The Hook: Why Doesn’t Your Blood pH Change When You Drink Lemon Juice?
Your blood pH stays rock-solid at 7.35-7.45, whether you eat acidic lemons (pH ~2) or alkaline spinach (pH ~8). Deviate by just 0.4 units, and you could die! The secret? Buffer systems—the body’s shock absorbers for pH.
Similarly:
- Swimming pool water needs buffers to resist pH changes from chlorine
- Shampoo pH ~5.5 matches your scalp (too basic = dry hair!)
- Ocean pH ~8.1 is buffered by carbonate system (climate change threatens this!)
Real-world question: How do chemists design solutions that refuse to change pH?
JEE Reality: pH and buffers appear in 3-5 questions every year. Master the Henderson-Hasselbalch equation, and you’ll breeze through these high-scoring numericals!
The Core Concept
The pH Scale
pH is a measure of hydrogen ion concentration in solution.
$$\boxed{pH = -\log[H^+]}$$Alternative definition (using hydronium ion):
$$pH = -\log[H_3O^+]$$In simple terms: pH is the negative power of 10 for H⁺ concentration.
If [H⁺] = 10⁻⁷ M → pH = 7 If [H⁺] = 10⁻³ M → pH = 3
The pOH Scale
Similarly, for hydroxide ions:
$$\boxed{pOH = -\log[OH^-]}$$Relationship Between pH and pOH
From the ionic product of water:
$$K_w = [H^+][OH^-] = 1.0 \times 10^{-14} \text{ at 25°C}$$Taking negative logarithm:
$$\boxed{pH + pOH = 14}$$(at 25°C)
The pH Range
| pH | [H⁺] | [OH⁻] | Nature |
|---|---|---|---|
| 0 | 10⁰ = 1 M | 10⁻¹⁴ M | Very acidic |
| 1 | 10⁻¹ M | 10⁻¹³ M | Strongly acidic |
| 3 | 10⁻³ M | 10⁻¹¹ M | Acidic |
| 7 | 10⁻⁷ M | 10⁻⁷ M | Neutral |
| 10 | 10⁻¹⁰ M | 10⁻⁴ M | Basic |
| 13 | 10⁻¹³ M | 10⁻¹ M | Strongly basic |
| 14 | 10⁻¹⁴ M | 10⁰ = 1 M | Very basic |
Interactive Demo: Visualize the pH Scale
Explore how hydrogen ion concentration relates to pH. See the color changes of indicators across the pH range.
Historical reason: Makes the scale more convenient
- [H⁺] ranges from 1 M to 10⁻¹⁴ M (huge range!)
- pH ranges from 0 to 14 (manageable scale)
Mathematical convenience: Lower pH = More acidic (intuitive)
Calculating pH of Strong Acids and Bases
Strong Acids
Strong acids completely ionize: α = 1
$$HCl \xrightarrow{100\%} H^+ + Cl^-$$For monoprotic strong acid:
$$\boxed{[H^+] = C_{acid}}$$ $$\boxed{pH = -\log C_{acid}}$$Example: 0.01 M HCl
- [H⁺] = 0.01 M = 10⁻² M
- pH = -log(10⁻²) = 2
For diprotic strong acid (like H₂SO₄):
$$H_2SO_4 \rightarrow 2H^+ + SO_4^{2-}$$ $$[H^+] = 2 \times C_{acid}$$Strong Bases
Strong bases completely ionize: α = 1
$$NaOH \xrightarrow{100\%} Na^+ + OH^-$$For strong base:
$$\boxed{[OH^-] = C_{base}}$$ $$\boxed{pOH = -\log C_{base}}$$ $$\boxed{pH = 14 - pOH}$$Example: 0.01 M NaOH
- [OH⁻] = 0.01 M = 10⁻² M
- pOH = 2
- pH = 14 - 2 = 12
Calculating pH of Weak Acids
Weak acids partially ionize: α « 1
$$HA \rightleftharpoons H^+ + A^-$$Using Ka
From ionic equilibrium:
$$[H^+] = \sqrt{K_a \cdot C}$$(when α « 1)
