Chapter 10: Acids and Bases

There is not one definition of an acid - there are three. Each one expands on the last, casting a wider net over the kinds of reactions we call “acid-base.” The MCAT expects you to know all three, recognize when each applies, and understand why the broadest definition (Lewis) matters most in biochemistry.

Arrhenius Definition - The Narrow View

The Arrhenius definition is the simplest and oldest. It only works in water.

• Arrhenius acid: produces H⁺ ions in aqueous solution (e.g., HCl → H⁺ + Cl⁻)
• Arrhenius base: produces OH⁻ ions in aqueous solution (e.g., NaOH → Na⁺ + OH⁻)

The limitation is obvious: this definition requires water. It cannot explain why NH₃ acts as a base when dissolved in a non-aqueous solvent, or why BF₃ behaves as an acid despite having no hydrogen atoms at all.

Brønsted-Lowry Definition - The Proton Transfer View

The Brønsted-Lowry definition removes the requirement for water. It focuses on proton transfer.

• Brønsted-Lowry acid: a proton (H⁺) donor
• Brønsted-Lowry base: a proton (H⁺) acceptor

In the reaction HCl + H₂O → H₃O⁺ + Cl⁻, HCl donates a proton to water (acid), and water accepts it (base). But in NH₃ + H₂O → NH₄⁺ + OH⁻, water donates a proton to ammonia - so now water is the acid and ammonia is the base.

This reveals an important property: water is amphoteric - it can act as either an acid or a base depending on its reaction partner.

Lewis Definition - The Broadest View

The Lewis definition abandons protons entirely and focuses on electron pairs.

• Lewis acid: an electron pair acceptor (has an empty orbital)
• Lewis base: an electron pair donor (has a lone pair)

This is the most inclusive definition. Every Brønsted-Lowry acid is a Lewis acid, but Lewis acids also include species like BF₃, AlCl₃, and metal cations (Fe³⁺, Zn²⁺) - none of which have a proton to donate.

Example: When Fe³⁺ binds to water molecules in solution, Fe³⁺ is the Lewis acid (accepts electron pairs from water’s lone pairs) and H₂O is the Lewis base (donates its lone pairs). This is how hydration shells form around metal ions - and it is why transition metal chemistry is fundamentally Lewis acid-base chemistry.

How the Three Definitions Nest

Each definition contains the one above it. Every Arrhenius acid is a Brønsted-Lowry acid. Every Brønsted-Lowry acid is a Lewis acid. But the reverse is not true - BF₃ is a Lewis acid but not a Brønsted-Lowry acid (no proton to donate).

Every time a Brønsted-Lowry acid donates a proton, it becomes something new - something that could accept a proton back. That “something new” is its conjugate base. Every acid-base reaction creates two conjugate pairs, and understanding this relationship is the key to predicting the direction of proton transfer.

What Conjugate Pairs Are

When an acid (HA) donates a proton, it becomes its conjugate base (A⁻). When a base (B) accepts a proton, it becomes its conjugate acid (BH⁺).

Example: CH₃COOH + H₂O ⇌ CH₃COO⁻ + H₃O⁺

• Pair 1: CH₃COOH (acid) and CH₃COO⁻ (conjugate base) - differ by one H⁺
• Pair 2: H₂O (base) and H₃O⁺ (conjugate acid) - differ by one H⁺

Every conjugate pair differs by exactly one proton. To find a conjugate base, remove one H⁺. To find a conjugate acid, add one H⁺.

The Inverse Strength Relationship

This is the single most important rule about conjugate pairs:

A strong acid has a very weak conjugate base. A weak acid has a relatively stronger conjugate base.

The same applies in reverse: a strong base has a very weak conjugate acid.

Predicting the Direction of Proton Transfer

In any acid-base equilibrium, the proton transfers from the stronger acid to the stronger base. The equilibrium favors the side with the weaker acid and weaker base.

Rule: Equilibrium favors the formation of the weaker acid-base pair.

Example: Will HF donate a proton to CN⁻?

• HF has $K_a = 6.8 \times 10^{-4}$ (weak acid)
• HCN has $K_a = 6.2 \times 10^{-10}$ (much weaker acid)

Since HF is the stronger acid and CN⁻ is the stronger base, the proton transfers from HF to CN⁻. The equilibrium lies to the right: HF + CN⁻ ⇌ F⁻ + HCN.

