Equilibrium — Chemical and Ionic

Equilibrium is the chapter where chemistry stops being one-directional. A reaction at equilibrium has not stopped; both directions are running at identical rates, which is why adding a labelled reactant to an equilibrium mixture will put the label into the products even though no concentration changes. Two ideas carry almost the whole chapter. First, K is a constant only with respect to concentration and pressure — it changes with temperature and with nothing else. Second, Le Chatelier's principle is a consequence, not an extra rule: it simply says a disturbed system moves in whichever direction restores Q to K. Every prediction about pressure, dilution, catalysts and inert gases follows from those two sentences.

K, Q and What K Does Not Depend On

For aA + bB ⇌ cC + dD the equilibrium constant is K_c = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ, with the products always on top. In terms of partial pressures, K_p = K_c (RT)^Δn, where Δn is the change in the number of moles of gas. Pure solids and pure liquids are omitted from the expression because their concentration is fixed by their density — this is why the K_p for CaCO₃(s) ⇌ CaO(s) + CO₂(g) is simply p(CO₂).

How K behaves when you rewrite the equation
Change to the equationNew K
Reverse the reaction1/K
Multiply all coefficients by nKⁿ
Halve all coefficients√K
Add two reactionsK₁ × K₂
🧠 Compare Q with K to find the direction without any calculation
The reaction quotient Q has exactly the same algebraic form as K but uses the current concentrations, whatever they are. If Q < K there is too little product, so the reaction runs forward. If Q > K there is too much product, so it runs backward. If Q = K the system is already at equilibrium. This single comparison replaces all guesswork about direction, and it is how the paper expects you to answer "in which direction will the reaction proceed" questions — compute Q, compare, done.
⚠️ A catalyst never changes K, and neither does adding an inert gas at constant volume
A catalyst lowers the activation energy of the forward and reverse reactions by the same amount, so both rates rise equally and equilibrium arrives sooner at exactly the same composition. It changes the time, never the destination. Adding an inert gas at constant volume changes the total pressure but leaves every partial pressure and every concentration untouched, so nothing shifts. Adding an inert gas at constant pressure does shift the equilibrium, because the total volume must expand and that dilutes all the reacting species — equivalent to a pressure drop, so the side with more gas moles is favoured. The distinction between constant volume and constant pressure is the whole question here.

Only temperature changes K, and the direction follows from the sign of ΔH through the van't Hoff relation. For an exothermic reaction, heating shifts the equilibrium backward and lowers K; for an endothermic reaction, heating shifts it forward and raises K. This is why the Haber process, which is exothermic, is run at a compromise temperature of about 700 K — hot enough for a workable rate, cool enough that K has not collapsed. High pressure is used because ammonia synthesis reduces the gas mole count from 4 to 2.

Acids, Bases and pH

On the Brønsted–Lowry definition an acid donates a proton and a base accepts one, which makes acidity a relationship rather than a property — HCO₃⁻ is an acid towards OH⁻ and a base towards H₃O⁺. On the Lewis definition an acid accepts an electron pair, which extends the idea to species with no proton at all, so BF₃ and AlCl₃ are acids. Every conjugate pair differs by exactly one H⁺, and the stronger the acid, the weaker its conjugate base: Cl⁻ is a negligible base precisely because HCl is a strong acid.

The pH toolkit
SituationFormulaNote
Strong acidpH = −log CFully ionised, so [H⁺] equals the concentration
Weak acid[H⁺] = √(K_a·C), pH = ½(pK_a − log C)Valid while ionisation is under about 5 %
Degree of ionisationα = √(K_a/C)Ostwald dilution law — α rises on dilution
BufferpH = pK_a + log([salt]/[acid])Henderson equation; pH = pK_a at equal amounts
WaterK_w = [H⁺][OH⁻] = 10⁻¹⁴ at 298 KpH + pOH = 14 only at 298 K
⚠️ Neutral water is not pH 7 at every temperature
The ionisation of water is endothermic, so heating increases K_w. At 100 °C, K_w rises to about 5.5 × 10⁻¹³, which makes [H⁺] = [OH⁻] ≈ 7.4 × 10⁻⁷ and the pH of pure water about 6.14. The water is still perfectly neutral, because neutrality means [H⁺] = [OH⁻], not pH = 7. So a question describing water at 100 °C with pH 6.14 and asking whether it is acidic is testing this exact point — the answer is neutral. Likewise pH + pOH = 14 is a 298 K statement, not a universal law.
🎯 Why a buffer resists pH change
A buffer holds a weak acid and its conjugate base side by side in comparable amounts. Added H⁺ is mopped up by the base, added OH⁻ is neutralised by the acid, and in both cases the ratio [salt]/[acid] changes only slightly — and pH depends on the logarithm of that ratio, so a small ratio change moves pH very little. Two consequences follow. Buffering is strongest when [salt] = [acid], where pH = pK_a and the ratio is least sensitive. And dilution does not change a buffer's pH, because diluting divides both salt and acid by the same factor and leaves the ratio untouched — which is exactly why blood plasma holds pH 7.4 despite constant fluid intake.

