🎯⭐ INTERACTIVE LESSON

Buffer Solutions and Henderson-Hasselbalch

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Buffer Solutions and Henderson-Hasselbalch - Complete Interactive Lesson

Part 1: What Is a Buffer?

🛡️ What Is a Buffer?

Part 1 of 7 — Resisting pH Change

Buffer solutions are among the most important concepts in AP Chemistry and biochemistry. They maintain a nearly constant pH even when small amounts of acid or base are added. Your blood, for example, is buffered at pH 7.4!

📖 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:

Buffer Type	Component 1	Component 2	Example
Acidic buffer	Weak acid ( $HA$ )	Conjugate base ( $A^-$ )	$CH_3COOH / CH_3COO^-$
Basic buffer	Weak base ( $B$ )	Conjugate acid ( $BH^+$ )	$NH_3 / NH_4^+$

Why Two Components?

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

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

What Does NOT Make 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 ( $CH_3COOH / CH_3COO^-$ ):

Buffer Concept Check 🎯

🛡️ Common Buffer Systems

Buffer System	Weak Acid	Conjugate Base	Approximate pH Range
Acetic acid/Acetate	$CH_3COOH$	$CH_3COO^-$

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+)

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

Exit Quiz — What Is a Buffer? ✅

Part 2: Henderson-Hasselbalch Equation

⚔️ How Buffers Work — Neutralizing Added Acid or Base

Part 2 of 7 — Quantitative Buffer Calculations

Now we'll calculate exactly how much the pH changes when acid or base is added to a buffer. The key is treating the neutralization as a stoichiometry problem first, then an equilibrium problem.

📋 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)

Part 3: Preparing Buffers

📐 The Henderson-Hasselbalch Equation

Part 3 of 7 — The Master Buffer Equation

The Henderson-Hasselbalch equation is the most important formula for buffer calculations. It directly relates pH to the $pK_a$ and the ratio of conjugate base to acid.

📌 Derivation

Starting from the $K_a$ expression:

Part 4: Buffer Capacity

🎯 Buffer Capacity and Effective Range

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

A buffer doesn't have unlimited ability to resist pH changes. Its buffer capacity tells us how much strong acid or base it can neutralize before the pH changes significantly. Its effective range tells us the pH range where it works well.

🛡️ 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

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
Ratio of components: Buffers work best when

Part 5: Adding Acid or Base to Buffers

🧪 Preparing Buffers

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

In the lab, you need to prepare buffers at specific pH values with specific capacities. This requires choosing the right weak acid and calculating the correct amounts of acid and conjugate base.

🧪 Step 1: Choose the Right Weak Acid

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

Why?

The buffer is most effective at $pH = pK_a$ (equal concentrations of and )

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

Part 6 of 7 — Buffer Solutions & Henderson-Hasselbalch

This workshop features multi-step problems combining buffer identification, pH calculation, capacity analysis, and preparation — the types of questions that appear on AP Chemistry free-response sections.

🛡️ Problem 1: Buffer After Multiple Additions

A 1.0 L buffer contains 0.30 mol $HCOOH$ (formic acid, $pK_a = 3.75$ ) and 0.30 mol (sodium formate).

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

Part 7 of 7 — Buffer Solutions & Henderson-Hasselbalch

This comprehensive review covers all buffer concepts: composition, mechanism, Henderson-Hasselbalch calculations, capacity, effective range, preparation, and multi-step problems.

📋 Complete Summary

Buffer Essentials

Concept	Key Point
Composition	Weak acid + conjugate base (or weak base + conjugate acid)
Mechanism	$HA$ neutralizes added $OH^-$ ; neutralizes added

The conjugate base neutralizes added acid:

A^- + H^+ \rightarrow HA

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.

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.

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

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

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 $HCl$ ?

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

Before Addition

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

Step 1: Stoichiometry

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

	$CH_3COO^-$	$H^+$

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!

Buffer Calculation Concepts 🎯

🧪 Worked Example: Adding Strong Base

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 $NaOH$ ?

Step 1: Stoichiometry

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

	$CH_3COOH$	$OH^-$	$C$

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

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 remains → no longer a buffer

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

Buffer Reaction Reasoning 🔍

Exit Quiz — How Buffers Work ✅

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

K_{a} = \frac{[ H ^{+} ] [ A ^{-} ]}{[ H A ]}

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

Take $-\log$ of both sides:

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

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

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

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 $pK_a$

📌 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 $pK_a$ !

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.

Henderson-Hasselbalch Reasoning 🔍

Exit Quiz — Henderson-Hasselbalch ✅

[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

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.

📊 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

Buffer System	$pK_a$	Effective Range
$HF/F^-$	3.17	pH 2.17 – 4.17

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.

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)

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 ✅

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

Target pH	Best Acid	$pK_a$
3 – 4	Formic acid ( $HCOOH$ )	3.75
4 – 5	Acetic acid ( $CH_3COOH$ )	4.74
6 – 8	$H_2PO_4^-$ (phosphate)	7.21
7 – 8	Tris buffer	8.07
9 – 10	$NH_4^+$ (ammonium)	9.25
9 – 11	$HCO_3^-$ (bicarbonate)	10.33

🔢 Step 2: Calculate the Required Ratio

From the Henderson-Hasselbalch equation:

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

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

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

Worked Example

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.