$$\boxed{pH = \frac{1}{2}(pK_a - \log C)}$$Derivation:
$$[H^+] = \sqrt{K_a \cdot C}$$Taking log:
$$\log[H^+] = \frac{1}{2}(\log K_a + \log C)$$ $$-\log[H^+] = -\frac{1}{2}(\log K_a + \log C)$$ $$pH = \frac{1}{2}(-\log K_a - \log C)$$ $$\boxed{pH = \frac{1}{2}(pK_a - \log C)}$$Example: 0.1 M CH₃COOH (Ka = 1.8×10⁻⁵, pKa = 4.74)
$$pH = \frac{1}{2}(4.74 - \log 0.1) = \frac{1}{2}(4.74 + 1) = \frac{5.74}{2} = 2.87$$Calculating pH of Weak Bases
Weak bases partially ionize: α « 1
$$BOH \rightleftharpoons B^+ + OH^-$$Using Kb
$$[OH^-] = \sqrt{K_b \cdot C}$$ $$\boxed{pOH = \frac{1}{2}(pK_b - \log C)}$$ $$\boxed{pH = 14 - pOH}$$Example: 0.1 M NH₄OH (Kb = 1.8×10⁻⁵, pKb = 4.74)
$$pOH = \frac{1}{2}(4.74 - \log 0.1) = 2.87$$ $$pH = 14 - 2.87 = 11.13$$Buffer Solutions: The pH Guardians
What is a Buffer?
Buffer solution: A solution that resists changes in pH upon addition of small amounts of acid or base.
How it works: Contains both:
- Weak acid (to neutralize added base)
- Conjugate base (to neutralize added acid)
Or:
- Weak base (to neutralize added acid)
- Conjugate acid (to neutralize added base)
Types of Buffers
1. Acidic Buffer (pH < 7)
Composition: Weak acid + Salt of weak acid with strong base
Example: CH₃COOH + CH₃COONa (acetic acid + sodium acetate)
$$CH_3COOH \rightleftharpoons H^+ + CH_3COO^-$$ $$CH_3COONa \xrightarrow{100\%} CH_3COO^- + Na^+$$When acid (H⁺) is added:
$$CH_3COO^- + H^+ \rightarrow CH_3COOH$$(Conjugate base neutralizes H⁺)
When base (OH⁻) is added:
$$CH_3COOH + OH^- \rightarrow CH_3COO^- + H_2O$$(Weak acid neutralizes OH⁻)
2. Basic Buffer (pH > 7)
Composition: Weak base + Salt of weak base with strong acid
Example: NH₄OH + NH₄Cl (ammonium hydroxide + ammonium chloride)
$$NH_4OH \rightleftharpoons NH_4^+ + OH^-$$ $$NH_4Cl \xrightarrow{100\%} NH_4^+ + Cl^-$$When acid (H⁺) is added:
$$NH_4OH + H^+ \rightarrow NH_4^+ + H_2O$$(Weak base neutralizes H⁺)
When base (OH⁻) is added:
$$NH_4^+ + OH^- \rightarrow NH_4OH$$(Conjugate acid neutralizes OH⁻)
Henderson-Hasselbalch Equation: The Buffer Formula
For Acidic Buffer
$$\boxed{pH = pK_a + \log\frac{[\text{salt}]}{[\text{acid}]}}$$Or more generally:
$$\boxed{pH = pK_a + \log\frac{[\text{conjugate base}]}{[\text{weak acid}]}}$$In terms of A⁻ and HA:
$$\boxed{pH = pK_a + \log\frac{[A^-]}{[HA]}}$$For Basic Buffer
$$\boxed{pOH = pK_b + \log\frac{[\text{salt}]}{[\text{base}]}}$$ $$\boxed{pH = 14 - pOH}$$Or more generally:
$$\boxed{pOH = pK_b + \log\frac{[\text{conjugate acid}]}{[\text{weak base}]}}$$Derivation of Henderson-Hasselbalch
For weak acid equilibrium:
$$K_a = \frac{[H^+][A^-]}{[HA]}$$ $$[H^+] = K_a \times \frac{[HA]}{[A^-]}$$Taking negative logarithm:
$$-\log[H^+] = -\log K_a - \log\frac{[HA]}{[A^-]}$$ $$pH = pK_a - \log\frac{[HA]}{[A^-]}$$ $$\boxed{pH = pK_a + \log\frac{[A^-]}{[HA]}}$$Henderson-Hasselbalch is the most important buffer equation!
