Chemical and Ionic Equilibrium

1The law of mass action and Kc

Equilibrium is not the reaction stopping. Both directions keep running; they have simply become equally fast, so nothing net changes. The law of mass action says that at that point one particular ratio is fixed: Kc = [products]/[reactants], each concentration raised to its stoichiometric coefficient. Pure solids and pure liquids never appear — their concentration cannot change.

Kc belongs to the balanced equation as written. Reverse the equation and K becomes 1/K; multiply every coefficient by n and K becomes Kⁿ. And Kc is a function of temperature alone: no amount of adding, removing or compressing moves it.

Figure. Both curves flatten, and from that moment the ratio 1.56²/(0.22 × 0.22) holds at 50.3 for as long as the temperature does. Flat does not mean stopped: molecules are still reacting both ways, just at matched rates.

Reading a value of K

1. K > 10³The products dominate at equilibrium; treat the reaction as essentially complete.
2. 10⁻³ < K < 10³Both sides are present in comparable amounts, and you must actually solve for the composition.
3. K < 10⁻³The reactants dominate; barely any product forms, however long you wait.

2Kp, Kc and Δn_g

For a gas-phase equilibrium you may write the constant in concentrations (Kc) or in partial pressures (Kp). Substituting p = (n/V)RT into the Kp expression gives Kp = Kc(RT)^Δn_g, where Δn_g counts gas moles on the right minus gas moles on the left. Solids and liquids are not counted, because they are not in the expression at all.

The exponent is the whole story. Kp and Kc are the same number only when Δn_g = 0, and then only by arithmetic accident of the exponent, not because pressure and concentration are the same thing. R must carry the units of the pressure you are quoting: 0.0821 L atm K⁻¹ mol⁻¹ for Kp in atm, 0.08314 L bar K⁻¹ mol⁻¹ for bar, 8.314 J K⁻¹ mol⁻¹ (that is Pa m³) for pascal.

Figure. Kp = Kc(RT)^Δn_g. The exponent is moles of gas products minus reactants: negative shrinks Kp relative to Kc, zero makes them equal, positive grows Kp. Read Δn_g from the balanced equation before converting.

3The reaction quotient and the direction of change

Q is written exactly like K — same expression, same exponents — but evaluated with whatever concentrations you happen to have, equilibrium or not. Comparing the two tells you which way the mixture must move, and it is the only tool you need for that question.

Q < K means there is too little product, so the reaction runs forward. Q > K means too much, so it runs in reverse. Q = K means you are already there. Every Le Chatelier prediction in the next two concepts is this same comparison in words rather than numbers.

Figure. The axis is Q on a linear scale from 0 to 100, with Kc = 50.3 marked at its true position. Whichever side of that mark the mixture starts on, it moves toward it — the arrows point at Kc from both directions, which is the whole content of the comparison.

4Le Chatelier: concentration, temperature, catalyst

Impose a change on a system at equilibrium and it responds in the direction that partly undoes the change. Add a reactant and Q drops below K, so the reaction consumes some of what you added. Remove a product and Q drops again, so more product forms. Both of these leave K untouched: the mixture moves, the constant does not.

Temperature is the exception, and the only one. Treat heat as a reagent: for an endothermic forward reaction heat sits on the left, so heating pushes forward and K rises; for an exothermic one heat sits on the right, so heating pushes back and K falls. Quantitatively that is the van 't Hoff equation, ln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁). A catalyst changes neither: it speeds both directions by the same factor, so equilibrium arrives sooner at exactly the same place.

Figure. Both curves are the van 't Hoff equation over 400–700 K, plotted relative to their own value at 500 K so they can share an axis. The exothermic case is ΔH° = −92.4 kJ mol⁻¹ (ammonia synthesis) and the endothermic one ΔH° = +180.5 kJ mol⁻¹ (N₂ + O₂ ⇌ 2NO). They cross where they are defined to cross, and nowhere else — the sign of ΔH° fixes which way each one goes for good. Read heights against the horizontal dashed ln = 0 line, not against the bottom of the frame: this quantity is negative below that line, so each curve spends half its length with K under its own 500 K value.

5Volume, pressure and the inert-gas trap

There is no such thing as a rule for "increasing the pressure". There are three different experiments that all raise the pressure gauge, and they have three different answers. Get at all of them the same way: work out what happens to the concentrations, put them into Q, and compare Q with K.

Squeeze the vessel and every concentration is multiplied by the same factor, so Q is multiplied by that factor raised to Δn_g — the mixture shifts toward the side with fewer gas moles, and does not move at all if Δn_g = 0. Pump in argon while holding the volume fixed and no concentration and no partial pressure changes, so Q is untouched and nothing shifts, however high the total pressure reads. Pump in argon while holding the total pressure fixed and the vessel must expand, every concentration falls, and the shift is toward the side with more gas moles — the opposite of squeezing.

Figure. The three boxes are the same gas mixture in three volumes, drawn to scale: 0.5, 1.0 and 1.5 times the original. The middle box is the inert-gas-at-fixed-volume case, which is why it is the same size as the original and carries no arrow. Volume is the variable that matters; total pressure by itself is not.

6Kw, pH and pOH

Water ionises very slightly into H⁺ and OH⁻, and the product of the two is a constant: Kw = [H⁺][OH⁻] = 1.0 × 10⁻¹⁴ at 25 °C. Taking negative logarithms turns that into pH + pOH = 14 — a relation that holds at 25 °C and at no other temperature, because Kw itself changes with temperature.

