Acid-Base Equilibria

Acid-base chemistry bridges general chemistry and biochemistry on the MCAT: oxygen delivery, enzyme catalysis, kidney function, and protein folding all depend on proton control. Recognize its fingerprints everywhere.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Brønsted-Lowry Acids and Bases

Must know

The MCAT uses the Brønsted-Lowry definition almost exclusively: an acid is a proton donor ($\ce{H+}$ donor), a base is a proton acceptor. This is broader than the older Arrhenius model — water itself can act as either, depending on its reaction partner. Every acid-base reaction is a competition for a proton between two bases; the stronger base "wins."

Passage-level

The Lewis definition: a Lewis acid is an electron-pair acceptor, a Lewis base an electron-pair donor. Every Brønsted base is a Lewis base, but the Lewis picture also covers reactions with no proton transfer (e.g., a metal cation accepting ligand lone pairs). Reach for it only when there is no proton to track.

Conjugate Acid-Base Pairs

Must know

When an acid donates a proton, what remains is its conjugate base; when a base accepts one, the result is its conjugate acid. The pair differs by exactly one $\ce{H+}$.

$\underbrace{\ce{CH3COOH}}_{\text{acid}} + \underbrace{\ce{H2O}}_{\text{base}} \rightleftharpoons \underbrace{\ce{H3O+}}_{\text{conj. acid}} + \underbrace{\ce{CH3COO-}}_{\text{conj. base}}$

A strong acid has a weak conjugate base, and a weak acid has a stronger conjugate base. This governs whether a salt makes a solution acidic or basic.

Quick check: In the reaction $\ce{NH3 + H2O <=> NH4+ + OH-}$, identify the two conjugate pairs.

Answer: Pair 1: $\ce{NH3}$ (base) / $\ce{NH4+}$ (conjugate acid). Pair 2: $\ce{H2O}$ (acid) / $\ce{OH-}$ (conjugate base).

A species that can act as either donor or acceptor — $\ce{H2O}$, $\ce{HCO3-}$, $\ce{H2PO4-}$ — is amphoteric (amphiprotic). These are the intermediate species between equivalence points in polyprotic titrations.

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The Autoionization of Water and $K_w$

Must know

Water molecules donate and accept protons from each other:

$\ce{H2O + H2O <=> H3O+ + OH-}$

The equilibrium expression is the ion-product constant of water:

$K_w = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14} \quad \text{at 25°C}$

So pure water has equal $[\ce{H+}] = [\ce{OH-}] = 1.0 \times 10^{-7}$ M. $K_w$ rises with temperature (autoionization is endothermic), but for the MCAT assume $K_w = 10^{-14}$.

The Complementary Relationship Between $[\ce{H+}]$ and $[\ce{OH-}]$

Must know

Because $K_w$ is constant, knowing one ion gives the other:

$[\ce{OH-}] = \frac{K_w}{[\ce{H+}]} = \frac{10^{-14}}{[\ce{H+}]}$

Acidic: $[\ce{H+}] > [\ce{OH-}]$; basic: $[\ce{OH-}] > [\ce{H+}]$; neutral: equal. The product is always $10^{-14}$.

Quick check: If $[\ce{H+}] = 10^{-3}$ M, what is $[\ce{OH-}]$?

Answer: $[\ce{OH-}] = \frac{10^{-14}}{10^{-3}} = 10^{-11}$ M. The solution is acidic.

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pH, pOH, and the p-Notation System

Must know

The pH scale is a logarithmic compression of $[\ce{H+}]$:

$\text{pH} = -\log[\ce{H+}] \qquad \text{pOH} = -\log[\ce{OH-}]$

Taking the negative log of $K_w$:

$\text{pH} + \text{pOH} = 14 \quad \text{(at 25°C)}$

Pure water: $\text{pH} = 7.0$ (neutral). Acidic < 7; basic > 7. The "p" operator means $-\log_{10}$, applying also to $\text{p}K_a = -\log K_a$, $\text{p}K_b$, etc. Because of the negative sign, a smaller pKa means a stronger acid.

