Chemical Equilibrium

Every reaction is secretly two reactions running in opposite directions. Equilibrium is the point where they tie. Once you can write down an equilibrium constant, read an ICE table, and apply Le Chatelier's principle, a huge slice of chemistry stops feeling like magic and starts feeling like bookkeeping.

1. Why equilibrium matters

You dissolve an aspirin tablet in water. Some of it dissociates into ions, some stays intact. You add a drop of acid to a buffer and the pH barely moves. A pharmaceutical company spends years tuning a synthesis so that an expensive precursor actually converts into product instead of sitting around unreacted. A cement kiln burns limestone to make quicklime, and the industrial engineers care intensely about the exact temperature because it decides how far the reaction runs. Your blood holds its pH at 7.4 within a hundredth of a unit even while you exercise and dump CO₂ into it.

Every one of those stories is the same story: a reaction that goes forward and backward at the same time, and the balance between the two directions decides what you end up with. That balance is chemical equilibrium. It is not "the reaction stopped." The reaction never stops. At equilibrium, forward and reverse are running at identical rates, so the net concentrations sit still. It's dynamic, not dead.

Why should you care, whether you build drugs, batteries, fertilizers, fuel cells, climate models, or just drink coffee in the morning? A few reasons in increasing order of how directly they touch your day job:

• Drug design. Binding affinity of a drug to a receptor is an equilibrium constant. The whole pharmacology vocabulary — $K_d$, $K_i$, IC₅₀ — is equilibrium language in disguise.
• Batteries and fuel cells. The voltage you get depends on how far a redox reaction is from equilibrium. That's the Nernst equation, which is equilibrium plus a logarithm.
• Climate. CO₂ dissolving in the ocean, then hydrolyzing to carbonic acid, then to bicarbonate and carbonate, is a coupled set of equilibria. Ocean acidification is a shift in that set.
• Life itself. Every enzyme, every pH buffer, every ion pump is a controlled equilibrium. Homeostasis is engineering equilibrium.

This page will teach you equilibrium from zero. You need to remember what moles and molarity mean, and how to read a balanced equation. Everything else we will build.

2. Vocabulary cheat sheet

Glance at this table now. Each symbol gets a full treatment below.

One convention warning before we start: concentrations are dimensionless in the formal definition of $K$ (they're implicitly divided by a 1 M reference state), which is why $K$ comes out as a pure number. In practice you just write $[\text{A}]$ in molarity and don't worry about the units — they cancel out if you're careful.

3. The equilibrium constant

Consider a generic reversible reaction:

At equilibrium, the law of mass action says the concentrations obey:

A concrete example: the Haber process, which makes nearly all of the world's fertilizer nitrogen:

Gases: $K_p$ vs $K_c$

When your species are gases, you can use partial pressures instead of concentrations:

What doesn't go in the expression

Pure solids and pure liquids don't appear in $K$ expressions. Their "concentration" is a constant — they have a fixed density — so the usual convention is to fold those constants into $K$ itself. Example: for limestone decomposition

4. The reaction quotient $Q$ — which way does it go?

$K$ tells you the ratio at equilibrium. But what if you mix some $A$, $B$, $C$, and $D$ together and want to know which way the reaction will run? That's what $Q$ is for. $Q$ has the same formula as $K$, but you plug in the current concentrations, not the equilibrium ones:

This is genuinely useful. Imagine you have a reaction vessel with $[A] = 0.5$ M, $[B] = 0.5$ M, and $[C] = 1.0$ M for $A + B \rightleftharpoons C$ with $K_c = 2$. Compute $Q = 1.0 / (0.5 \cdot 0.5) = 4.0$. Since $4 > 2$, the system has too many products. Net reverse. You'll watch $[C]$ drop and $[A], [B]$ rise until $Q$ hits 2.

5. Le Chatelier's principle

Here's a rule that looks like magic and is actually just $Q$-vs-$K$ logic in plain English.

The principle sounds almost like personality advice, and it is often taught as a handwave. But every case is just "what did the disturbance do to $Q$, and what does that imply about direction?"

• Add more reactant. $[A]$ goes up, so $Q$ goes down, so $Q < K$, so the reaction shifts forward to consume the added reactant. Industry uses this to drive reactions to higher yield — keep dumping in one reactant so $Q$ is always low.
• Remove product. $[C]$ drops, $Q$ drops, reaction shifts forward. This is "distillation trick": boil off the product as it forms and you push the equilibrium.
• Compress the volume (gases). All concentrations rise by the same factor, but the side with more gas moles rises faster in the $Q$ expression (more exponents). So the system shifts toward the side with fewer gas moles. Haber synthesis exploits this: $4 \to 2$ mol of gas on forming ammonia, so pressure pushes forward.
• Raise temperature. This is the tricky one. Temperature changes $K$ itself, not just $Q$. If the forward reaction is exothermic (releases heat), higher $T$ decreases $K$ — the equilibrium shifts backward. If endothermic, higher $T$ increases $K$. Dye equilibria in chemistry demos show this beautifully: some go pink at room temperature and blue on heating.