Salts are not automatically neutral, because the ion of a weak parent hydrolyses. A salt of a strong acid and strong base (NaCl) is neutral. A salt of a weak acid and strong base (CH₃COONa) is basic, since acetate takes a proton from water. A salt of a strong acid and weak base (NH₄Cl) is acidic, since ammonium donates one. A salt of a weak acid and weak base (CH₃COONH₄) has pH = 7 + ½(pK_a − pK_b), so the sign depends on which parent is weaker — and its pH is independent of concentration, an unusual result that questions like to test.

Solubility Product and the Common Ion Effect

For a sparingly soluble salt A_xB_y, K_sp = [Aʸ⁺]ˣ[Bˣ⁻]ʸ, and the relation to solubility s depends on the stoichiometry: K_sp = s² for AgCl, 4s³ for Ag₂CrO₄ or CaF₂, and 27s⁴ for Ca₃(PO₄)₂ in its simplest treatment. Precipitation occurs when the ionic product exceeds K_sp; the solution is unsaturated below that and exactly saturated at equality.

⚠️ Smaller K_sp does not always mean smaller solubility
You may compare K_sp values directly only when the salts have the same stoichiometry. AgCl has K_sp = 1.8 × 10⁻¹⁰, giving s = √K_sp = 1.3 × 10⁻⁵ M. Ag₂CrO₄ has the smaller K_sp of 1.1 × 10⁻¹², yet s = ∛(K_sp/4) = 6.5 × 10⁻⁵ M — about five times more soluble, because its cube-root relation is far more forgiving than a square root. So a question offering "the salt with the lowest K_sp is least soluble" as an option is testing whether you noticed the formula types differ. Always convert to s before ranking.

The common ion effect is Le Chatelier applied to a sparingly soluble salt: adding one of its own ions from another source pushes the dissolution equilibrium backward and lowers solubility. AgCl is markedly less soluble in 0.1 M NaCl than in water, and in that case the added chloride dominates so completely that s ≈ K_sp/0.1 rather than √K_sp. The same principle explains why H₂S is passed through acidified solution in qualitative analysis — the added H⁺ suppresses the ionisation of H₂S, keeping [S²⁻] low enough that only the least soluble sulphides precipitate, which is how group II is separated from group IV.

ℹ️ Indicators and the choice of end point
An indicator is itself a weak acid whose colour changes over roughly pK_a ± 1, so it must be chosen to match the pH at the equivalence point of the particular titration. A strong acid against a strong base has its equivalence point at pH 7 and either phenolphthalein or methyl orange works. A weak acid against a strong base ends up basic, near pH 8–9, so phenolphthalein (range 8.3–10) is correct and methyl orange would change colour far too early. A weak base against a strong acid ends up acidic, so methyl orange (range 3.1–4.4) is the right choice. Getting this backwards is a favourite one-mark question.

Key takeaways

  • K_p = K_c(RT)^Δn; pure solids and liquids are left out of the expression entirely.
  • Reversing gives 1/K, scaling coefficients by n gives Kⁿ, and adding reactions multiplies their Ks.
  • Q < K runs forward, Q > K runs backward — this replaces all guesswork about direction.
  • Only temperature changes K. A catalyst changes the time to equilibrium, never the composition.
  • Inert gas at constant volume shifts nothing; at constant pressure it acts like a dilution.
  • Pure water is neutral at every temperature, but its pH is 7 only at 298 K — 6.14 at 100 °C.
  • Buffer pH = pK_a + log([salt]/[acid]) and is unchanged by dilution, since the ratio is unchanged.
  • Compare K_sp values only for salts of the same formula type; otherwise convert to solubility first.

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Adding a catalyst to a system at equilibrium:

    1. Shifts the equilibrium towards the products
    2. Increases the value of K
    3. Changes nothing about the equilibrium composition
    4. Shifts the equilibrium towards the reactants
    Show answer

    Answer: C — Changes nothing about the equilibrium composition

    A catalyst lowers the activation energy of the forward and reverse reactions by exactly the same amount, so both rates increase in the same proportion and the ratio that defines K is untouched. Equilibrium is reached sooner but at an identical composition. Only a change of temperature can alter K.

