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Environmental Chemistry

7% of exam

Stoichiometry and chemical equilibrium, acid-base and oxidation-reduction reactions, pC-pH diagrams, reaction kinetics, organic chemistry, and multimedia partitioning by Henry's law and the octanol-water coefficient.

4 concepts

A. Stoichiometry and chemical reactions

Chemical Equilibrium, Acids–Bases, and pH

Balance reactions and find limiting reactants, then use the equilibrium constant, K_w, and pK_a to predict pH, buffering, and as-CaCO3 equivalents.

Almost every quantitative idea in environmental chemistry rides on two skills: counting moles correctly through a balanced reaction, and finding how far that reaction proceeds at equilibrium. The FE exam tests both relentlessly — a chlorine dose, a lime requirement, the pH of a treated stream, the alkalinity that protects a river from an acid spill all reduce to stoichiometry plus an equilibrium constant. Points are lost not in the chemistry but in the bookkeeping: forgetting that sulfuric acid is diprotic, mixing molarity with normality, or reporting hardness in

\text{mg/L}

of calcium when the problem wanted

\text{mg/L as CaCO}_3

. Everything here lives in the FE Reference Handbook's Chemistry and Biology chapter (equilibrium constant,

K_{sp}

, acids/bases/pH) and the equivalent-weight table in Environmental Engineering.

Moles, stoichiometry, and the limiting reactant

A balanced equation is a recipe in moles, so the first move on any reaction problem is to convert every mass or concentration to moles with

n = m/M

, where

M

is the molar mass read off the periodic table. The reactant that runs out first — the one giving the fewest moles of product when divided by its stoichiometric coefficient — is the limiting reactant, and it fixes the yield; everything else is in excess. In water treatment this is how you size a chemical dose: the dose must at least meet the stoichiometric demand of whatever it reacts with, and any surplus simply remains as residual.

n = \dfrac{m}{M}, \qquad aA + bB \rightarrow cC + dD

The equilibrium constant

For a reversible reaction at equilibrium, the ratio of product activities to reactant activities — each raised to its stoichiometric coefficient — is a constant

K_{eq}

at a given temperature. In dilute aqueous work we use molar concentrations for dissolved species and partial pressures for gases, and we take the activity of any pure solid or of the water solvent as

1

. A large

K_{eq}

means products are favored; a small one means the reaction barely proceeds. The same structure reappears as the acid dissociation constant

K_a

, the solubility product

K_{sp}

, and the ion product of water

K_w

— they are all just

K_{eq}

for a particular kind of reaction.

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

Water, K_w, and the pH scale

Water self-ionizes,

\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-

, with

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

at

25^\circ\text{C}

. Because the product is fixed, knowing one ion fixes the other, and pH and pOH are simply the negative base-ten logarithms of those concentrations. Neutral water has

[\text{H}^+]=[\text{OH}^-]=10^{-7}

, so

\text{pH}=7

; acids drive

\text{pH}<7

and bases

\text{pH}>7

. The single most useful identity is that pH and pOH always sum to

14

at

25^\circ\text{C}

.

K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}, \qquad \text{pH} + \text{pOH} = 14

Strong versus weak acids and bases

A strong acid (HCl,

\text{HNO}_3

,

\text{H}_2\text{SO}_4

) dissociates essentially completely, so

[\text{H}^+]

equals the acid normality and pH follows directly — but mind the protons per molecule:

\text{H}_2\text{SO}_4

delivers two. A weak acid only partly dissociates, governed by

K_a

, so

[\text{H}^+]

is far smaller than the analytical concentration. For a weak monoprotic acid at concentration

C_a

where dissociation is slight, the hydrogen-ion concentration is well approximated by

[\text{H}^+]\approx\sqrt{K_a C_a}

, which you can derive in seconds from the

K_a

expression by neglecting the small amount that dissociates.

K_a = \dfrac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}, \qquad [\text{H}^+] \approx \sqrt{K_a\,C_a}

Buffers and the Henderson–Hasselbalch equation

A buffer — a weak acid together with its conjugate base — resists pH change because added acid or base is absorbed by the conjugate pair rather than changing

[\text{H}^+]

directly. Rearranging the

K_a

expression and taking logarithms gives the Henderson–Hasselbalch relation: pH equals

\text{p}K_a

plus the log of the base-to-acid ratio. The practical lesson is that a buffer is strongest when

[\text{A}^-]\approx[\text{HA}]

, i.e. when

\text{pH}\approx\text{p}K_a

— exactly the carbonate buffering that holds most natural waters near

\text{pH}\,6.3

to

8.3

and keeps treatment processes stable.

\text{pH} = \text{p}K_a + \log_{10}\!\left(\dfrac{[\text{A}^-]}{[\text{HA}]}\right)

As-CaCO3 equivalents, hardness, and alkalinity

Environmental chemists add up ions of different charge by converting each to a common currency: milligrams per liter as

\text{CaCO}_3

, whose equivalent weight is exactly

50\ \text{g/eq}

. You convert any species by multiplying its

\text{mg/L}

by the ratio of

50

to that species' own equivalent weight,

M/z

. Total hardness is the sum of

\text{Ca}^{2+}

and

\text{Mg}^{2+}

expressed this way; alkalinity is the acid-neutralizing capacity, dominated in most waters by bicarbonate and carbonate. Working as

\text{CaCO}_3

lets you compare and combine hardness, alkalinity, and lime dose on one scale — the backbone of softening and stabilization calculations.