Identifying Conjugate Pairs in Complex Reactions

For any Brønsted-Lowry reaction, use this method:

1. Find the species that lost a proton - it became a conjugate base
2. Find the species that gained a proton - it became a conjugate acid
3. Match each acid with its conjugate base (they differ by one H⁺)

Salt Hydrolysis - Conjugate Pairs in Action

When a salt dissolves in water, its ions may act as acids or bases through hydrolysis:

For the last case (both ions hydrolyze), compare Ka of the conjugate acid to Kb of the conjugate base. Whichever is larger determines the pH.

A strong acid or strong base dissociates completely in water. There is no equilibrium, no Ka, no partial reaction. Every molecule breaks apart into ions. This makes pH calculations for strong acids and bases trivially easy - but first, you need to memorize which ones are strong.

The Seven Strong Acids

There are exactly seven strong acids. Everything else is weak.

Why HF Is Weak

Students are often surprised that HF is not a strong acid. Fluorine is the most electronegative element - shouldn’t it pull the proton away most easily?

The answer lies in bond strength, not electronegativity. The H-F bond is extremely short and strong (bond energy 570 kJ/mol vs. 431 kJ/mol for H-Cl). This bond is so tough to break that HF only partially dissociates in water. Electronegativity determines polarity, but bond strength determines how easily the proton leaves.

The Strong Bases

Strong bases are the hydroxides of Group 1 metals and the heavier Group 2 metals. They dissociate completely in water.

What “Complete Dissociation” Means for Calculations

For a strong acid like HCl at concentration c:

• HCl → H⁺ + Cl⁻ (100% dissociation)
• [H⁺] = c
• No Ka needed, no ICE table needed

For a strong base like NaOH at concentration c:

• NaOH → Na⁺ + OH⁻ (100% dissociation)
• [OH⁻] = c

For Ca(OH)₂ at concentration c:

• Ca(OH)₂ → Ca²⁺ + 2 OH⁻
• [OH⁻] = 2c (two hydroxides per formula unit)

Sulfuric Acid - A Special Case

H₂SO₄ is diprotic. Its first dissociation is strong (complete), but its second dissociation is weak ($K_{a2} = 1.2 \times 10^{-2}$).

• H₂SO₄ → H⁺ + HSO₄⁻ (strong, 100%)
• HSO₄⁻ ⇌ H⁺ + SO₄²⁻ (weak, Ka₂ = 0.012)

For dilute H₂SO₄ solutions on the MCAT, you can usually approximate that both protons dissociate fully, giving [H⁺] ≈ 2c. But for more concentrated solutions or precise calculations, only the first dissociation is truly complete.

Most acids and bases are weak. They do not hand over all their protons or accept all available protons. Instead, they reach an equilibrium - and the position of that equilibrium is described by Ka or Kb.

Weak Acid Equilibrium

A weak acid HA partially dissociates in water:

HA + H₂O ⇌ H₃O⁺ + A⁻

The equilibrium constant for this reaction is Ka (the acid dissociation constant):

Common weak acids and their Ka values:

Weak Base Equilibrium

A weak base B accepts a proton from water:

B + H₂O ⇌ BH⁺ + OH⁻

The equilibrium constant for this reaction is Kb (the base dissociation constant):

The Relationship Between Ka, Kb, and Kw

For any conjugate acid-base pair, the product of Ka and Kb always equals Kw:

Example: Acetic acid has $K_a = 1.8 \times 10^{-5}$. What is $K_b$ for the acetate ion (CH₃COO⁻)?

$K_b = K_w / K_a = (1.0 \times 10^{-14}) / (1.8 \times 10^{-5}) = 5.6 \times 10^{-10}$

This confirms that acetate is a much weaker base than acetic acid is an acid.

pKa and pKb - The Log Scale

Just as pH = -log[H⁺], we define:

• pKa = -log(Ka) - a smaller pKa means a stronger acid
• pKb = -log(Kb) - a smaller pKb means a stronger base

The “p” operator simply converts to a more convenient scale. An acid with $K_a = 10^{-5}$ has $pK_a = 5$. An acid with $K_a = 10^{-10}$ has $pK_a = 10$. The smaller the $pK_a$, the stronger the acid.