  2. An inert gas is added to N₂ + 3H₂ ⇌ 2NH₃ at equilibrium at constant volume. The equilibrium:

    1. Shifts forward
    2. Shifts backward
    3. Does not shift
    4. Shifts forward only if the gas is monatomic
    Show answer

    Answer: C — Does not shift

    At constant volume, adding an inert gas raises the total pressure but leaves every individual partial pressure and concentration exactly as it was, so Q is unchanged and still equals K. Had the addition been at constant pressure, the volume would have to expand, diluting all species and shifting the equilibrium towards the side with more gas moles — here backwards.

  3. At 100 °C the pH of pure water is about 6.14. The water is therefore:

    1. Acidic
    2. Basic
    3. Neutral
    4. Impossible — pure water is always pH 7
    Show answer

    Answer: C — Neutral

    Neutrality means [H⁺] = [OH⁻], which remains true in pure water at any temperature. Because ionisation of water is endothermic, heating raises K_w, so both ion concentrations rise together and the pH falls below 7 while the water stays neutral. The rule pH + pOH = 14 and the value 7 for neutrality both apply only at 298 K.

  4. A buffer is made from equal moles of CH₃COOH and CH₃COONa. Its pH is:

    1. 7
    2. Equal to pK_a of acetic acid
    3. Equal to pK_b of acetate
    4. Dependent on how much water is added
    Show answer

    Answer: B — Equal to pK_a of acetic acid

    By the Henderson equation pH = pK_a + log([salt]/[acid]), and with equal moles the ratio is 1 and its logarithm zero, so pH = pK_a — about 4.74 for acetic acid, not 7. Adding water divides salt and acid by the same factor, leaving the ratio and hence the pH unchanged, which is precisely the property that makes buffers useful.

  5. For the salt Ag₂CrO₄, the relation between solubility s and K_sp is:

    1. K_sp = s²
    2. K_sp = s³
    3. K_sp = 4s³
    4. K_sp = 27s⁴
    Show answer

    Answer: C — K_sp = 4s³

    Ag₂CrO₄ dissolves to give 2Ag⁺ and one CrO₄²⁻, so [Ag⁺] = 2s and [CrO₄²⁻] = s, giving K_sp = (2s)²(s) = 4s³. The answer s² applies to a 1 : 1 salt such as AgCl, and 27s⁴ to a 1 : 3 salt. Because the exponent differs, K_sp values of salts with different formula types can never be compared directly to rank their solubilities.

  6. For an exothermic reaction, raising the temperature will:

    1. Increase K and favour products
    2. Decrease K and favour reactants
    3. Leave K unchanged but speed up the reaction
    4. Decrease K only if Δn is negative
    Show answer

    Answer: B — Decrease K and favour reactants

    Heat is a product of an exothermic reaction, so supplying more of it shifts the equilibrium backward and lowers K. The rate does also increase, but that is a kinetic effect and the third option wrongly claims K is unchanged — temperature is the one variable that genuinely alters K. Δn governs the response to pressure, not to temperature.

  7. An aqueous solution of NH₄Cl is:

    1. Neutral, since it is a salt
    2. Acidic, because NH₄⁺ hydrolyses
    3. Basic, because Cl⁻ hydrolyses
    4. Basic, because ammonia is a base
    Show answer

    Answer: B — Acidic, because NH₄⁺ hydrolyses

    NH₄Cl comes from a strong acid (HCl) and a weak base (NH₃). The chloride ion is the conjugate of a strong acid and therefore a negligible base that does not hydrolyse, while ammonium is the conjugate of a weak base and donates a proton to water, producing H₃O⁺ and an acidic solution. The last option confuses the parent base with the salt actually present in solution.

  8. Which indicator is appropriate for titrating acetic acid against NaOH?

    1. Methyl orange, range 3.1–4.4
    2. Phenolphthalein, range 8.3–10.0
    3. Either works equally well
    4. Neither, since a weak acid cannot be titrated
    Show answer

    Answer: B — Phenolphthalein, range 8.3–10.0

    A weak acid titrated with a strong base leaves acetate in solution at the equivalence point, which hydrolyses to give a basic pH of roughly 8–9, so an indicator changing colour in that window is needed and phenolphthalein fits. Methyl orange would change colour around pH 4, well before the equivalence point, giving a badly low titre. Weak acids are routinely titrated; only the indicator choice changes.