MCAT Log Estimation Trick

Must know

No calculator on test day, so estimate logs: $-\log(10^{-n}) = n$; and for a leading coefficient, subtract $-\log(\text{coeff})$, e.g. $-\log(2\times10^{-n}) \approx n - 0.3$, $-\log(5\times10^{-n}) \approx n - 0.7$. So $[\ce{H+}] = 3 \times 10^{-4} \Rightarrow$ pH $\approx 4 - 0.5 = 3.5$.

Quick check: What is the pH of a solution with $[\ce{H+}] = 5 \times 10^{-9}$ M?

Answer: pH $\approx 9 - 0.7 = 8.3$. Basic, as expected since $[\ce{H+}]$ is below $10^{-7}$.

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Strong Acids and Bases

Must know

Strong acids and bases dissociate essentially completely (>99%). There is no meaningful equilibrium — set $[\ce{H+}]$ (or $[\ce{OH-}]$) equal to the formal concentration directly.

The strong acids: $\ce{HCl}$, $\ce{HBr}$, $\ce{HI}$, $\ce{HNO3}$, $\ce{HClO4}$, and $\ce{H2SO4}$ (first dissociation). $\ce{H2SO4}$ is diprotic: the first proton is strong, the second is weak ($K_{a2} \approx 0.012$); dilute $\ce{H2SO4}$ is usually treated as giving 2 $\ce{H+}$.

Strong bases: Group I hydroxides (NaOH, KOH, LiOH) and the heavy Group II hydroxides $\ce{Ca(OH)2}$, $\ce{Sr(OH)2}$, $\ce{Ba(OH)2}$. ($\ce{Mg(OH)2}$ is only sparingly soluble — not a strong base for MCAT purposes.)

Worked Example: pH of a Strong Acid

Must know

Calculate the pH of $0.010$ M $\ce{HCl}$. It dissociates completely, so $[\ce{H+}] = 1.0 \times 10^{-2}$ M:

$\text{pH} = -\log(1.0 \times 10^{-2}) = 2.0$

Each factor-of-10 change in concentration shifts pH by 1 unit (1 M → pH 0; 0.001 M → pH 3).

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Weak Acids and Bases: $K_a$, $K_b$, p$K_a$, p$K_b$

Must know

Most acids and bases (acetic acid, carbonic acid, ammonia, amines) are weak — only partially dissociated, governed by $K_a$ and $K_b$.

Acid Dissociation Constant $K_a$

Must know

For a weak acid $\ce{HA}$:

$\ce{HA + H2O <=> H3O+ + A-} \qquad K_a = \frac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$

A large $K_a$ (small p$K_a$) means a stronger acid.

Know the logic

What makes an acid strong: acid strength tracks conjugate-base stability. Anything that stabilizes $\ce{A-}$ favors dissociation: higher electronegativity ($\ce{HF} > \ce{H2O} > \ce{NH3}$ across a period); larger atomic size / weaker bond down a group ($\ce{HF} < \ce{HCl} < \ce{HBr} < \ce{HI}$); electron-withdrawing inductive effects (trichloroacetic ≫ acetic); and resonance delocalization (carboxylic acids ≫ alcohols).

Passage-level

Typical p$K_a$ values (provided on the exam; don't memorize the table): acetic acid ≈ 4.75, carbonic acid p$K_{a1}$ ≈ 6.1, $\ce{H2PO4-}$ ≈ 7.2, $\ce{NH4+}$ ≈ 9.25, $\ce{HCO3-}$ ≈ 10.3.

Base Dissociation Constant $K_b$

Must know

For a weak base $\ce{B}$:

$\ce{B + H2O <=> BH+ + OH-} \qquad K_b = \frac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}$

Conjugate pair relationship:

$K_a \times K_b = K_w = 10^{-14} \qquad \text{p}K_a + \text{p}K_b = 14$

Once you know $K_a$ of acetic acid, you know $K_b$ of acetate — the backbone of salt hydrolysis.