Percent Dissociation

For weak acids, percent dissociation tells you what fraction of the acid molecules have donated their proton:

% dissociation = ([H⁺] at equilibrium / initial [HA]) x 100

Key facts about percent dissociation:

• Diluting a weak acid increases percent dissociation (Le Chatelier’s principle - adding water shifts equilibrium right)
• Ka does not change with dilution (it is a constant at a given temperature)
• A stronger weak acid (larger Ka) has a higher percent dissociation at the same concentration

The hydrogen ion concentration in biological systems spans an enormous range - from about 10 M in concentrated stomach acid to $10^{-15}$ M in strong drain cleaner. Working with numbers that vary over 16 orders of magnitude is clumsy, so chemists use a logarithmic scale: pH.

Autoionization of Water

Pure water is not truly “neutral” in the sense that it has no ions. Water molecules constantly transfer protons to each other:

H₂O + H₂O ⇌ H₃O⁺ + OH⁻

This is the autoionization (or self-ionization) of water. The equilibrium constant for this process is Kw:

Defining pH and pOH

The “p” operator means “negative log”:

• pH = -log[H⁺]
• pOH = -log[OH⁻]

And from Kw, we get the critical relationship:

The pH Scale Mapped Out

Converting Between pH and [H⁺]

Going from [H⁺] to pH:

• If [H⁺] = 10⁻ˣ, then pH = x (exact power of 10 - easy)
• If $[\text{H}^+] = 3.5 \times 10^{-4}$, then $\text{pH} = -\log(3.5 \times 10^{-4}) \approx 3.46$

Going from pH to [H⁺]:

• $[\text{H}^+] = 10^{-\text{pH}}$
• If pH = 5, then $[\text{H}^+] = 10^{-5}$ M
• If pH = 3.46, then $[\text{H}^+] = 10^{-3.46} \approx 3.5 \times 10^{-4}$ M

Quick Estimation Tricks for the MCAT

The MCAT does not give you a calculator, so you need mental math shortcuts:

1. Exact powers of 10: If $[\text{H}^+] = 10^{-3}$, pH = 3. Done.
2. Leading coefficient between 1 and 10: If $[\text{H}^+] = 2 \times 10^{-5}$, the pH is between 4 and 5 (closer to 5 since 2 is small). Specifically: $-\log(2) \approx 0.3$, so $\text{pH} \approx 5 - 0.3 = 4.7$.
3. Know your logs: log(2) ≈ 0.3, log(3) ≈ 0.5, log(5) ≈ 0.7. These three values handle most MCAT pH calculations.

Temperature Dependence of Kw

$K_w = 10^{-14}$ is only valid at 25°C. At higher temperatures, $K_w$ increases (autoionization is endothermic). At 37°C (body temperature), $K_w \approx 2.4 \times 10^{-14}$, meaning the neutral pH is slightly below 7 (about 6.8).

This means “neutral” does not always mean pH 7 - neutral means [H⁺] = [OH⁻], which only corresponds to pH 7 at 25°C.

Calculating pH for strong acids and bases is the easiest type of acid-base problem. Because strong acids and bases dissociate completely, there is no equilibrium to solve - just arithmetic.

Strong Acid pH

For a monoprotic strong acid (like HCl, HBr, HI, HNO₃, HClO₃, HClO₄) at concentration c:

1. HCl → H⁺ + Cl⁻ (complete dissociation)
2. [H⁺] = c
3. pH = -log(c)

Example: What is the pH of 0.0020 M HCl?

• $[\text{H}^+] = 0.0020 \text{ M} = 2.0 \times 10^{-3}$ M
• $\text{pH} = -\log(2.0 \times 10^{-3}) = 3 - \log(2) = 3 - 0.3 =$ 2.7

Strong Base pH

For a monobasic strong base (like NaOH, KOH) at concentration c:

1. NaOH → Na⁺ + OH⁻ (complete dissociation)
2. [OH⁻] = c
3. pOH = -log(c)
4. pH = 14 - pOH

Example: What is the pH of 0.050 M NaOH?

