MCAT General Chemistry: Acid-base, Electrochemistry

Last updated: May 2, 2026

General Chemistry: Acid-base, Electrochemistry questions are one of the highest-leverage areas to study for the MCAT. This guide breaks down the rule, the elements you need to recognize, the named traps that catch most students, and a memory aid that scales to test day. Read it once, then practice the same sub-topic adaptively in the app.

The rule

Acid-base and electrochemistry questions both reduce to the same move: identify what is at equilibrium, then plug the correct concentrations into the correct relationship. For acid-base, that relationship is the Henderson-Hasselbalch form $pH = pK_a + \log\frac{[A^-]}{[HA]}$ for buffers, or $K_a = \frac{[H^+][A^-]}{[HA]}$ for an isolated weak acid. For an electrochemical cell, it is the Nernst equation $E = E^{\circ} - \frac{0.0592}{n}\log Q$ at $25^{\circ}\text{C}$, where $Q$ is products-over-reactants for the cell reaction as written. Get the expression right, the sign right, and the units right — the number falls out.

Elements breakdown

Identify the species at equilibrium

Decide whether you have a strong acid/base (full dissociation), a weak acid/base (partial dissociation governed by $K_a$ or $K_b$), or a buffered conjugate pair.

  • Strong acid: $[H^+] = C_{acid}$
  • Weak acid alone: solve $K_a = \frac{x^2}{C_0 - x}$
  • Buffer: use Henderson-Hasselbalch
  • Salt of weak acid: hydrolyze with $K_b = \frac{K_w}{K_a}$

Common examples:

  • $0.10\,\text{M}$ HCl $\rightarrow$ pH = 1
  • $0.10\,\text{M}$ HF $\rightarrow$ pH from $K_a$

Use Henderson-Hasselbalch correctly

For any buffer, $pH = pK_a + \log\frac{[\text{conjugate base}]}{[\text{weak acid}]}$. The base goes on top.

  • Confirm both members of conjugate pair are present
  • Compute $pK_a = -\log K_a$
  • Plug ratio with base in numerator
  • Check: $pH > pK_a$ when base dominates

Build a cell from standard reduction potentials

$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$, where both are looked up as reduction potentials. Do NOT flip the sign of the anode value before subtracting — the formula already does it.

  • Cathode = species more easily reduced
  • Anode = species being oxidized
  • Subtract anode reduction potential
  • If $E^{\circ}_{\text{cell}} > 0$, reaction is spontaneous

Apply Nernst at non-standard conditions

$E = E^{\circ} - \frac{0.0592}{n}\log Q$ at $25^{\circ}\text{C}$. $Q$ is written for the cell reaction as it proceeds spontaneously.

  • Determine $n$ = electrons transferred per reaction
  • Write $Q$ = products / reactants (aqueous only)
  • If $Q < 1$, $E > E^{\circ}$
  • If $Q > 1$, $E < E^{\circ}$

Connect $\Delta G$, $E$, and $K$

Three thermodynamic statements of the same fact: $\Delta G^{\circ} = -nFE^{\circ}$ and $\Delta G^{\circ} = -RT\ln K$.

  • $E^{\circ} > 0 \Leftrightarrow \Delta G^{\circ} < 0 \Leftrightarrow K > 1$
  • $F = 96{,}485\,\text{C/mol}$
  • Use SI units (J, V, mol) — no kJ

Common patterns and traps

The Inverted Henderson-Hasselbalch Ratio

Henderson-Hasselbalch puts the conjugate BASE in the numerator: $pH = pK_a + \log\frac{[A^-]}{[HA]}$. Students under time pressure often write $\log\frac{[HA]}{[A^-]}$ instead, producing a pH on the wrong side of $pK_a$. The wrong choice will be the mirror image of the right one across $pK_a$.

If the correct answer is $pH = 7.40$ and $pK_a = 7.21$, the trap choice is $pH = 7.02$ — same distance from $pK_a$, opposite direction.

Sign Flip on the Anode Potential

$E^{\circ}_{\text{cell}} = E^{\circ}_{\text{cathode}} - E^{\circ}_{\text{anode}}$ uses both values as TABULATED REDUCTION potentials. Test-takers who 'flip the anode to oxidation' before subtracting end up double-correcting and get the wrong magnitude or sign. Watch for choices that differ from the correct answer by exactly twice the anode's potential.