Calculating pH of a Weak Acid: The ICE Table

Must know

For initial concentration $C$ and constant $K_a$:

	$\ce{HA}$	$\ce{H+}$	$\ce{A-}$
I	$C$	$\approx 0$	$0$
C	$-x$	$+x$	$+x$
E	$C-x$	$x$	$x$

$K_a = \frac{x^2}{C - x}$

Simplifying assumption: when $K_a$ is small and $C$ not too dilute, $x \ll C$, so $C - x \approx C$ and $x \approx \sqrt{K_a \cdot C}$. Valid when $x < 5\%$ of $C$ — always check.

Percent Ionization

Know the logic

$\%\text{ ionization} = \frac{[\ce{H+}]}{C} \times 100\% = \frac{x}{C} \times 100\%$

Since $x \approx \sqrt{K_a C}$, percent ionization $\approx \sqrt{K_a/C}$, so it increases on dilution (Ostwald dilution law) — counterintuitive: the fraction ionized rises even as $[\ce{H+}]$ falls.

Worked Example: pH of a Weak Acid

Must know

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

$x = \sqrt{K_a \cdot C} = \sqrt{1.8 \times 10^{-5} \times 0.10} = \sqrt{1.8 \times 10^{-6}} \approx 1.34 \times 10^{-3} \text{ M}$

Check: $1.34\%$ of $C$ ✓ (under 5%). Then $\text{pH} = -\log(1.34 \times 10^{-3}) \approx 2.87$ — higher (less acidic) than a 0.1 M strong acid (pH 1), as expected.

Quick check: If you double a weak acid from 0.1 M to 0.2 M, does the pH halve?

Answer: No. Since $[\ce{H+}] \approx \sqrt{K_a \cdot C}$, doubling $C$ raises $[\ce{H+}]$ by $\sqrt{2} \approx 1.4$-fold, dropping pH only $\approx 0.15$ units. pH is logarithmic — doubling concentration ≠ halving pH.

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Dissociation with Added Salt: The Common Ion Effect

Must know

Dissolving a weak acid in a solution already containing its conjugate base shifts the equilibrium left, suppressing dissociation — the common ion effect, a direct application of Le Chatelier. Adding sodium acetate to acetic acid raises the pH. The ICE table still applies; you just start with non-zero $[\ce{A-}]$. This is the chemistry behind buffers.

Quick check: You dissolve 0.1 mol acetic acid in 1 L of 0.5 M sodium acetate. Higher or lower pH than pure 0.1 M acetic acid?

Answer: Higher. Excess acetate suppresses dissociation via the common ion effect, so $[\ce{H+}]$ is lower.

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Hydrolysis of Salts of Weak Acids and Bases

Must know

When a salt dissolves, its ions may react with water (hydrolysis) and shift pH. The ion from the weak partner hydrolyzes; the ion from the strong partner is a spectator:

• Strong acid + weak base (e.g., $\ce{NH4Cl}$): cation hydrolyzes, releases $\ce{H+}$ → acidic.
• Weak acid + strong base (e.g., $\ce{CH3COONa}$): anion hydrolyzes, grabs $\ce{H+}$ → basic.
• Strong acid + strong base (e.g., $\ce{NaCl}$): neither hydrolyzes → neutral, pH 7.
• Weak acid + weak base: both hydrolyze; pH depends on relative $K_a$ vs. $K_b$.

Calculating pH of a Salt Solution

Must know

Identify the hydrolyzing ion, find its $K$ (from $K_w/K_a$ or $K_w/K_b$), and run ICE.

Worked Example: pH of 0.10 M sodium acetate. Acetate is the conjugate base of acetic acid ($K_a = 1.8 \times 10^{-5}$):

$K_b = \frac{K_w}{K_a} = \frac{10^{-14}}{1.8 \times 10^{-5}} = 5.6 \times 10^{-10}$

$\ce{CH3COO- + H2O <=> CH3COOH + OH-}, \quad x = [\ce{OH-}] = \sqrt{K_b \cdot C} \approx 7.5 \times 10^{-6} \text{ M}$

$\text{pOH} \approx 5.12 \Rightarrow \text{pH} = 14 - 5.12 = 8.88$

pH > 7 for a basic salt solution ✓.