• $[\text{OH}^-] = 0.050 \text{ M} = 5.0 \times 10^{-2}$ M
• $\text{pOH} = -\log(5.0 \times 10^{-2}) = 2 - \log(5) = 2 - 0.7 = 1.3$
• pH = 14 - 1.3 = 12.7

For dibasic strong bases (like Ba(OH)₂, Ca(OH)₂):

• [OH⁻] = 2c (two hydroxide ions per formula unit)

Example: What is the pH of 0.0010 M Ba(OH)₂?

• $[\text{OH}^-] = 2(0.0010) = 0.0020 \text{ M} = 2.0 \times 10^{-3}$ M
• pOH = 3 - log(2) = 3 - 0.3 = 2.7
• pH = 14 - 2.7 = 11.3

The Very Dilute Acid Trap

What is the pH of $10^{-8}$ M HCl?

If you naively calculate: $\text{pH} = -\log(10^{-8}) = 8$.

But wait - pH 8 is basic! A solution of acid cannot be basic, no matter how dilute.

The problem: at $10^{-8}$ M, the contribution of H⁺ from the acid is comparable to the H⁺ from water’s autoionization ($10^{-7}$ M). You must add both sources:

$[\text{H}^+]_{\text{total}} = 10^{-8} + 10^{-7} = 1.1 \times 10^{-7}$ M

$\text{pH} = -\log(1.1 \times 10^{-7}) \approx$ 6.96 (just barely acidic, as expected)

Mixing Strong Acids and Strong Bases

When a strong acid and strong base are mixed, the net reaction is neutralization:

H⁺ + OH⁻ → H₂O

After mixing, determine which species is in excess:

1. Calculate moles of H⁺ = $M_{\text{acid}}$ x $V_{\text{acid}}$
2. Calculate moles of OH⁻ = $M_{\text{base}}$ x $V_{\text{base}}$ (multiply by 2 for dibasic)
3. The excess determines pH

Example: 50.0 mL of 0.10 M HCl is mixed with 30.0 mL of 0.10 M NaOH. What is the pH?

• Moles H⁺ = (0.10)(0.050) = 0.0050 mol
• Moles OH⁻ = (0.10)(0.030) = 0.0030 mol
• Excess H⁺ = 0.0050 - 0.0030 = 0.0020 mol
• Total volume = 80.0 mL = 0.080 L
• $[\text{H}^+] = 0.0020 / 0.080 = 0.025 \text{ M} = 2.5 \times 10^{-2}$ M
• pH = 2 - log(2.5) ≈ 2 - 0.4 = 1.6

Calculating pH for a weak acid or base requires an equilibrium calculation. The acid does not fully dissociate, so you need to solve for the equilibrium concentration of H⁺ using Ka and an ICE table.

Weak Acid pH - The ICE Table Method

For a weak acid HA with initial concentration c and acid dissociation constant Ka:

Step 1: Write the equilibrium and set up the ICE table.

HA ⇌ H⁺ + A⁻

Step 2: Write the Ka expression and substitute.

Ka = [H⁺][A⁻] / [HA] = (x)(x) / (c - x) = x² / (c - x)

Step 3: Apply the x-is-small approximation if c/Ka > 100.

If c >> x (the acid barely dissociates), then c - x ≈ c:

Ka ≈ x² / c

x = √(Ka x c)

Step 4: Find pH = -log(x).

Worked Example - Weak Acid

What is the pH of 0.10 M acetic acid ($K_a = 1.8 \times 10^{-5}$)?

1. Check the approximation: $c/K_a = 0.10 / (1.8 \times 10^{-5}) = 5{,}600$. Since 5,600 >> 100, the approximation is valid.

2. $[\text{H}^+] = \sqrt{K_a \times c} = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}}$

3. $\sqrt{1.8 \times 10^{-6}} = \sqrt{18 \times 10^{-7}} \approx 4.2 \times 10^{-7}$… wait, let us redo more carefully:

   • $\sqrt{1.8 \times 10^{-6}} = \sqrt{1.8} \times 10^{-3} \approx 1.34 \times 10^{-3}$

4. $\text{pH} = -\log(1.34 \times 10^{-3}) = 3 - \log(1.34) \approx 3 - 0.13 =$ 2.87

Weak Base pH - Same Method, Different Constant

For a weak base B with initial concentration c and Kb:

B + H₂O ⇌ BH⁺ + OH⁻

[OH⁻] = √(Kb x c) (same approximation, same conditions)

Then: pOH = -log[OH⁻], and pH = 14 - pOH.