For acidic buffer:
$$pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}$$Special case: When [Salt] = [Acid]
$$pH = pK_a + \log(1) = pK_a + 0 = pK_a$$Maximum buffer capacity when pH = pKa!
Buffer Capacity
Buffer capacity (β): The amount of acid or base a buffer can neutralize before significant pH change occurs.
$$\beta = \frac{\text{Amount of acid/base added}}{\text{Change in pH}}$$Factors Affecting Buffer Capacity
Concentration of components
- Higher concentrations → Greater capacity
- 1 M buffer > 0.1 M buffer (same ratio)
Ratio of [Salt]/[Acid]
- Maximum capacity when ratio = 1 (equal concentrations)
- When [Salt] = [Acid], pH = pKa
- Effective range: 0.1 < [Salt]/[Acid] < 10
- Effective pH range: pKa ± 1
Nature of acid/base
- Polyprotic acids can have multiple buffer regions
Buffer Range
Effective buffer range: pH = pKa ± 1
Why?
- If pH = pKa + 1: [Salt]/[Acid] = 10 (still reasonably balanced)
- If pH = pKa - 1: [Salt]/[Acid] = 0.1 (still reasonably balanced)
- Beyond this range: Buffer loses effectiveness
Example: Acetic acid buffer (pKa = 4.74)
- Effective range: pH 3.74 to 5.74
Memory Tricks & Patterns
Mnemonic for Henderson-Hasselbalch
“pH Keeps Adding Logs of Salt/Acid”
- pH = pH
- Keeps = K
- Adding = a (Ka)
- Logs = log
- Salt/Acid = [Salt]/[Acid]
→ pH = pKa + log([Salt]/[Acid])
Pattern Recognition: pH Calculations
| Solution Type | pH Formula | Example |
|---|---|---|
| Strong acid | $pH = -\log C$ | 0.01 M HCl → pH = 2 |
| Weak acid | $pH = \frac{1}{2}(pK_a - \log C)$ | 0.1 M CH₃COOH → pH = 2.87 |
| Strong base | $pH = 14 + \log C$ | 0.01 M NaOH → pH = 12 |
| Weak base | $pH = 14 - \frac{1}{2}(pK_b - \log C)$ | 0.1 M NH₃ → pH = 11.13 |
| Acidic buffer | $pH = pK_a + \log\frac{[Salt]}{[Acid]}$ | CH₃COOH/CH₃COONa |
| Basic buffer | $pH = 14 - (pK_b + \log\frac{[Salt]}{[Base]})$ | NH₃/NH₄Cl |
Mnemonic for Buffer Effectiveness
“Plus/Minus One from pKa, Buffer’s Having Fun”
Effective buffer range: pKa - 1 to pKa + 1
Common Mistakes to Avoid
Wrong: Mixing up initial concentrations with equilibrium concentrations
Correct: For buffers (high [Salt] suppresses acid ionization), use initial concentrations directly in Henderson-Hasselbalch
Why? Common ion effect ensures weak acid barely ionizes, so [HA]initial ≈ [HA]equilibrium
Example: 0.1 M CH₃COOH + 0.15 M CH₃COONa
$$pH = pK_a + \log\frac{0.15}{0.1}$$✓ (Use these directly)
Wrong: Not adjusting volumes when mixing solutions
Correct: When mixing, use final concentrations after dilution
Example: Mix 50 mL of 0.2 M CH₃COOH with 50 mL of 0.2 M CH₃COONa
Total volume = 100 mL
Final [CH₃COOH] = (0.2 × 50)/100 = 0.1 M Final [CH₃COONa] = (0.2 × 50)/100 = 0.1 M
$$pH = pK_a + \log\frac{0.1}{0.1} = pK_a$$✓
Common Error: Using pKa of acid when you need pKb of conjugate base (or vice versa)
Correct: For conjugate pairs:
$$pK_a + pK_b = 14$$Example: Given pKa of NH₄⁺ = 9.26, find pKb of NH₃
$$pK_b = 14 - 9.26 = 4.74$$✓
JEE Tip: They might give you pKa when you need pKb (or vice versa) to test if you know the relationship!