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: mol

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 $p H = p_{}$

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 are needed? (3 decimal places)

Buffer Design Reasoning 🔍

Exit Quiz — Preparing Buffers ✅

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 ✅

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 ❌ (only 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)

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 ✅

Problem-Solving Strategy

Identify if a buffer exists (weak acid + conjugate base?)
Stoichiometry first — react any added strong acid/base completely
Check if buffer survives (both components still present?)
Henderson-Hasselbalch to find new pH using remaining moles
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 $CH_3COO^-$ (). 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 ✅

CH_3COOH/CH_3COO^-

C H_{3} COO H / C H_{3} CO O^{-}

Ratio = 1:1 →

pH = pK_a = 4.74

✓

p H = p K_{a} (N H_{4}^{+}) = 9.25

Buffer Solutions and Henderson-Hasselbalch

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

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

Key Insight

Biological Buffers

Step 2: Equilibrium (Henderson-Hasselbalch)

🧪 Worked Example: Adding Strong Acid

Before Addition

Step 1: Stoichiometry

Step 2: Henderson-Hasselbalch

🧪 Worked Example: Adding Strong Base

Step 1: Stoichiometry

Step 2: Henderson-Hasselbalch

🛡️ When Is a Buffer Destroyed?

Example

Key Features

📌 Using the Henderson-Hasselbalch Equation

Example 1: Find pH

Example 2: Find Ratio

Example 3: Equal Concentrations

🛡️ Henderson-Hasselbalch for Basic Buffers

Maximum Capacity

📊 Effective Buffer Range

The Rule

Examples

Why $pK_a \pm 1$ ?

📊 Effect of Dilution on Buffers

Example

Common Buffer Acids and Their $pK_a$ Values

🔢 Step 2: Calculate the Required Ratio

Worked Example

📌 Alternative Preparation Methods

Method 2: Partial Neutralization

Method 3: Partial Neutralization of Base

🛡️ Problem 2: Buffer or Not?

Problem-Solving Strategy

Before	0.15 mol	0.020 mol	0.15 mol
Change	-0.020	-0.020	+0.020
After	0.13 mol	0	0.17 mol

Before	0.15 mol	0.030 mol	0.15 mol
Change	-0.030	-0.030	+0.030
After	0.12 mol	0	0.18 mol

Buffer Solutions and Henderson-Hasselbalch

Buffer Solutions and Henderson-Hasselbalch - Complete Interactive Lesson

Part 1: What Is a Buffer?

🛡️ What Is a Buffer?

📖 Buffer Definition

Composition

Why Two Components?

What Does NOT Make a Buffer?

🔧 How Buffers Maintain pH

🛡️ Common Buffer Systems

Part 2: Henderson-Hasselbalch Equation

⚔️ How Buffers Work — Neutralizing Added Acid or Base

📋 The Two-Step Method

Step 1: Stoichiometry (Neutralization)

Part 3: Preparing Buffers

📐 The Henderson-Hasselbalch Equation

📌 Derivation

Part 4: Buffer Capacity

🎯 Buffer Capacity and Effective Range

🛡️ Buffer Capacity

Factors That Affect Capacity

Part 5: Adding Acid or Base to Buffers

🧪 Preparing Buffers

🧪 Step 1: Choose the Right Weak Acid

Why?

Part 6: Problem-Solving Workshop

🛠️ Problem-Solving Workshop

🛡️ Problem 1: Buffer After Multiple Additions

Part 7: Synthesis & AP Review

🎓 Synthesis & AP Review

📋 Complete Summary

Buffer Essentials

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

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

Key Insight

Biological Buffers

Step 2: Equilibrium (Henderson-Hasselbalch)

🧪 Worked Example: Adding Strong Acid

Before Addition

Step 1: Stoichiometry

Step 2: Henderson-Hasselbalch

🧪 Worked Example: Adding Strong Base

Step 1: Stoichiometry

Step 2: Henderson-Hasselbalch

🛡️ When Is a Buffer Destroyed?

Example

Key Features

📌 Using the Henderson-Hasselbalch Equation

Example 1: Find pH

Example 2: Find Ratio

Example 3: Equal Concentrations

🛡️ Henderson-Hasselbalch for Basic Buffers

Maximum Capacity

📊 Effective Buffer Range

The Rule

Examples

Why pKa±1pK_a \pm 1pKa​±1?

📊 Effect of Dilution on Buffers

Example

Common Buffer Acids and Their pKapK_apKa​ Values

🔢 Step 2: Calculate the Required Ratio

Worked Example

📌 Alternative Preparation Methods

Method 2: Partial Neutralization

Method 3: Partial Neutralization of Base

🛡️ Problem 2: Buffer or Not?

Problem-Solving Strategy

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

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

Why $pK_a \pm 1$ ?

Common Buffer Acids and Their $pK_a$ Values