The temperature dependence is quantified by the van't Hoff equation, which is essentially the Clausius-Clapeyron relation applied to $K$:

6. ICE tables — the algebra machine for equilibrium

Most equilibrium problems reduce to this question: "I start with these amounts, and I know $K$. What are the concentrations at equilibrium?" The universal solving technique is the ICE table. ICE stands for Initial, Change, Equilibrium. Make three rows, fill them in, and your unknowns line up on a silver platter.

Example. Consider $N_2O_4(g) \rightleftharpoons 2 NO_2(g)$ with $K_c = 4.6 \times 10^{-3}$ at 25°C. You start with $[N_2O_4]_0 = 0.10$ M and no $NO_2$. Find the equilibrium concentrations.

The change row reflects the stoichiometry: for every $x$ moles per liter of $N_2O_4$ that disappear, $2x$ moles per liter of $NO_2$ appear (coefficient of 2). Now plug equilibrium values into the $K$ expression:

Solving exactly: $(2x)^2 = 4.6 \times 10^{-3} (0.10 - x)$ gives $4x^2 + 4.6 \times 10^{-3} x - 4.6 \times 10^{-4} = 0$. Quadratic formula gives $x \approx 0.0102$ M. So at equilibrium, $[N_2O_4] \approx 0.090$ M and $[NO_2] \approx 0.020$ M. About 10% of the $N_2O_4$ dissociated — which is why you see brown $NO_2$ in a flask that was initially colorless $N_2O_4$.

7. Acid-base equilibria

Acid-base chemistry is just equilibrium chemistry applied to proton transfers. The two big constants are $K_a$ (how strongly an acid gives up its proton) and $K_b$ (how strongly a base grabs one).

pH, pOH, and $K_w$

Pure water autoionizes a little bit: $H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-$. The equilibrium constant for this is

Define $pH = -\log_{10}[H^+]$ and $pOH = -\log_{10}[OH^-]$. Taking $-\log$ of both sides of $K_w$:

Henderson-Hasselbalch and buffers

Rearrange the $K_a$ expression by taking $-\log$ of both sides:

Worked example. You want a buffer at pH 4.8 using acetic acid ($pK_a = 4.76$). Then $\log([A^-]/[HA]) = 4.8 - 4.76 = 0.04$, so $[A^-]/[HA] = 10^{0.04} \approx 1.10$. You need 1.10 moles of sodium acetate per mole of acetic acid. If you mix 0.100 mol acetic acid with 0.110 mol sodium acetate in 1 L of water, you have a pH 4.8 buffer that will hold its pH against modest acid/base additions.

8. Interactive: titration curve plotter

A titration is: you have a weak acid of known concentration, you add strong base slowly, and you watch the pH. The curve has four interesting features — the initial pH, the buffer region near $pH = pK_a$, the steep equivalence point, and the final plateau. Slide $pK_a$ and see how the curve morphs; slide the initial acid concentration and watch the equivalence-point pH rise.

Things to look for as you drag:

• At the half-equivalence point (half the base needed to fully neutralize has been added), $[A^-] = [HA]$ and Henderson-Hasselbalch gives $pH = pK_a$. The pink dot always sits exactly at that height.
• At the equivalence point, all the acid is neutralized. The solution is basic (because you've made the conjugate base $A^-$ which hydrolyzes back). Stronger acids push the equivalence-point pH closer to 7; weaker acids push it higher.
• Dilute acids ($[HA]_0$ small) have flatter curves and less distinct equivalence points — which is why titration accuracy breaks down at very low concentrations.

9. Solubility equilibria — $K_{sp}$

Insoluble salts aren't totally insoluble. A tiny amount dissolves, and the dissolved fraction sits at equilibrium with the solid. For $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$:

Common-ion effect

Add a source of $Cl^-$ to that silver chloride solution — say, sodium chloride at 0.10 M. Now $[Cl^-]$ is fixed at 0.10, so $[Ag^+]$ has to shrink to $K_{sp} / 0.10 = 1.8 \times 10^{-9}$ M to maintain the product. The silver chloride is less soluble in salt water than in pure water. This is the common-ion effect. It's Le Chatelier's principle applied to dissolution: you added a product, so the equilibrium shifts toward the solid.

Practical use: in water treatment, adding excess phosphate precipitates out calcium. In kidney-stone pathology, a high-oxalate diet raises $[C_2O_4^{2-}]$ and shifts $CaC_2O_4$ out of solution inside your kidneys. The same math, different context.

Prediction rule. If you mix ionic solutions and wonder whether a precipitate will form, compute $Q_{sp}$ from the initial concentrations. If $Q_{sp} > K_{sp}$, precipitation will occur until $Q$ drops to $K$. If $Q_{sp} < K_{sp}$, nothing precipitates — the solution is undersaturated.