Buffer: Weak acid + Its conjugate base (both present) Neutralization: Acid + Base react completely (one or both consumed)
Example: CH₃COOH + NaOH
- If NaOH < CH₃COOH: Buffer forms (excess acid remains)
- If NaOH = CH₃COOH: Only salt (neutralization, NOT a buffer!)
- If NaOH > CH₃COOH: Excess base (NOT a buffer!)
For buffer, you need BOTH components present!
Practice Problems
Level 1: Foundation (NCERT)
Calculate pH of:
- 0.001 M HCl
- 0.01 M NaOH
Solution:
1. 0.001 M HCl (strong acid)
$$[H^+] = 0.001 = 10^{-3} \text{ M}$$ $$pH = -\log(10^{-3}) = 3$$2. 0.01 M NaOH (strong base)
$$[OH^-] = 0.01 = 10^{-2} \text{ M}$$ $$pOH = -\log(10^{-2}) = 2$$ $$pH = 14 - 2 = 12$$Calculate pH of 0.1 M acetic acid (CH₃COOH). Given: Ka = 1.8×10⁻⁵
Solution:
First, calculate pKa:
$$pK_a = -\log(1.8 \times 10^{-5}) = 4.74$$Using weak acid formula:
$$pH = \frac{1}{2}(pK_a - \log C) = \frac{1}{2}(4.74 - \log 0.1)$$ $$pH = \frac{1}{2}(4.74 - (-1)) = \frac{1}{2}(4.74 + 1) = \frac{5.74}{2}$$ $$pH = 2.87$$Calculate pH of a buffer containing 0.1 M CH₃COOH and 0.1 M CH₃COONa. (pKa = 4.74)
Solution:
Using Henderson-Hasselbalch:
$$pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}$$ $$pH = 4.74 + \log\frac{0.1}{0.1} = 4.74 + \log(1) = 4.74 + 0$$ $$pH = 4.74$$Key insight: When [Salt] = [Acid], pH = pKa!
Level 2: JEE Main
Calculate pH of a buffer solution containing 0.2 M CH₃COOH and 0.3 M CH₃COONa. (pKa of CH₃COOH = 4.74)
Solution:
$$pH = pK_a + \log\frac{[\text{Salt}]}{[\text{Acid}]} = 4.74 + \log\frac{0.3}{0.2}$$ $$pH = 4.74 + \log(1.5) = 4.74 + 0.176 = 4.92$$1 L of buffer contains 0.1 M CH₃COOH and 0.1 M CH₃COONa (pKa = 4.74).
If 0.01 mol HCl is added, what is the new pH?
Solution:
Initial moles:
- CH₃COOH: 0.1 mol
- CH₃COONa: 0.1 mol
After adding 0.01 mol HCl:
HCl reacts with acetate ion:
$$H^+ + CH_3COO^- \rightarrow CH_3COOH$$- CH₃COO⁻ consumed: 0.01 mol → Remaining: 0.1 - 0.01 = 0.09 mol
- CH₃COOH formed: 0.01 mol → New amount: 0.1 + 0.01 = 0.11 mol
New concentrations (volume still 1 L):
- [CH₃COOH] = 0.11 M
- [CH₃COO⁻] = 0.09 M
Change: 4.74 → 4.65 (only 0.09 units!)
Compare: If 0.01 M HCl added to pure water → pH = 2! Buffer works!
