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Chemistry and Biology

7% of exam

Inorganic chemistry (molarity, normality, acids and bases, redox, solubility product, pH and pK, electrochemistry), organic chemistry (nomenclature, structure, reactions, synthesis), analytical chemistry, biochemistry and microbiology (cell function, glycolysis and the Krebs cycle, enzymes, genetics), and bioprocessing (fermentation and aerobic/anaerobic treatment).

5 concepts

A. Inorganic chemistry

Solutions, pH, Buffers and Electrochemistry

Concentration units, weak-acid and buffer pH, the solubility product, and the Nernst equation — the aqueous-chemistry workhorses the FE rewards for speed.

Aqueous inorganic chemistry is where the FE Chemical exam hands out fast points to anyone who is fluent — and quietly takes them back from anyone who confuses a concentration unit, drops a logarithm, or sets up a half-cell backwards. Almost every question in this area is one of five moves: convert a concentration, find the pH of a weak acid or buffer, decide whether a salt precipitates, or compute a cell potential. The relations below follow the FE Reference Handbook — Chemistry and Biology (equilibrium constant, solubility product, acids/bases/pH, and the Nernst equation), and they are short enough to recall cold. What earns the points is keeping the bookkeeping straight: molarity versus normality, concentration versus activity, and reduction-direction versus oxidation-direction potentials.

Molarity, normality, and the equivalent

Molarity

M

is moles of solute per liter of solution; it is the unit nearly all equilibrium expressions want. Normality

N

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is molarity multiplied by the number of equivalents per mole — protons donated for an acid, electrons transferred for a redox couple, or charge for an ion. The handbook defines normality as the product of molarity and the number of valence changes, so a

1\,\text{M}

solution of

\mathrm{H_2SO_4}

is

2\,\text{N}

in acid-base titrations because each molecule releases two protons. Working in normality, reactions of any stoichiometry balance directly through

N_1 V_1 = N_2 V_2

at the equivalence point.

N = M \times (\text{equivalents per mole}), \qquad N_1 V_1 = N_2 V_2

pH, pOH, and the water equilibrium

pH is the negative base-ten logarithm of the hydrogen-ion molarity. In any aqueous solution at

25\,^\circ\mathrm{C}

the ion product of water fixes

[\mathrm{H^+}][\mathrm{OH^-}] = K_w = 1.0\times10^{-14}

, so

\mathrm{pH} + \mathrm{pOH} = 14

. A strong acid or base dissociates completely, so its pH comes straight from its formal concentration; the only subtlety is remembering that a

1\times10^{-8}\,\text{M}

strong acid is not pH 8 — water's own autoionization keeps it just below 7.

\mathrm{pH} = -\log_{10}[\mathrm{H^+}], \qquad [\mathrm{H^+}][\mathrm{OH^-}] = K_w = 1.0\times10^{-14}

Weak acids and the pK approximation

A weak acid

\mathrm{HA}

only partially dissociates, governed by

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

, with

\mathrm{p}K_a = -\log_{10}K_a

. When dissociation is small (less than about 5 % of the formal concentration

C

), the produced

[\mathrm{H^+}]

and

[\mathrm{A^-}]

are equal and the undissociated acid is still roughly

C

, giving the compact estimate below. For a weak base, work the parallel

K_b

expression and convert with

\mathrm{p}K_a + \mathrm{p}K_b = 14

. Always sanity-check that the approximation held by confirming the dissociated fraction is indeed small.

[\mathrm{H^+}] \approx \sqrt{K_a\,C} \quad\Rightarrow\quad \mathrm{pH} \approx \tfrac{1}{2}\left(\mathrm{p}K_a - \log_{10}C\right)

Buffers and Henderson-Hasselbalch

A buffer is a weak acid and its conjugate base together; it resists pH change because added acid or base is mostly absorbed by converting one form to the other. Rearranging the

K_a

expression and taking logs gives the Henderson-Hasselbalch equation, in which the pH depends only on the

\mathrm{p}K_a

and the ratio of conjugate base to acid — not on dilution, since both species dilute equally. Buffer capacity is greatest when that ratio is one, i.e., at

\mathrm{pH} = \mathrm{p}K_a

, which is exactly the half-equivalence point of a titration.