10. Complex-ion equilibria

Transition metal cations love to grab ligands — small molecules or anions that donate electron pairs to the metal. The equilibrium constant for complex formation is $K_f$:

Biological ligands work the same way. Heme in hemoglobin is an $Fe^{2+}$ porphyrin complex. $O_2$ binding to heme is a complex-formation equilibrium with a $K_f$ that's sharply tuned by surrounding protein to grab oxygen in lungs and release it in tissues. Carbon monoxide poisoning is this equilibrium with a far larger $K_f$ for CO than for $O_2$ — CO wins the binding competition irreversibly.

11. Equilibrium in code

Three small routines: solving for $x$ in an ICE table using the quadratic formula, computing pH of a weak acid, and simulating a titration curve.

import numpy as np

# ---------- 1. Solve (2x)^2 / (C0 - x) = K for x, via quadratic ----------
def ice_solve_dissociation(C0, K):
    # N2O4 -> 2 NO2 style: 4 x^2 + K x - K C0 = 0
    a, b, c = 4.0, K, -K * C0
    disc = b * b - 4 * a * c
    x = (-b + np.sqrt(disc)) / (2 * a)
    return x

# Sanity check against worked example in the text
x = ice_solve_dissociation(0.10, 4.6e-3)
print(f"x = {x:.5f} M,  [N2O4] = {0.10 - x:.4f},  [NO2] = {2*x:.4f}")

# ---------- 2. pH of a weak acid at concentration C, given Ka ----------
def weak_acid_pH(C, Ka):
    # HA -> H+ + A-,  K = x^2 / (C - x)
    # x^2 + K x - K C = 0
    disc = Ka * Ka + 4 * Ka * C
    x = (-Ka + np.sqrt(disc)) / 2
    return -np.log10(x)

print(f"pH of 0.10 M acetic acid: {weak_acid_pH(0.10, 1.8e-5):.2f}")
# Expected: ~2.87

# ---------- 3. Titration curve: weak acid + strong base ----------
def titration_curve(C_HA, V_HA, C_NaOH, pKa, V_base):
    # V_base is an array of NaOH volumes added (mL).
    # Returns pH at each volume.
    Ka = 10 ** (-pKa)
    n_HA = C_HA * V_HA / 1000                      # moles of acid
    n_OH = C_NaOH * V_base / 1000                   # moles base added
    V_tot = (V_HA + V_base) / 1000                    # total volume L

    pH = np.zeros_like(V_base, dtype=float)
    for i, (nb, V) in enumerate(zip(n_OH, V_tot)):
        if nb < n_HA:
            # Buffer region: Henderson-Hasselbalch
            nHA = n_HA - nb
            nA  = nb
            pH[i] = pKa + np.log10(max(nA, 1e-12) / max(nHA, 1e-12))
        elif np.isclose(nb, n_HA):
            # Equivalence: pure A- hydrolysis, pH > 7
            C_A = n_HA / V
            Kb  = 1e-14 / Ka
            x   = np.sqrt(Kb * C_A)
            pH[i] = 14 + np.log10(max(x, 1e-14))
        else:
            # Excess strong base
            C_OH = (nb - n_HA) / V
            pH[i] = 14 + np.log10(C_OH)
    return pH

V = np.linspace(0.1, 100, 200)
curve = titration_curve(0.10, 50, 0.10, 4.76, V)
print("half-equiv pH =", float(curve[49]))     # ~ pKa = 4.76

import math

# Same ideas, plain Python.

def pH_weak_acid(C, Ka):
    disc = Ka * Ka + 4 * Ka * C
    x = (-Ka + math.sqrt(disc)) / 2
    return -math.log10(x)

def pH_buffer(pKa, ratio_A_to_HA):
    return pKa + math.log10(ratio_A_to_HA)

def ksp_molar_solubility(Ksp, n_cation, n_anion):
    # e.g. Ca3(PO4)2: n_cat=3, n_an=2
    # Ksp = (n_cat s)^n_cat * (n_an s)^n_an
    prefactor = (n_cation ** n_cation) * (n_anion ** n_anion)
    total_order = n_cation + n_anion
    return (Ksp / prefactor) ** (1 / total_order)

print(f"0.10 M acetic pH: {pH_weak_acid(0.10, 1.8e-5):.2f}")
print(f"pH 4.8 buffer ratio: {pH_buffer(4.76, 1.10):.3f}")
print(f"Ca3(PO4)2 solubility: {ksp_molar_solubility(2e-29, 3, 2):.2e} M")

A few practical notes:

• Real chemistry packages like PHREEQC and CHEMKIN do not solve one equilibrium at a time — they solve large coupled systems of equilibria, often hundreds of species, by minimizing total free energy under mass-balance constraints. The underlying math is the same. Your toy routines above are pedagogical, not production.
• The quadratic approach breaks down if $K$ is comparable to $C_0$ or if multiple equilibria interact (e.g. polyprotic acids). Use iterative solvers for those.
• Activity corrections (finite-concentration effects) become important above ~0.01 M. At high ionic strength, plug in activity coefficients from Debye-Hückel or specific-ion theory.

12. Cheat sheet

See also