Correct $E^{\circ}_{\text{cell}} = +1.10\,\text{V}$ from Cu/Zn; trap choice is $+0.42\,\text{V}$ or $-0.42\,\text{V}$ from incorrectly adding the negated anode value.

Strong-Acid Treatment of a Weak Acid

For a $0.10\,\text{M}$ weak acid with $K_a = 10^{-5}$, only about 1% dissociates, so $[H^+] \approx 10^{-3}\,\text{M}$ and $pH \approx 3$. Treating it as a strong acid would give $pH = 1$. Watch for any pH that exactly matches $-\log C_{acid}$ when the species is clearly weak.

Question gives $0.10\,\text{M}$ formic acid ($K_a = 1.8 \times 10^{-4}$). Trap choice: $pH = 1.0$. Correct answer is near $pH = 2.4$.

The Nernst Direction Check

After Nernst, ask: did $E$ move the way it should? If reactants increased ($Q$ decreased), $E$ should rise above $E^{\circ}$. If products built up ($Q > 1$), $E$ should fall. A wrong-sign error in the Nernst term reverses this and is the most common Nernst mistake.

$E^{\circ} = 1.10\,\text{V}$, $Q = 100$. Correct $E = 1.04\,\text{V}$ (lower). Trap choice: $1.16\,\text{V}$ (higher — sign error).

Confusing Concentration with Moles in a Buffer

Henderson-Hasselbalch uses the RATIO of conjugate base to weak acid; because both are in the same solution, that ratio equals either the mole ratio OR the molarity ratio — they are interchangeable. But if a question changes total volume by adding water, the ratio (and pH) is unchanged. Distractors lean on the false intuition that dilution shifts buffer pH.

A buffer at $pH = 4.74$ is diluted from $1.0\,\text{L}$ to $2.0\,\text{L}$. Trap choices show pH shifted by $\log 2$; the correct answer is unchanged at $4.74$.

How it works

Imagine a buffer made by mixing $0.10\,\text{mol}$ of acetic acid ($pK_a = 4.74$) with $0.10\,\text{mol}$ of sodium acetate in $1\,\text{L}$ of water. Henderson-Hasselbalch gives $pH = 4.74 + \log\frac{0.10}{0.10} = 4.74$. Now drop in $0.01\,\text{mol}$ of NaOH: it converts $0.01\,\text{mol}$ of acid into base, so the new ratio is $\frac{0.11}{0.09}$ and $pH = 4.74 + \log(1.22) \approx 4.83$. The pH barely moves — that is what a buffer does. Now switch to a galvanic cell of $\text{Cu}^{2+}/\text{Cu}$ at the cathode and $\text{Zn}^{2+}/\text{Zn}$ at the anode. $E^{\circ}_{\text{cell}} = 0.34 - (-0.76) = +1.10\,\text{V}$, positive, so it runs spontaneously and lights a bulb. If $[\text{Cu}^{2+}]$ drops to $0.010\,\text{M}$ while $[\text{Zn}^{2+}]$ stays at $1.0\,\text{M}$, then $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = 100$ and $E$ falls by $\frac{0.0592}{2}\log(100) = 0.059\,\text{V}$, giving $1.04\,\text{V}$. Same logic both times: identify the equilibrium expression, write Q, plug in.

Worked examples

Worked Example 1
Dr. Marta Reyes is preparing an in vitro buffer to mimic intracellular pH for an enzyme assay. She uses the dihydrogen phosphate / monohydrogen phosphate pair: $\text{H}_2\text{PO}_4^- \rightleftharpoons \text{HPO}_4^{2-} + \text{H}^+$, with $pK_{a2} = 7.21$ at $37^{\circ}\text{C}$. Her target pH is $7.40$, matching cytosolic conditions for a fibroblast cell line. She prepares a $1.00\,\text{L}$ stock by combining solid $\text{KH}_2\text{PO}_4$ and $\text{K}_2\text{HPO}_4$. Total phosphate concentration is fixed at $0.100\,\text{M}$. After dissolving the salts, she measures the pH with a calibrated electrode and confirms $7.40 \pm 0.02$. She then dilutes a $10.0\,\text{mL}$ aliquot to $50.0\,\text{mL}$ in a separate flask for kinetic runs and remeasures the pH.

What ratio of $[\text{HPO}_4^{2-}]$ to $[\text{H}_2\text{PO}_4^-]$ did Dr. Reyes need to achieve pH 7.40 in the stock solution?