\mathrm{pH} = \mathrm{p}K_a + \log_{10}\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}

Solubility product

For a sparingly soluble salt

\mathrm{A}_m\mathrm{B}_n

dissolving to

m\,\mathrm{A}^{n+} + n\,\mathrm{B}^{m-}

, the equilibrium

K_{sp}

is the product of ion activities each raised to its stoichiometric coefficient. Solving for the molar solubility

s

requires writing each ion concentration in terms of

s

before substituting — the coefficients ride into both the exponent and the multiplier. A common-ion already present in solution suppresses solubility, and comparing the reaction quotient

Q

to

K_{sp}

tells you whether a precipitate forms (

Q > K_{sp}

) or dissolves (

Q < K_{sp}

).

K_{sp} = [\mathrm{A}^{n+}]^{m}\,[\mathrm{B}^{m-}]^{n}, \qquad \text{e.g. } \mathrm{M(OH)_2}:\; K_{sp} = (s)(2s)^2 = 4s^3

Electrochemistry and the Nernst equation

A galvanic cell pairs an oxidation at the anode with a reduction at the cathode; oxidation is loss of electrons, reduction is gain. Tabulated standard potentials let you compute

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

using reduction potentials for both, and a positive

E^\circ_{cell}

means the reaction is spontaneous as written. Away from standard conditions the Nernst equation corrects the potential for the reaction quotient

Q

; at

25\,^\circ\mathrm{C}

the prefactor

RT/F\ln(10)

collapses to the famous

0.0592/n

. Faraday's law then converts the charge passed into mass deposited or dissolved at an electrode.

E = E^\circ - \dfrac{RT}{nF}\ln Q = E^\circ - \dfrac{0.0592}{n}\log_{10}Q \;\;(25\,^\circ\mathrm{C})

Exam strategy

Read the unit first: 'M' is molarity, 'N' is normality, and titration shortcuts (

N_1V_1=N_2V_2

) only work in normality. For weak-acid pH use

[\mathrm{H^+}]=\sqrt{K_aC}

and only refine if dissociation exceeds about 5 %. For buffers go straight to Henderson-Hasselbalch — never recompute from

K_a

when a ratio is given. For cells, write both half-reactions as reductions, subtract, and remember the Nernst sign: making the reactant side more concentrated raises

E

. Keep

0.0592/n

memorized so a concentration cell never costs more than thirty seconds.

Key equations

Molarity and normality

M = \dfrac{n_{solute}}{V_{soln}(\text{L})}, \qquad N = M\,z

$M$ in mol/L; $z$ = equivalents per mole (protons for acids, electrons for redox). Normality is solution-context dependent.

Titration equivalence

N_1 V_1 = N_2 V_2

pH and water ion product

\mathrm{pH} = -\log_{10}[\mathrm{H^+}], \quad \mathrm{pH}+\mathrm{pOH}=14, \quad K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 10^{-14}

Weak-acid dissociation

K_a = \dfrac{[\mathrm{H^+}][\mathrm{A^-}]}{[\mathrm{HA}]}, \qquad \mathrm{p}K_a = -\log_{10}K_a

Weak-acid pH approximation

[\mathrm{H^+}] \approx \sqrt{K_a\,C}

Henderson-Hasselbalch

\mathrm{pH} = \mathrm{p}K_a + \log_{10}\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}

Solubility product

\mathrm{A}_m\mathrm{B}_n \rightleftharpoons m\mathrm{A}^{n+}+n\mathrm{B}^{m-}, \quad K_{sp} = [\mathrm{A}^{n+}]^m[\mathrm{B}^{m-}]^n

Equilibrium constant (general)

K_{eq} = \dfrac{[\mathrm{C}]^c[\mathrm{D}]^d}{[\mathrm{A}]^a[\mathrm{B}]^b}

Standard cell potential

E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode}

Nernst equation (25 °C)

E = E^\circ - \dfrac{0.0592}{n}\log_{10}Q

Faraday's law of electrolysis

m = \dfrac{Q\,M}{zF}

$m$ = mass deposited (g); $Q$ = charge (C = A·s); $M$

Worked examples

pH of a weak acid

Problem. Find the pH of $0.10\,\text{M}$ acetic acid ( $K_a = 1.8\times10^{-5}$ ).

Solution. Acetic acid is weak, so use the small-dissociation estimate.

