎯

Comprehensive Review 🔍

Final Exit Quiz — Buffers & Henderson-Hasselbalch ✅