Worked Example - Weak Base

What is the pH of 0.20 M ammonia ($K_b = 1.8 \times 10^{-5}$)?

1. $[\text{OH}^-] = \sqrt{K_b \times c} = \sqrt{1.8 \times 10^{-5} \times 0.20} = \sqrt{3.6 \times 10^{-6}} = 1.9 \times 10^{-3}$

2. $\text{pOH} = -\log(1.9 \times 10^{-3}) = 3 - \log(1.9) \approx 3 - 0.28 = 2.72$

3. pH = 14 - 2.72 = 11.28

When the Approximation Fails

If c/Ka < 100, the approximation c - x ≈ c is not valid, and you need the quadratic formula:

Ka = x² / (c - x) → Ka(c - x) = x² → x² + Ka·x - Ka·c = 0

x = [-Ka + √(Ka² + 4Ka·c)] / 2

On the MCAT, this rarely comes up because they typically give you concentrations and $K_a$ values where the approximation works. But if you are told that $K_a$ is $10^{-2}$ and c is 0.10, then $c/K_a = 10$ - the approximation fails, and you need the quadratic (or they will give you enough information to avoid it).

Checking Your Answer

Always verify that your answer makes sense:

• Weak acid solution: pH must be between 0 and 7 (acidic but not as low as a strong acid at the same concentration)
• Weak base solution: pH must be between 7 and 14
• Percent dissociation should be small (less than 5% if you used the approximation)
• Higher concentration = lower pH for acids (more acid = more H⁺)

Polyprotic acids have more than one proton to donate, but they do not give them all up at once. Each proton leaves in a separate step, and each step has its own Ka - getting progressively smaller as it becomes harder to pull a proton from an already-negative species.

Stepwise Dissociation

A diprotic acid like H₂CO₃ dissociates in two steps:

Step 1: H₂CO₃ ⇌ H⁺ + HCO₃⁻ ($K_{a1} = 4.3 \times 10^{-7}$)

Step 2: HCO₃⁻ ⇌ H⁺ + CO₃²⁻ ($K_{a2} = 4.7 \times 10^{-11}$)

A triprotic acid like H₃PO₄ dissociates in three steps:

Step 1: H₃PO₄ ⇌ H⁺ + H₂PO₄⁻ ($K_{a1} = 7.5 \times 10^{-3}$)

Step 2: H₂PO₄⁻ ⇌ H⁺ + HPO₄²⁻ ($K_{a2} = 6.2 \times 10^{-8}$)

Step 3: HPO₄²⁻ ⇌ H⁺ + PO₄³⁻ ($K_{a3} = 4.8 \times 10^{-13}$)

Why Each Ka Gets Smaller

Each successive $K_a$ is always smaller than the previous one - typically by a factor of $10^4$ to $10^6$. The reason is electrostatic:

• Removing the first proton from a neutral molecule (H₃PO₄) is relatively easy.
• Removing the second proton from a negatively charged ion (H₂PO₄⁻) is harder because you are pulling a positive proton away from a negative species.
• Removing the third proton from a doubly negative ion (HPO₄²⁻) is even harder.

Common Polyprotic Acids on the MCAT

Calculating pH of a Polyprotic Acid

Because $K_{a1} >> K_{a2}$ (typically by $10^4$ or more), the first dissociation produces nearly all of the H⁺ in solution. The second and third dissociations contribute negligible additional H⁺.

The shortcut: To find the pH of a polyprotic acid, use only Ka₁. Treat it as a monoprotic weak acid.

Example: What is the pH of 0.10 M H₂CO₃?

Use $K_{a1} = 4.3 \times 10^{-7}$ only:

$[\text{H}^+] = \sqrt{K_{a1} \times c} = \sqrt{4.3 \times 10^{-7} \times 0.10} = \sqrt{4.3 \times 10^{-8}} \approx 2.1 \times 10^{-4}$

$\text{pH} \approx 4 - \log(2.1) \approx 4 - 0.3 =$ 3.7

(The second dissociation would add roughly $4.7 \times 10^{-11}$ M more H⁺ - completely negligible.)