[\mathrm{H^+}] \approx \sqrt{K_a C} = \sqrt{(1.8\times10^{-5})(0.10)} = \sqrt{1.8\times10^{-6}} = 1.34\times10^{-3}\,\text{M}

[\mathrm{H^+}] = \sqrt{(1.8\times10^{-5})(0.10)} = 1.34\times10^{-3}\,\text{M}

Buffer pH from Henderson-Hasselbalch

Problem. An acetate buffer contains $0.10\,\text{M}$ acetic acid and $0.15\,\text{M}$ sodium acetate. With $\mathrm{p}K_a = 4.74$ , find the pH.

Solution. Henderson-Hasselbalch with

Molar solubility from Ksp

Problem. Estimate the molar solubility of $\mathrm{Mg(OH)_2}$ in pure water given $K_{sp} = 5.6\times10^{-12}$

Cell potential by Nernst

Problem. For the Daniell cell, $\mathrm{Cu^{2+}}+2e^-\to\mathrm{Cu}$ ( $E^\circ=+0.337\,\text{V}$

Common pitfalls

•Confusing molarity and normality: $1\,\text{M}\ \mathrm{H_2SO_4}$ is $2\,\text{N}$ in acid-base work. The shortcut $N_1V_1=N_2V_2$ only works in equivalents — using molarities for a polyprotic acid doubles your error.
•Forgetting the stoichiometric coefficient inside $K_{sp}$ : for $\mathrm{M(OH)_2}$ it is $4s^3$
•Computing weak-acid pH as $-\log C$ . That is the strong-acid answer; a weak acid is far less dissociated, so use $\sqrt{K_aC}$
•Sign errors in the Nernst term: writing $Q$ upside down or using $+0.0592/n$ . $Q$ is products over reactants, and raising reactant concentration lowers $Q$ and raises $E$ .
•Mixing oxidation and reduction potentials: $E^\circ_{cell}=E^\circ_{cathode}-E^\circ_{anode}$
•Treating $\mathrm{p}K_a+\mathrm{p}K_b$ as anything but 14 at $25\,^\circ\mathrm{C}$
•Assuming a buffer's pH changes on dilution. Henderson-Hasselbalch depends only on the mole ratio, which dilution preserves — pH barely moves until a component is exhausted.

References

NCEES FE Reference Handbook — Chemistry and Biology
NCEES FE Reference Handbook — Environmental Engineering — Water-chemistry context for pH, alkalinity, and solubility applications
Felder & Rousseau, Elementary Principles of Chemical Processes — Concentration units and solution stoichiometry

B. Organic chemistry

Organic Chemistry: Functional Groups and Reactions

Functional-group recognition, the carbon oxidation ladder, common reaction types, and the isomerism counting that the FE tests for fast structural reasoning.

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C. Analytical chemistry

Analytical Chemistry and Instrumental Methods

Titration equivalence, gravimetric and volumetric stoichiometry, the Beer-Lambert law, and how to choose among the common instrumental methods of analysis.

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D. Biochemistry, microbiology, and molecular biology

Biochemistry, Microbiology and Molecular Biology

Cell structure, the glycolysis-Krebs energy pathway and its ATP yield, Michaelis-Menten enzyme kinetics, and the central dogma of molecular biology.

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E. Bioprocessing

Bioprocessing: Fermentation and Biological Treatment

Microbial growth and yield, batch versus fed-batch versus continuous (chemostat) operation, aerobic and anaerobic treatment, and oxygen transfer in bioreactors.

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Equal equivalents at the endpoint. Volumes in any consistent unit; valid only in normality, not molarity (unless 1:1).

At $25\,^\circ\mathrm{C}$ . $[\mathrm{H^+}]$ in mol/L. Acids pH < 7, bases pH > 7.

Equilibrium constant for $\mathrm{HA}\rightleftharpoons \mathrm{H^+}+\mathrm{A^-}$ . Smaller $K_a$ (larger $\mathrm{p}K_a$ ) = weaker acid.

Valid when dissociation $<5\%$ of formal concentration $C$ (mol/L). Verify the fraction $[\mathrm{H^+}]/C$ afterward.

Buffer pH from the conjugate-base to acid ratio. Independent of dilution; $\mathrm{pH}=\mathrm{p}K_a$ when the ratio is 1.

Solids and pure liquids omitted (activity = 1). Compare $Q$ to $K_{sp}$ to test precipitation.