50 mL of 0.2 M CH₃COOH is mixed with 50 mL of 0.1 M NaOH. Calculate pH of the resulting solution. (pKa = 4.74)
Solution:
Initial moles:
- CH₃COOH: 0.2 × 0.05 = 0.01 mol
- NaOH: 0.1 × 0.05 = 0.005 mol
Reaction:
$$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$$NaOH is limiting (0.005 mol)
After reaction:
- CH₃COOH remaining: 0.01 - 0.005 = 0.005 mol
- CH₃COONa formed: 0.005 mol
- NaOH: 0 mol
This is a buffer! (Weak acid + its conjugate base both present)
Total volume = 100 mL = 0.1 L
Final concentrations:
- [CH₃COOH] = 0.005/0.1 = 0.05 M
- [CH₃COONa] = 0.005/0.1 = 0.05 M
Level 3: JEE Advanced
A buffer is prepared by mixing 0.5 M CH₃COOH and 0.5 M CH₃COONa. Another buffer is prepared with 0.05 M of each. Both have pH = 4.74.
If 0.01 M HCl is added to 1 L of each buffer, calculate the pH change in both cases. (pKa = 4.74)
Solution:
Buffer 1: 0.5 M each
Initial: pH = 4.74 (equal concentrations)
After adding 0.01 mol HCl:
- CH₃COOH: 0.5 + 0.01 = 0.51 M
- CH₃COO⁻: 0.5 - 0.01 = 0.49 M
Change: 0.02 units
Buffer 2: 0.05 M each
After adding 0.01 mol HCl:
- CH₃COOH: 0.05 + 0.01 = 0.06 M
- CH₃COO⁻: 0.05 - 0.01 = 0.04 M
Change: 0.18 units
Conclusion: Higher concentration buffer (Buffer 1) has greater capacity (smaller pH change)!
Calculate pH of a buffer containing 0.1 M H₂PO₄⁻ and 0.15 M HPO₄²⁻.
Given: For H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻, pKa₂ = 7.21
Solution:
This is an acidic buffer using the second ionization of phosphoric acid:
- Weak acid: H₂PO₄⁻
- Conjugate base: HPO₄²⁻
Using Henderson-Hasselbalch:
$$pH = pK_{a2} + \log\frac{[HPO_4^{2-}]}{[H_2PO_4^-]}$$ $$pH = 7.21 + \log\frac{0.15}{0.1} = 7.21 + \log(1.5)$$ $$pH = 7.21 + 0.176 = 7.39$$JEE Insight: This is similar to blood phosphate buffer system! pH ~7.4
Blood pH is maintained by carbonic acid buffer:
$$H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$$Given: pKa = 6.1, [HCO₃⁻]/[H₂CO₃] = 20 in normal blood
Questions:
- Calculate normal blood pH
- In acidosis, ratio becomes 10. What is new pH?
- Why can we breathe out to increase pH?
Solution:
1. Normal blood pH:
$$pH = pK_a + \log\frac{[HCO_3^-]}{[H_2CO_3]} = 6.1 + \log(20)$$ $$pH = 6.1 + 1.3 = 7.4$$✓
2. During acidosis (ratio = 10):
$$pH = 6.1 + \log(10) = 6.1 + 1 = 7.1$$(Dangerous! Below 7.35 = acidosis)
3. Why breathing out increases pH:
$$CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$$- Exhaling removes CO₂
- Equilibrium shifts left (Le Chatelier)
- [H₂CO₃] decreases
- Ratio [HCO₃⁻]/[H₂CO₃] increases
- pH increases!
This is why hyperventilating can cause alkalosis (pH too high)!
JEE Connection: Biology meets chemistry—prepare for such interdisciplinary questions in JEE Advanced!