\text{mg/L as CaCO}_3 = \text{mg/L as species} \times \dfrac{50}{\,M/z\,}

Exam strategy

Start every reaction problem by balancing and converting to moles, then test for the limiting reactant before computing any yield. For pH, decide first whether the acid or base is strong (use

[\text{H}^+]=

normality directly) or weak (use

[\text{H}^+]\approx\sqrt{K_a C_a}

, and remember

\text{pH}=\text{p}K_a

at the half-equivalence point). Keep

K_w

handy to flip between pH and pOH. When a problem mixes ions — hardness, alkalinity, dosing — convert everything to

\text{mg/L as CaCO}_3

using equivalent weights (

\text{CaCO}_3 = 50

,

\text{Ca}^{2+}=20

,

\text{HCO}_3^- = 61

) before adding. Watch diprotic acids and the difference between molarity and normality.

Key equations

Moles from mass

n = \dfrac{m}{M}

$n$ = moles, $m$ = mass (g), $M$ = molar mass (g/mol). The starting point of every stoichiometry problem.

Equilibrium constant

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

Ion product of water

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

pH and pOH

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

Acid dissociation constant

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

Weak-acid pH approximation

[\text{H}^+] \approx \sqrt{K_a\,C_a}

Henderson–Hasselbalch

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

Solubility product

K_{sp} = [\text{A}^{+}]^{m}[\text{B}^{-}]^{n}

Equivalent weight and normality

EW = \dfrac{M}{z}, \qquad N = M\,z

$M$

Conversion to as-CaCO3

C_{\text{as CaCO}_3} = C_{species}\times\dfrac{50}{EW_{species}}

Alkalinity (carbonate system)

\text{Alk} = [\text{HCO}_3^-] + 2[\text{CO}_3^{2-}] + [\text{OH}^-] - [\text{H}^+]

Worked examples

Limiting reactant and the resulting pH

Problem. A $1.0\ \text{L}$ acidic waste contains $0.10\ \text{mol}$ of sulfuric acid ( $\text{H}_2\text{SO}_4$ ). It is dosed with $0.15\ \text{mol}$ of sodium hydroxide ( $\text{NaOH}$ ). Treating both as strong and the reaction as complete, identify the limiting reactant and find the final pH.

Solution. Sulfuric acid is diprotic, so

0.10\ \text{mol}

supplies

0.20\ \text{mol}

of

\text{H}^+

. Each

\text{OH}^-

neutralizes one

\text{H}^+

\text{pH} = -\log_{10}(0.20-0.15) = 1.30

pH of a weak acid

Problem. Find the pH of a $0.10\ \text{M}$ acetic acid solution. The acid dissociation constant is $K_a = 1.8\times10^{-5}$ ( $\text{p}K_a = 4.74$

Total hardness as CaCO3

Problem. A groundwater contains $\text{Ca}^{2+} = 60\ \text{mg/L}$ and $\text{Mg}^{2+} = 18\ \text{mg/L}$ . Express the total hardness in mg/L as $\text{CaCO}_3$

Common pitfalls

•Forgetting diprotic/polyprotic acids: $\text{H}_2\text{SO}_4$ gives two moles of $\text{H}^+$ per mole, $\text{H}_3\text{PO}_4$ three. Use normality, not molarity, for neutralization.
•Treating a weak acid as strong: a $0.1\ \text{M}$ weak acid is nowhere near $\text{pH}=1$ . Use $[\text{H}^+]=\sqrt{K_aC_a}$
•Confusing $\text{p}K_a$ with pH. They coincide only at the half-equivalence point, where $[\text{HA}]=[\text{A}^-]$ .
•Reporting hardness or alkalinity as mg/L of the actual ion instead of mg/L as CaCO3, or forgetting the $50\ \text{g/eq}$ equivalent weight of CaCO3 when converting.
•Misreading hardness classification: on the Sawyer/McCarty/Parkin scale the very-hard cutoff is $>300\ \text{mg/L as CaCO}_3$ , not $150$ ; $150$ – $300$ is merely hard. Use the correct band before labeling a water.
•Sign/log slips: $\text{pH}=-\log[\text{H}^+]$ , so a 10-fold rise in $[\text{H}^+]$ lowers pH by 1 unit. Confirm acids give pH < 7.
•Mixing molarity and normality when computing as-CaCO3: $N = M\times z$ , and the conversion factor is $50/EW$ , not $50/M$ .

References

NCEES FE Reference Handbook — Chemistry and Biology
NCEES FE Reference Handbook — Environmental Engineering (equivalent-weight / common radicals table)
Davis & Cornwell, Introduction to Environmental Engineering — acid-base chemistry, hardness and alkalinity
Sawyer, McCarty & Parkin, Chemistry for Environmental Engineering and Science — hardness classification scale and as-CaCO3 equivalents
Snoeyink & Jenkins, Water Chemistry — carbonate system and buffering background

Redox Reactions and pC–pH Diagrams

Assign oxidation numbers, balance redox half-reactions and count electron equivalents, read a pC–pH log-concentration diagram, and apply redox to nitrification and iron oxidation.