For $a\mathrm{A}+b\mathrm{B}\rightleftharpoons c\mathrm{C}+d\mathrm{D}$ . Brackets are activities; = molar concentration in dilute solution, partial pressure for gases, 1 for pure solids/liquids.

Both values are reduction potentials (V). $E^\circ_{cell}>0$ means spontaneous as written.

$n$ = electrons transferred; $Q$ = reaction quotient (products over reactants). Full form: $E=E^\circ-(RT/nF)\ln Q$ , $F=96{,}485\,\mathrm{C/mol}$ .

= molar mass (g/mol);

z

= electrons per ion;

F=96{,}485\,\mathrm{C/mol}

.

\mathrm{pH} = -\log_{10}(1.34\times10^{-3}) = 2.87

. Validity check: dissociated fraction

= 1.34\times10^{-3}/0.10 = 1.3\% < 5\%

, so the approximation holds. Sanity check: a weak

0.1\,\text{M}

acid should land between the strong-acid pH 1.0 and neutral 7 — pH

2.87

is reasonable. Final:

\mathrm{pH} = 2.87

.

[\mathrm{A^-}]=0.15

,

[\mathrm{HA}]=0.10

:

\mathrm{pH} = 4.74 + \log_{10}(0.15/0.10) = 4.74 + \log_{10}(1.5) = 4.74 + 0.176 = 4.92

. Sanity check: more conjugate base than acid pushes pH above

\mathrm{p}K_a

, and indeed

4.92 > 4.74

. Final:

\mathrm{pH} = 4.92

.

\mathrm{pH} = 4.74 + \log_{10}\dfrac{0.15}{0.10} = 4.92

.

Solution. Dissolution:

\mathrm{Mg(OH)_2} \rightleftharpoons \mathrm{Mg^{2+}} + 2\mathrm{OH^-}

. If

s

mol/L dissolves,

[\mathrm{Mg^{2+}}]=s

and

[\mathrm{OH^-}]=2s

.

K_{sp} = (s)(2s)^2 = 4s^3 = 5.6\times10^{-12}

, so

s = \left(\dfrac{5.6\times10^{-12}}{4}\right)^{1/3} = (1.4\times10^{-12})^{1/3} = 1.12\times10^{-4}\,\text{M}

. Sanity check:

[\mathrm{OH^-}]=2s=2.24\times10^{-4}

gives

\mathrm{pOH}=3.65

,

\mathrm{pH}=10.4

— appropriately basic for a dissolving hydroxide. Final:

s = 1.12\times10^{-4}\,\text{M}

.

s = \left(\dfrac{K_{sp}}{4}\right)^{1/3} = 1.12\times10^{-4}\,\text{M}

) and

\mathrm{Zn^{2+}}+2e^-\to\mathrm{Zn}

(

E^\circ=-0.763\,\text{V}

). Find

E

when

[\mathrm{Cu^{2+}}]=0.10\,\text{M}

and

[\mathrm{Zn^{2+}}]=1.0\,\text{M}

.

Solution. Copper has the higher reduction potential, so it is the cathode:

E^\circ_{cell}=0.337-(-0.763)=1.100\,\text{V}

. Net reaction

\mathrm{Zn}+\mathrm{Cu^{2+}}\to\mathrm{Zn^{2+}}+\mathrm{Cu}

, so

Q=[\mathrm{Zn^{2+}}]/[\mathrm{Cu^{2+}}]=1.0/0.10=10

, with

n=2

.

E = 1.100 - \dfrac{0.0592}{2}\log_{10}(10) = 1.100 - 0.0296 = 1.070\,\text{V}

. Sanity check: depleting the cathode reactant (

\mathrm{Cu^{2+}}

) lowers

E

below

E^\circ

, as found. Final:

E = 1.07\,\text{V}

.

E = 1.100 - \dfrac{0.0592}{2}\log_{10}(10) = 1.07\,\text{V}

, not

s^2

. The 2 enters both as the exponent on

[\mathrm{OH^-}]

and as the multiplier

2s

.

uses reduction potentials for BOTH electrodes. The handbook corrosion table is written as oxidation half-cells — flip the sign before combining.

, or forgetting

\mathrm{pH}+\mathrm{pOH}=14

when a base is given by

[\mathrm{OH^-}]

.