  • A 0.65
  • B 1.55 ✓ Correct
  • C 0.19
  • D 3.10

Why B is correct: Apply Henderson-Hasselbalch with the conjugate base on top: $7.40 = 7.21 + \log\frac{[\text{HPO}_4^{2-}]}{[\text{H}_2\text{PO}_4^-]}$. Solving, $\log(\text{ratio}) = 0.19$, so the ratio is $10^{0.19} \approx 1.55$. Because pH > $pK_a$, the base form must dominate — confirming the answer is greater than 1.

Why each wrong choice fails:

  • A: This is the inverted ratio $\frac{1}{1.55} = 0.65$, obtained by writing Henderson-Hasselbalch with the acid on top instead of the base. Since pH > $pK_a$, the answer must be greater than 1. (The Inverted Henderson-Hasselbalch Ratio)
  • C: This is the value of $\log(\text{ratio})$, not the ratio itself. The student stopped one step early and forgot to take $10^{0.19}$.
  • D: This is roughly $2 \times 1.55$, the result of doubling the antilog or computing $10^{0.19} \times 2$. There is no algebraic justification — it represents an arithmetic slip after the correct setup.
Worked Example 2
Dr. Fei Liu builds a galvanic cell to demonstrate the Nernst equation to a class. The cell uses a zinc electrode in $1.00\,\text{M}\,\text{ZnSO}_4$ and a copper electrode in a solution whose $[\text{Cu}^{2+}]$ can be varied. The half-reactions and standard reduction potentials at $25^{\circ}\text{C}$ are: $\text{Cu}^{2+} + 2e^- \rightarrow \text{Cu}$, $E^{\circ} = +0.34\,\text{V}$; $\text{Zn}^{2+} + 2e^- \rightarrow \text{Zn}$, $E^{\circ} = -0.76\,\text{V}$. With $[\text{Cu}^{2+}] = 1.00\,\text{M}$, the voltmeter reads $1.10\,\text{V}$, matching the calculated $E^{\circ}_{\text{cell}}$. Dr. Liu then dilutes the copper compartment so that $[\text{Cu}^{2+}] = 0.010\,\text{M}$, while $[\text{Zn}^{2+}]$ remains at $1.00\,\text{M}$. The temperature is held at $25^{\circ}\text{C}$ throughout.

What cell potential does the voltmeter read after the dilution?

  • A 1.04 V ✓ Correct
  • B 1.10 V
  • C 1.16 V
  • D 0.42 V

Why A is correct: The cell reaction is $\text{Zn} + \text{Cu}^{2+} \rightarrow \text{Zn}^{2+} + \text{Cu}$, so $Q = \frac{[\text{Zn}^{2+}]}{[\text{Cu}^{2+}]} = \frac{1.00}{0.010} = 100$, with $n = 2$. Apply Nernst: $E = 1.10 - \frac{0.0592}{2}\log(100) = 1.10 - 0.0296 \times 2 = 1.10 - 0.0592 \approx 1.04\,\text{V}$. Because product built up relative to reactant, $E$ drops below $E^{\circ}$ — directionally correct.

Why each wrong choice fails:

  • B: This is $E^{\circ}_{\text{cell}}$ — the answer if you ignore the Nernst correction entirely. The dilution changes $Q$, which must be reflected in the calculated potential.
  • C: This adds the Nernst term instead of subtracting, giving $1.10 + 0.0592$. Since $Q > 1$, the potential must drop below $E^{\circ}$, not rise above it. (The Nernst Direction Check)
  • D: This results from incorrectly flipping the anode's reduction potential before subtracting: computing $0.34 - (+0.76)$ instead of $0.34 - (-0.76)$ as the standard cell potential, then applying Nernst to that wrong baseline. (Sign Flip on the Anode Potential)
Worked Example 3

A student dissolves enough formic acid (HCOOH, $K_a = 1.8 \times 10^{-4}$) in pure water at $25^{\circ}\text{C}$ to make a $0.20\,\text{M}$ solution. No conjugate base is added.

What is the approximate pH of the solution?