Neutral means [H⁺] = [OH⁻], which is pH = ½pKw. That happens to be 7.00 at 25 °C and is something else anywhere else. Ionising water absorbs heat, so warming water raises Kw, raises both ion concentrations together and lowers the neutral pH below 7 — without making the water acidic, because the two ions are still equal.

Figure. One position, two readings: the upper rail is pH running 0 to 14 left to right, the lower rail is pOH running 14 to 0 the other way. Any vertical line crosses both at values summing to 14 — which is exactly the content of Kw at 25 °C. The worked example's mixture sits at pH 12.70, and therefore at pOH 1.30.

7Weak acids, Ka and Ostwald's dilution law

A weak acid HA only partly ionises, to a degree α set by Ka = Cα²/(1 − α). While α is small, 1 − α ≈ 1 and this collapses to the two expressions worth memorising: α = √(Ka/C) and [H⁺] = Cα = √(KaC). That is Ostwald's dilution law — α is inversely proportional to √C, so diluting a weak acid ionises a larger fraction of it.

The trap is what happens to the acidity. α rises on dilution but [H⁺] = √(KaC) falls, because C falls faster than α rises. A more completely ionised acid is not a stronger one. The approximation itself holds while α < 5%, which means C/Ka > 400; below that you must go back to the quadratic.

Figure. α for acetic acid against log C from 10⁻⁴ to 1 M. The dashed Ostwald line lies on top of the exact curve across the right-hand half, where C/Ka is 560 or more, and peels away from it below about 10⁻³ M — at 10⁻⁴ M it claims 42% against a true 34%. The 5% rule of thumb bites at 7 × 10⁻³ M, most of a decade before any gap is visible: the arithmetic notices long before the eye does.

8Buffers and Henderson–Hasselbalch

A buffer is a weak acid sitting alongside a decent reservoir of its own conjugate base. Added H⁺ is mopped up by the base, added OH⁻ by the acid, and the ratio of the two barely moves — so neither does the pH. Rearranging Ka = [H⁺][A⁻]/[HA] gives the Henderson–Hasselbalch equation, pH = pKa + log([salt]/[acid]).

Read that equation twice. Only the ratio appears, so diluting a buffer does not change its pH, only how much abuse it can absorb. And the ratio is 1 exactly when pH = pKa, which is the centre of the plateau and where the buffer is at its most resistant. In practice a buffer works over roughly pKa ± 1, so you choose the acid to match the pH you want, not the other way round.

Figure. Three buffer systems placed on a pH axis running 3 to 11, each bar spanning pKa ± 1 and each tick dropping from the bar to the pKa itself. Choosing a buffer is choosing a bar that covers your target pH: CH₃COOH/CH₃COO⁻ reaches 3.74–5.74, H₂PO₄⁻/HPO₄²⁻ covers 6.21–8.21 from Ka₂ = 6.2 × 10⁻⁸, and NH₄⁺/NH₃ covers 8.26–10.26 from Kb = 1.8 × 10⁻⁵.

9Titration curves and the equivalence point

Plot pH against added titrant and a weak acid gives four regions in order: a modest initial pH, a long flat buffer stretch, a near-vertical jump, and a tail governed by excess base. The flat stretch is the previous concept in graphical form — half way to equivalence, exactly half the acid has been converted, so [salt] = [acid] and pH = pKa. Reading pKa straight off the half-equivalence point is a standard exam move.

The equivalence point is where stoichiometrically equivalent amounts have been added — not where the pH is 7, and not where the indicator turns. For a weak acid against a strong base the solution at equivalence is a solution of the conjugate base, which hydrolyses, so the pH is above 7. It is 7.00 only when the acid was strong too. The first derivative of the curve peaks exactly at the equivalence point, since that is where the curve is steepest, which is how an autotitrator finds it without an indicator at all.

Figure. Both curves are computed from the exact charge balance, not sketched. Both reach equivalence at the same 25.0 mL, but at different heights — 8.72 against 7.00 — and the weak acid starts far higher (about pH 2.9, not 1.00) and climbs through a buffer plateau the strong acid does not have. Past equivalence the two coincide exactly, because from there on both solutions are simply excess NaOH.

10Solubility product and the common-ion effect

A sparingly soluble salt in contact with its saturated solution is an equilibrium like any other, and its constant is Ksp. For AₓBᵧ dissolving to give xs of one ion and ys of the other, Ksp = (xs)ˣ(ys)ʸ. The solid itself never appears in the expression, which is why an excess of undissolved salt at the bottom of the beaker changes nothing.

Two consequences carry most of the marks. First, comparing solubilities means comparing s, not Ksp — the exponent linking them depends on the stoichiometry, so a salt with the smaller Ksp can easily be the more soluble one. Second, an ion already in solution suppresses dissolution violently: this is the common-ion effect, and it is again just Q pushed above Ksp and having to come back down.

Figure. Solubility of AgCl against the concentration of chloride already present, both on log scales. Below about 10⁻⁵ M the added chloride is swamped by what the salt itself supplies and the curve is flat at √Ksp; above it the line falls with a slope of exactly −1, which is s = Ksp/[Cl⁻] made visible. The knee sits where the added ion first matches the salt's own contribution.