Finding the Concentration of the Fully Deprotonated Species

Although Ka₂ does not affect pH significantly, MCAT questions sometimes ask for the concentration of the second or third conjugate base.

For a diprotic acid: the concentration of the fully deprotonated species (A²⁻) numerically equals Ka₂ (when the x-is-small approximation applies to the first dissociation).

This is because at equilibrium: Ka₂ = [H⁺][A²⁻] / [HA⁻]. Since [H⁺] ≈ [HA⁻] from the first dissociation, Ka₂ ≈ [A²⁻].

Amphoteric Intermediate Species

The intermediate species of a polyprotic acid (like HCO₃⁻ or H₂PO₄⁻) are amphoteric - they can act as either an acid or a base. The pH of a solution of the amphoteric species is approximated by:

pH ≈ (pKa₁ + pKa₂) / 2

This averages the two relevant pKa values, giving the pH at which the amphoteric species is most stable (neither donating nor accepting protons significantly).

A titration curve is a graph of pH versus the volume of titrant (acid or base) added. It tells you everything: the strength of the acid or base being titrated, the pKa, the equivalence point, and the buffer region. Reading titration curves quickly is one of the most tested skills in MCAT general chemistry.

The Anatomy of a Titration Curve

Every titration curve has these regions:

1. Initial point - pH before any titrant is added
2. Buffer region - the gradual rise (or fall) where both HA and A⁻ coexist
3. Half-equivalence point - exactly halfway to the equivalence point, where [HA] = [A⁻] and pH = pKa
4. Equivalence point - where moles of acid = moles of base (neutralization complete)
5. Post-equivalence region - excess titrant dominates the pH

Strong Acid + Strong Base Titration

Example: Titrating HCl with NaOH

Key features:

• Initial pH: low (determined by [HCl])
• Equivalence point pH: exactly 7.0 (only water and a neutral salt remain)
• Curve shape: steep, dramatic jump centered at pH 7
• No buffer region (strong acids have no conjugate base that buffers)

At the equivalence point: moles HCl = moles NaOH. The solution contains only NaCl and water - both neutral.

Weak Acid + Strong Base Titration

Example: Titrating CH₃COOH with NaOH

Key features:

• Initial pH: higher than the strong acid case (weak acid, less H⁺)
• Buffer region: the flat area before the equivalence point where both CH₃COOH and CH₃COO⁻ coexist
• Half-equivalence point: pH = pKa (this is how you find pKa experimentally)
• Equivalence point pH: above 7 (the solution contains only the conjugate base CH₃COO⁻, which is a weak base)
• Curve shape: gentler rise, equivalence point shifted above pH 7

Weak Base + Strong Acid Titration

Example: Titrating NH₃ with HCl

Key features:

• Initial pH: above 7 (basic solution)
• Buffer region: where both NH₃ and NH₄⁺ coexist
• Half-equivalence point: pOH = pKb, so pH = 14 - pKb = pKa of the conjugate acid
• Equivalence point pH: below 7 (the solution contains NH₄⁺, a weak acid)
• Curve shape: pH decreases as acid is added, steep drop at equivalence

Summary: Equivalence Point pH

Reading a Titration Curve on the MCAT

When you see a titration curve, extract this information immediately:

1. Initial pH - read from the y-axis at volume = 0
2. Equivalence point - the steepest part of the curve (the inflection point of the steep rise)
3. Volume at equivalence - read from the x-axis at the equivalence point
4. Half-equivalence volume - half of the equivalence volume
5. pH at half-equivalence - this equals pKa (for weak acid titrated with strong base)
6. pH at equivalence - above 7 (weak acid) or below 7 (weak base) or exactly 7 (strong-strong)

Calculating the Equivalence Point

At the equivalence point:

• moles of acid = moles of base
• $M_{\text{acid}}$ x $V_{\text{acid}}$ = $M_{\text{base}}$ x $V_{\text{base}}$ (for monoprotic acids and monobasic bases)

Solve for the unknown concentration or volume.

pH at the Equivalence Point of a Weak Acid Titration

At the equivalence point of a weak acid-strong base titration, all the HA has been converted to A⁻. The solution is equivalent to a solution of the sodium salt of the weak acid.

To find the pH: treat A⁻ as a weak base with Kb = Kw/Ka, and use [OH⁻] = √(Kb x c), where c is the concentration of A⁻ at the equivalence point (accounting for dilution).