  • A 0.70
  • B 2.22 ✓ Correct
  • C 3.74
  • D 4.44

Why B is correct: Only the weak acid is present, so set up the equilibrium $K_a = \frac{x^2}{C_0 - x} \approx \frac{x^2}{0.20}$. Solving, $x = [\text{H}^+] = \sqrt{1.8 \times 10^{-4} \times 0.20} = \sqrt{3.6 \times 10^{-5}} \approx 6.0 \times 10^{-3}\,\text{M}$. Therefore $pH = -\log(6.0 \times 10^{-3}) \approx 2.22$. Because formic acid is weak, do NOT use Henderson-Hasselbalch — there is no conjugate base in solution.

Why each wrong choice fails:

  • A: This treats formic acid as if it fully dissociates, giving $[\text{H}^+] = 0.20\,\text{M}$ and $pH = -\log(0.20) \approx 0.70$. Formic acid is weak ($K_a \sim 10^{-4}$); only a small fraction ionizes. (Strong-Acid Treatment of a Weak Acid)
  • C: This is simply $pK_a = -\log(1.8 \times 10^{-4}) \approx 3.74$, which is the pH of a buffer with equal acid and conjugate base — not the pH of the pure weak acid. Henderson-Hasselbalch does not apply here because no formate has been added. (The Inverted Henderson-Hasselbalch Ratio)
  • D: This corresponds to $-\log(K_a)$ shifted by $\frac{1}{2}\log(0.20)$ in the wrong direction, an algebraic slip when applying $pH \approx \frac{1}{2}(pK_a - \log C_0)$. Performed correctly, that approximation also yields $\approx 2.22$.

Memory aid

Two-step check for any acid-base/electrochem item: (1) What is at equilibrium? (2) Which equation links the unknown to what I'm given? For buffers think 'BASE/ACID, base on top.' For cells think 'CATHODE minus ANODE — never flip the sign yourself.'

Key distinction

A weak acid alone uses $K_a = \frac{x^2}{C_0 - x}$ and gives $pH \approx \frac{1}{2}(pK_a - \log C_0)$. A buffer uses Henderson-Hasselbalch and gives $pH \approx pK_a$. Confusing the two — using H-H when only the acid is present, or vice versa — is the most common acid-base error on the MCAT.

Summary

Acid-base and electrochemistry are equilibrium problems wearing different costumes; identify the equilibrium, write the right expression, watch the signs, and the algebra is trivial.

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Frequently asked questions

What is general chemistry: acid-base, electrochemistry on the MCAT?

Acid-base and electrochemistry questions both reduce to the same move: identify what is at equilibrium, then plug the correct concentrations into the correct relationship. For acid-base, that relationship is the Henderson-Hasselbalch form $pH = pK_a + \log\frac{[A^-]}{[HA]}$ for buffers, or $K_a = \frac{[H^+][A^-]}{[HA]}$ for an isolated weak acid. For an electrochemical cell, it is the Nernst equation $E = E^{\circ} - \frac{0.0592}{n}\log Q$ at $25^{\circ}\text{C}$, where $Q$ is products-over-reactants for the cell reaction as written. Get the expression right, the sign right, and the units right — the number falls out.

How do I practice general chemistry: acid-base, electrochemistry questions?

The fastest way to improve on general chemistry: acid-base, electrochemistry is targeted, adaptive practice — working questions that focus on your specific weak spots within this sub-topic, getting immediate feedback, and revisiting items you missed on a spaced-repetition schedule. Neureto's adaptive engine does this automatically across the MCAT; start a free 7-day trial to see your sub-topic mastery climb in real time.

What's the most important distinction to remember for general chemistry: acid-base, electrochemistry?

A weak acid alone uses $K_a = \frac{x^2}{C_0 - x}$ and gives $pH \approx \frac{1}{2}(pK_a - \log C_0)$. A buffer uses Henderson-Hasselbalch and gives $pH \approx pK_a$. Confusing the two — using H-H when only the acid is present, or vice versa — is the most common acid-base error on the MCAT.

Is there a memory aid for general chemistry: acid-base, electrochemistry questions?

Two-step check for any acid-base/electrochem item: (1) What is at equilibrium? (2) Which equation links the unknown to what I'm given? For buffers think 'BASE/ACID, base on top.' For cells think 'CATHODE minus ANODE — never flip the sign yourself.'

What's a common trap on general chemistry: acid-base, electrochemistry questions?

Inverting the Henderson-Hasselbalch ratio (acid on top instead of base)

What's a common trap on general chemistry: acid-base, electrochemistry questions?

Flipping the anode's reduction potential sign before subtracting

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