Quick check: Approximate pH of 0.1 M $\ce{NH4Cl}$? (p$K_a$ of $\ce{NH4+}$ = 9.25, so $K_a = 5.6 \times 10^{-10}$)

Answer: $x = \sqrt{K_a \cdot C} = \sqrt{5.6 \times 10^{-11}} \approx 7.5 \times 10^{-6}$ M; pH $\approx 5.12$. Acidic, as expected for the salt of a weak base and strong acid.

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Buffers

The Concept

Must know

A buffer resists pH change on adding small amounts of strong acid or base. It contains both a weak acid (to neutralize added base) and its conjugate base (to neutralize added acid), in significant concentrations. It acts as a proton reservoir: added $\ce{H+}$ is grabbed by the conjugate base; added $\ce{OH-}$ is neutralized by the weak acid. The equilibrium shifts only slightly, so pH barely moves.

The Henderson-Hasselbalch Equation

Must know

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

where $[\text{A}^-]$ is the conjugate base and $[\text{HA}]$ the weak acid. Key insights:

1. When $[\text{A}^-] = [\text{HA}]$, $\log(1) = 0$, so pH = p$K_a$ (buffer midpoint).
2. Valid when both species are significant (neither near zero).
3. pH depends on the ratio, not absolute concentrations — diluting a buffer doesn't change pH (to first approximation).

Buffer Range and Capacity

Must know

A buffer works best within ±1 unit of the p$K_a$ (ratio $[\text{A}^-]/[\text{HA}]$ between 0.1 and 10) — the effective buffer range. Buffer capacity (how much acid/base it absorbs before pH shifts) increases with total buffer concentration.

Physiologically Important Buffer Systems

Must know

Bicarbonate ($\ce{HCO3-/H2CO3}$, p$K_a$ ≈ 6.1) is the dominant blood-plasma buffer; phosphate ($\ce{H2PO4-/HPO4^{2-}}$, ≈ 7.2) matters intracellularly and in urine; proteins/hemoglobin also buffer blood.

The bicarbonate buffer: despite p$K_a$ 6.1 being far from blood pH 7.4, it is enormously effective because it is an open system — $\ce{CO2}$ is volatile and continuously regulated by the lungs (ventilation), and $\ce{HCO3-}$ by the kidneys, so the body shifts the equilibrium by changing breathing rate.

Worked Example: H-H Calculation

Must know

A buffer of 0.20 M acetic acid + 0.30 M sodium acetate (p$K_a$ = 4.75):

$\text{pH} = 4.75 + \log\frac{0.30}{0.20} = 4.75 + \log(1.5) \approx 4.75 + 0.18 = 4.93$

pH > p$K_a$ because there is more conjugate base than acid ✓.

Quick check: Adding a small amount of $\ce{NaOH}$ to this buffer — which component is consumed and which produced?

Answer: $\ce{OH-}$ reacts with the weak acid: $\ce{CH3COOH + OH- -> CH3COO- + H2O}$. Acid decreases, conjugate base increases, the ratio rises — pH rises modestly.

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Titration Curves

Must know

A titration curve plots pH vs. volume of titrant. Knowing the shape and what each region means is essential.

Strong Acid–Strong Base Titration

Must know

• Start: low pH (set by $[\text{acid}]$).
• Near equivalence: pH rises steeply.
• Equivalence point: pH = 7.0 exactly (only water + neutral salt).
• After: pH > 7 from excess $\ce{OH-}$.

A large near-vertical jump at equivalence makes indicator choice easy.

Weak Acid–Strong Base Titration

Must know

The more MCAT-relevant case (e.g., acetic acid + NaOH). Key landmarks:

1. Initial pH: higher than a strong acid of equal concentration (partial dissociation).
2. Buffer region: both $\ce{CH3COOH}$ and $\ce{CH3COO-}$ present; the curve is relatively flat.
3. Half-equivalence point: $[\text{A}^-] = [\text{HA}]$, so pH = p$K_a$ — read p$K_a$ directly off the curve here.
4. Equivalence point: all acid → conjugate base, a basic salt solution (pH > 7, typically 8–10); solve by salt hydrolysis.
5. After: excess NaOH; pH rises steeply.

The vertical jump is smaller than for a strong acid because the starting pH is already elevated.