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C. Organic chemistry

Organic Chemistry for Environmental Engineering

Recognize functional groups, hydrocarbons, and VOCs, assign carbon oxidation states, judge biodegradability, and connect the TOC, COD, BOD, and ThOD surrogate measures.

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D. Multimedia equilibrium partitioning

Reaction Kinetics and Multimedia Partitioning

Use zero-, first-, and second-order rate laws with half-lives and Arrhenius/theta temperature corrections, then partition contaminants among air, water, and solids via H, K_ow, K_d, and BCF.

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Ratio of product to reactant activities for $aA+bB\rightleftharpoons cC+dD$ ; pure solids/liquids count as 1. Dimensionless in concentration form here.

At $25^\circ\text{C}$ . Fixes $[\text{OH}^-]$ once $[\text{H}^+]$ is known (mol/L).

$[\text{H}^+]$ in mol/L. Acids: pH < 7; bases: pH > 7 at $25^\circ\text{C}$ .

Strength of a weak acid HA. Larger $K_a$ (smaller $\text{p}K_a$ ) = stronger acid.

$C_a$ = analytical acid concentration (mol/L). Valid when dissociation is < ~5% of $C_a$ .

Buffer pH from the conjugate base-to-acid ratio. $\text{pH}=\text{p}K_a$ when the two are equal.

For $\text{A}_m\text{B}_n(s)\rightleftharpoons m\text{A}^{+}+n\text{B}^{-}$ . Precipitation occurs when the ion product exceeds $K_{sp}$ .

= molarity (mol/L),

z

= charge or protons exchanged. Normality

N = M\times z

; this is the basis for the as-CaCO3 system. (Do not confuse this

M

with molar mass.)

$EW(\text{CaCO}_3)=50$ g/eq. Use to add hardness/alkalinity ions on a common basis (mg/L as CaCO3).

Acid-neutralizing capacity in eq/L; multiply by $50{,}000$ to express as mg/L as CaCO3.

:

\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}

. Available

\text{OH}^- = 0.15\ \text{mol}

< available

\text{H}^+ = 0.20\ \text{mol}

, so

\textbf{NaOH is the limiting reactant}

and acid is in excess. Leftover

\text{H}^+ = 0.20 - 0.15 = 0.05\ \text{mol}

in

1.0\ \text{L}

, so

[\text{H}^+]=0.05\ \text{mol/L}

.

\text{pH} = -\log_{10}(0.05) = 1.30

. Sanity check: still strongly acidic, as expected when base is short of the stoichiometric demand. Final answer: NaOH limits;

\textbf{pH} = \mathbf{1.30}

.

).

Solution. Acetic acid is weak, so use

[\text{H}^+]\approx\sqrt{K_a C_a}

with

C_a = 0.10\ \text{M}

.

[\text{H}^+] = \sqrt{(1.8\times10^{-5})(0.10)} = \sqrt{1.8\times10^{-6}} = 1.34\times10^{-3}\ \text{mol/L}

.

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

. Sanity check: the fraction dissociated is

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

, well under

5\%

, so the approximation is valid; the exact quadratic gives

\text{pH}=2.88

. A weak acid sits well above the pH of a strong acid at the same concentration (

\text{pH}=1.0

), as it should. Final answer:

\textbf{pH} \approx \mathbf{2.87}

.

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

and classify the water.

Solution. Equivalent weights:

EW(\text{Ca}^{2+}) = 40.1/2 = 20.05\ \text{g/eq}

;

EW(\text{Mg}^{2+}) = 24.3/2 = 12.15\ \text{g/eq}

. Convert each to as-CaCO3 by multiplying by

50/EW

:

\text{Ca}: 60\times\dfrac{50}{20.05} = 149.6\ \text{mg/L as CaCO}_3

.

\text{Mg}: 18\times\dfrac{50}{12.15} = 74.1\ \text{mg/L as CaCO}_3

. Total hardness

= 149.6 + 74.1 = 223.7 \approx 224\ \text{mg/L as CaCO}_3

. Sanity check: equivalently using the standard multipliers

2.50

for Ca and

4.12

for Mg gives

60(2.50)+18(4.12) = 224\ \text{mg/L}

, confirming the result. On the standard Sawyer/McCarty/Parkin hardness scale (

0

–

75

soft,

75

–

150

moderately hard,

150

–

300

hard,

>300

very hard),

224\ \text{mg/L as CaCO}_3

falls in the

150

–

300

band, so the water is classified as hard (it would need to exceed

300\ \text{mg/L}

to be very hard). Final answer:

\mathbf{224\ mg/L\ as\ CaCO_3}

, classified as hard water.

TH = 60\cdot\tfrac{50}{20.05} + 18\cdot\tfrac{50}{12.15} = 224\ \text{mg/L as CaCO}_3

for weak acids and

[\text{H}^+]=

normality only for strong ones.