An acid-base indicator is a weak acid (or weak base) that changes color when it gains or loses a proton. The protonated form (HIn) has one color, and the deprotonated form (In⁻) has a different color. By choosing an indicator whose color change occurs near the equivalence point of your titration, you can visually detect when neutralization is complete.

How Indicators Work

An indicator is itself a weak acid in equilibrium:

HIn ⇌ H⁺ + In⁻

(Color A) (Color B)

The equilibrium constant for this dissociation is KIn:

KIn = [H⁺][In⁻] / [HIn]

• When [H⁺] is high (acidic solution), the equilibrium shifts left → more HIn → Color A dominates
• When [H⁺] is low (basic solution), the equilibrium shifts right → more In⁻ → Color B dominates

The color change occurs over a range of about 2 pH units centered on the pKIn (the pKa of the indicator).

Common Indicators

Choosing the Right Indicator

The indicator must change color at a pH close to the equivalence point of the titration:

The Endpoint vs. the Equivalence Point

These are not the same thing:

• Equivalence point: the theoretical point where moles of acid = moles of base (exact neutralization)
• Endpoint: the experimental point where the indicator changes color

A well-chosen indicator makes the endpoint approximately equal to the equivalence point. A poorly chosen indicator gives an inaccurate result because the color change occurs at the wrong pH.

Why the Color Change Spans About 2 pH Units

The indicator changes color gradually over a pH range of approximately pKIn ± 1. This is because:

• At pH = pKIn - 1: [HIn] / [In⁻] = 10 (10x more acid form - Color A dominates)
• At pH = pKIn: [HIn] / [In⁻] = 1 (equal amounts - mixed color)
• At pH = pKIn + 1: [HIn] / [In⁻] = 0.1 (10x more base form - Color B dominates)

The human eye can generally detect when one form is about 10x more abundant than the other, which is why the visible transition covers about 2 pH units.

A buffer is a solution that resists changes in pH when small amounts of acid or base are added. Buffers are everywhere in biology - your blood, every cell in your body, and every biochemistry experiment relies on them. On the MCAT, buffers are one of the most frequently tested topics in all of general chemistry.

What Makes a Buffer

A buffer requires two components:

1. A weak acid (HA) - to neutralize any added base
2. Its conjugate base (A⁻) - to neutralize any added acid

Both must be present in significant amounts. Common buffer systems:

• Acetic acid / sodium acetate (CH₃COOH / CH₃COO⁻)
• Carbonic acid / bicarbonate (H₂CO₃ / HCO₃⁻) - the blood buffer
• Dihydrogen phosphate / hydrogen phosphate (H₂PO₄⁻ / HPO₄²⁻) - intracellular buffer
• Ammonia / ammonium (NH₃ / NH₄⁺)

How a Buffer Works - The Mechanism

When acid (H⁺) is added:

A⁻ + H⁺ → HA

The conjugate base “soaks up” the added protons, converting to the weak acid. [A⁻] decreases slightly, [HA] increases slightly, but pH barely changes.

When base (OH⁻) is added:

HA + OH⁻ → A⁻ + H₂O

The weak acid neutralizes the added hydroxide, converting to the conjugate base. [HA] decreases slightly, [A⁻] increases slightly, but pH barely changes.

The Henderson-Hasselbalch Equation

This is the master equation for buffer chemistry:

Key Implications of Henderson-Hasselbalch

Making a Buffer at a Specific pH

To prepare a buffer at a desired pH:

1. Choose a weak acid whose pKa is close to the desired pH (within ±1)
2. Use Henderson-Hasselbalch to calculate the required ratio [A⁻]/[HA]
3. Mix the weak acid and its conjugate base (usually as a sodium or potassium salt) in that ratio

Example: You want a buffer at pH 5.00 using acetic acid (pKa = 4.74).

pH = pKa + log([A⁻]/[HA])

5.00 = 4.74 + log([A⁻]/[HA])

log([A⁻]/[HA]) = 0.26

$[\text{A}^-]/[\text{HA}] = 10^{0.26} \approx 1.8$

You need about 1.8 times as much acetate as acetic acid.