Quick Revision Box
| Solution Type | pH Formula | Key Point |
|---|---|---|
| Strong acid | $pH = -\log C$ | Complete ionization |
| Weak acid | $pH = \frac{1}{2}(pK_a - \log C)$ | Use when no salt present |
| Strong base | $pH = 14 + \log C$ | Complete ionization |
| Weak base | $pH = 14 - \frac{1}{2}(pK_b - \log C)$ | Use when no salt present |
| Acidic buffer | $pH = pK_a + \log\frac{[Salt]}{[Acid]}$ | Henderson-Hasselbalch |
| Basic buffer | $pOH = pK_b + \log\frac{[Salt]}{[Base]}$ | Then pH = 14 - pOH |
Key Relationships:
- pH + pOH = 14
- pKa + pKb = 14 (for conjugate pairs)
- Maximum buffer capacity: pH = pKa (when [Salt] = [Acid])
- Effective buffer range: pKa ± 1
When to Use This
Step 1: Identify the solution type
Is it a strong acid/base?
- YES → Direct formula: pH = -log[H⁺] or pH = 14 + log[OH⁻]
Is it a weak acid/base alone?
- YES → Use Ka/Kb formula: pH = ½(pKa - logC)
Is it a mixture of weak acid + conjugate base (or weak base + conjugate acid)?
- YES → Henderson-Hasselbalch equation
- Check: Both components present? → Buffer!
Are you mixing acid and base?
- Find limiting reagent
- Determine what’s left after reaction
- If both weak acid and conjugate base remain → Buffer (use Henderson-Hasselbalch)
- If only one component left → Not a buffer (use Ka/Kb)
Connection to Other Topics
pH and buffers connect to:
- Ionic Equilibrium - Foundation for Ka, Kb, and weak electrolytes
- Acids and Bases - Theoretical basis (Bronsted-Lowry conjugate pairs)
- Chemical Equilibrium - Uses same K, Q, Le Chatelier concepts
- Solubility Product - pH affects solubility of salts
- Electrochemistry - Nernst equation uses pH
- Organic Chemistry - Reaction conditions often require specific pH
- Coordination Chemistry - Complex formation pH-dependent
Biological Buffer Systems (JEE Advanced Special Topic)
1. Bicarbonate Buffer (Primary Blood Buffer)
$$CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$$- pKa = 6.1
- Normal ratio: [HCO₃⁻]/[H₂CO₃] = 20:1
- Normal pH: 7.4
Control mechanisms:
- Lungs: Regulate CO₂ (respiratory control)
- Kidneys: Regulate HCO₃⁻ (metabolic control)
2. Phosphate Buffer
$$H_2PO_4^- \rightleftharpoons H^+ + HPO_4^{2-}$$- pKa = 7.21 (close to blood pH!)
- Important in intracellular fluid
3. Protein Buffer (Hemoglobin)
- Amino acids have both -COOH (acidic) and -NH₂ (basic) groups
- Act as zwitterions: H₃N⁺-CHR-COO⁻
- Can accept or donate H⁺
Teacher’s Summary
- pH = -log[H⁺] - The fundamental definition (memorize this!)
- pH + pOH = 14 - Valid at 25°C (use for base calculations)
- Henderson-Hasselbalch is gold - pH = pKa + log([Salt]/[Acid])
- Buffer needs BOTH components - Weak acid AND conjugate base (or vice versa)
- Maximum buffer capacity at pH = pKa - When [Salt] = [Acid]
“A buffer is like a shock absorber for pH—it takes the hits from acids and bases so your solution doesn’t have to!”
JEE Strategy:
- Henderson-Hasselbalch appears in 70% of buffer questions—master it!
- Practice mixing problems (buffer formation from acid + base)
- Buffer capacity questions are rising in JEE Advanced
- Biological buffers (blood pH) are trendy—know carbonic acid system
- Don’t forget: Use initial concentrations in Henderson-Hasselbalch for buffers
- Link to real-world: Swimming pools, blood, ocean—buffers are everywhere!
What’s Next?
You’ve mastered pH and buffers! Next, explore ionic equilibrium in heterogeneous systems:
- Solubility Product - Ksp, common ion effect, and selective precipitation
- Chemical Kinetics - Rate vs equilibrium (how fast vs how far)
- Electrochemistry - Nernst equation uses pH extensively