Weak Base Titrated with Strong Acid

Must know

The mirror image: starts at high pH, half-equivalence gives pH = p$K_a$ (of the conjugate acid), and the equivalence point is acidic (pH < 7).

Polyprotic Acid Titrations

Must know

Polyprotic acids ($\ce{H3PO4}$, $\ce{H2CO3}$) show multiple equivalence points — one per dissociable proton — with a buffer region and half-equivalence point (pH = p$K_{a,n}$) between each. Recognize the staircase shape and read p$K_a$ values from the half-equivalence points.

Indicators

Must know

Acid-base indicators are weak acids whose protonated and deprotonated forms differ in color; the change occurs near the indicator's p$K_a$. Choose one whose p$K_a$ is near the equivalence-point pH:

• Strong/strong (pH 7): phenolphthalein or bromothymol blue (the jump spans pH 4–10).
• Weak acid/strong base (pH 8–10): phenolphthalein.
• Weak base/strong acid (pH 4–6): methyl orange or methyl red.

Quick check: Titrating ammonia (p$K_a$ of $\ce{NH4+} = 9.25$) with HCl — pH at the half-equivalence point?

Answer: $[\ce{NH3}] = [\ce{NH4+}]$, so pH = p$K_a$ of $\ce{NH4+}$ = 9.25.

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Common Confusions & Tricks

1. p$K_a$ and strength. Lower p$K_a$ = larger $K_a$ = more dissociation = stronger acid. Mnemonic: "lower pKa = louder acid."

2. Equivalence point ≠ pH 7 (unless strong/strong). Weak acid/strong base equivalence is above 7; weak base/strong acid is below 7. Only both-strong gives pH 7.

3. Half-equivalence point = p$K_a$. Memorize cold. To read p$K_a$ off a curve: find the equivalence volume, halve it, read the pH.

4. The H-H ratio is conjugate base over acid. $\text{pH} = \text{p}K_a + \log[\text{base}]/[\text{acid}]$. Inverting it makes pH wrong by twice the log term.

5. Diluting a buffer doesn't change pH — but over-diluting destroys capacity. H-H depends on the ratio; very dilute buffers have negligible capacity and $K_w$ effects creep in.

6. $K_a \times K_b = K_w$ only for a conjugate pair — an acid and its own conjugate base, not two unrelated species.

7. Salt of weak acid/strong base → basic; strong acid/weak base → acidic. The ion from the strong partner is inert; the ion from the weak partner hydrolyzes.

8. The "5% rule." If $x/C > 5\%$, the $\sqrt{K_a C}$ approximation fails — use the quadratic. Happens when $K_a$ is tiny or the solution very dilute.

9. Buffer range vs. capacity. Range (±1 pH unit of p$K_a$) = which pH it holds; capacity = how much acid/base it absorbs before breaking.

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Key Equations

Equation	Variables & When to Use
$K_w = [\ce{H+}][\ce{OH-}] = 1.0 \times 10^{-14}$	Ion product of water (25°C); find one ion from the other
$\text{pH} = -\log[\ce{H+}]$	Definition of pH
$\text{pOH} = -\log[\ce{OH-}]$	Definition of pOH
$\text{pH} + \text{pOH} = 14$	From $K_w$ at 25°C; links pH and pOH
$K_a = \dfrac{[\ce{H+}][\ce{A-}]}{[\ce{HA}]}$	Acid dissociation constant; ICE for weak acid pH
$K_b = \dfrac{[\ce{BH+}][\ce{OH-}]}{[\ce{B}]}$	Base dissociation constant; ICE for weak base pH
$K_a \times K_b = K_w$	Conjugate pair; find $K_b$ from $K_a$ for hydrolysis
$\text{p}K_a + \text{p}K_b = 14$	Log form of above
$[\ce{H+}] \approx \sqrt{K_a \cdot C}$	Weak acid approximation (valid when $x < 5\%$ of $C$)
$\text{pH} = \text{p}K_a + \log\dfrac{[\ce{A-}]}{[\ce{HA}]}$	Henderson-Hasselbalch; buffer pH, half-equivalence point
$\text{pH} = \text{p}K_a$ at half-equivalence point	Special case when $[\ce{A-}] = [\ce{HA}]$