The Bicarbonate Buffer System

The most important buffer in human physiology - and the foundation for understanding acid-base disorders in biology:

CO₂ + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻

At blood pH 7.40, using pKa₁ = 6.10 for the CO₂/HCO₃⁻ system:

pH = 6.10 + log([HCO₃⁻]/[CO₂])

7.40 = 6.10 + log([HCO₃⁻]/[CO₂])

log([HCO₃⁻]/[CO₂]) = 1.30

[HCO₃⁻]/[CO₂] = 20

Normal blood has about 20 times more bicarbonate than dissolved CO₂. This ratio maintains the pH at 7.40.

Using Henderson-Hasselbalch During a Titration

Henderson-Hasselbalch works at any point during a weak acid-strong base titration (except the initial point and the equivalence point):

1. Before any base is added: Use Ka and ICE table (no A⁻ yet)
2. Between initial and equivalence: Use Henderson-Hasselbalch (both HA and A⁻ present)
3. At half-equivalence: pH = pKa (simplest case)
4. At equivalence: Use Kb of the conjugate base (all HA converted to A⁻)
5. Past equivalence: Excess strong base determines pH

A buffer resists pH changes, but it cannot resist forever. Eventually, if you add enough acid or base, the buffer is overwhelmed and the pH changes dramatically. Buffer capacity tells you how much punishment the buffer can take before it breaks.

What Is Buffer Capacity?

Buffer capacity is the amount of strong acid or strong base that a buffer can absorb before its pH changes significantly (usually defined as more than 1 pH unit from pKa).

Buffer capacity depends on two factors:

1. Total concentration of the buffer components - more buffer = more capacity
2. The ratio of [A⁻]/[HA] - closest to 1:1 = maximum capacity

The Effective Buffer Range: pKa ± 1

A buffer works effectively only when the pH stays within approximately one pH unit of the pKa:

Effective buffer range = pKa ± 1

At the edges of this range:

• At pH = pKa + 1: [A⁻]/[HA] = 10 (10x more base form than acid form)
• At pH = pKa - 1: [HA]/[A⁻] = 10 (10x more acid form than base form)

Beyond these limits, one component is so depleted that the buffer can no longer resist pH changes effectively. The buffer “breaks.”

Why 1:1 Ratio Gives Maximum Capacity

When [A⁻] = [HA] (1:1 ratio, pH = pKa), the buffer has equal capacity to absorb both acid and base:

• Adding acid: uses up A⁻. Starting with lots of A⁻ means lots of capacity.
• Adding base: uses up HA. Starting with lots of HA means lots of capacity.

At a 1:1 ratio, both stockpiles are equal, giving balanced protection in both directions. If the ratio is 10:1, the buffer can absorb a lot of acid (plenty of A⁻) but very little base (almost no HA left).

How Concentration Affects Capacity

Compare two acetate buffers, both at pH = pKa = 4.74:

The concentrated buffer has 100 times more of each component, so it can neutralize 100 times more added acid or base before the ratio shifts beyond pKa ± 1.

What Happens When a Buffer Breaks

Once all of one component is consumed, the buffer ceases to function. For example, in an acetate buffer where you keep adding HCl:

1. While buffer works: H⁺ + CH₃COO⁻ → CH₃COOH (pH barely changes)
2. Buffer breaking point: all CH₃COO⁻ consumed
3. After buffer breaks: added H⁺ accumulates freely, pH plummets rapidly

This “breaking” is visible on a titration curve as the steep rise or fall beyond the buffer region.

Choosing a Buffer for the MCAT

When a question asks you to select the best buffer for a given pH:

1. Match pKa to desired pH (pKa should be within ±1 of target pH)
2. The buffer pair must be a weak acid and its conjugate base (never a strong acid or strong base)
3. Higher concentration = better buffer capacity

Preparing Buffers - Two Methods

Method 1: Mix a weak acid with the salt of its conjugate base directly.

• Example: mix CH₃COOH and CH₃COONa

Method 2: Start with excess weak acid and partially neutralize it with strong base.

• Example: add NaOH to CH₃COOH until half is converted to CH₃COO⁻
• This produces the same buffer (HA + A⁻) through a neutralization reaction

Both methods produce the same buffer. Method 2 is essentially what happens during the first half of a weak acid-strong base titration - which is why the buffer region appears on the titration curve.