Equilibrium

This chapter has two connected halves: general chemical equilibrium, and its most important special case, ionic equilibrium in aqueous solution. The guided map below follows the order the NCERT text builds in.

• Equilibrium in physical processes — solid-liquid, liquid-vapour, solid-vapour and dissolution equilibria, and the features common to all of them.
• Dynamic equilibrium in chemical reactions — reversible reactions, and how the Haber process and the deuterium-exchange experiment prove equilibrium is dynamic, not frozen.
• The law of chemical equilibrium — Guldberg and Waage's law of mass action, the equilibrium constant $K_c$, and its properties.
• $K_p$, homogeneous and heterogeneous equilibria — expressing $K$ in terms of partial pressure, relating $K_p$ to $K_c$, and why pure solids and liquids drop out of the expression.
• Applications of $K$ — using its magnitude to judge how far a reaction goes, and using the reaction quotient $Q$ to predict direction.
• $K$, $Q$ and Gibbs energy — the thermodynamic backbone, $\Delta G^\circ = -RT\ln K$.
• Factors affecting equilibria — Le Chatelier's principle applied to concentration, pressure/volume, inert gas addition, temperature and catalysts.
• Ionic equilibrium: acids, bases and salts — the Arrhenius, Bronsted-Lowry and Lewis definitions, and conjugate acid-base pairs.
• Ionization constants — $K_a$, $K_b$, degree of ionization, polyprotic acids, and what makes one acid stronger than another.
• Water, pH and the pH scale — $K_w$, pH, pOH, pKa, pKb, pKw.
• Common ion effect and hydrolysis of salts — how a shared ion suppresses ionization, and why salt solutions are not always neutral.
• Buffer solutions — the Henderson-Hasselbalch equation and how to design a buffer of a target pH.
• Solubility equilibria — the solubility product constant $K_{sp}$, and how a common ion suppresses solubility.

Equilibrium in physical processes

The cleanest place to see equilibrium is a phase change. Ice and water sealed in an insulated flask at $273\ \text{K}$ and atmospheric pressure show constant mass of each phase over time, yet at the ice-water boundary molecules are continuously crossing both ways — some liquid water molecules freeze onto the ice, and an equal number of ice molecules melt into the liquid. This is solid-liquid equilibrium, and the temperature at which it occurs (at $1$ atm) is the substance's normal melting or freezing point.

A sealed container holding liquid water and empty space above it shows the same pattern for liquid-vapour equilibrium: molecules with enough kinetic energy escape into the vapour phase (evaporation) while vapour molecules strike the surface and rejoin the liquid (condensation). Initially evaporation dominates and pressure rises, but condensation speeds up as vapour accumulates until the two rates match, at which point pressure stabilises at the liquid's equilibrium vapour pressure — higher for more volatile liquids and rising with temperature. This can only happen in a closed system; in an open watch glass, escaping vapour disperses into the room faster than it can return, so the liquid eventually evaporates completely with no equilibrium ever reached. The normal boiling point is the temperature at which vapour pressure equals $1.013\ \text{bar}$, which is why water boils below $100\,^{\circ}\text{C}$ at high altitude, where atmospheric pressure is lower.

Solids that sublime, like iodine, camphor or ammonium chloride, show solid-vapour equilibrium in a closed vessel: iodine's violet vapour intensifies and then holds constant once the rate of sublimation equals the rate of deposition. Dissolution shows the pattern twice over: a saturated solution of sugar in water is one where the rate of dissolution equals the rate of crystallisation (confirmed by tagging some sugar radioactively and watching the label spread evenly between solid and solution even after concentration stops changing); and dissolved gases obey Henry's law — the mass of gas dissolved in a solvent is proportional to the pressure of that gas above it — exactly why an opened soda bottle goes flat as dissolved $\text{CO}_2$ escapes toward the lower partial pressure outside. Five features are common to every physical equilibrium: it is possible only in a closed system at constant temperature; both opposing processes occur simultaneously at equal rates; all measurable macroscopic properties stay constant; a fixed quantity (vapour pressure, melting point, solubility, or the gas-to-dissolved ratio) characterises it at a given temperature; and the value of that quantity reflects how far the process has proceeded.

Dynamic chemical equilibrium

Chemical reactions that can run in both directions — reversible reactions — reach the same kind of dynamic balance. Consider the general case $A + B \rightleftharpoons C + D$. As the reaction proceeds, reactants deplete and products accumulate, which slows the forward reaction (less $A$ and $B$ available) and speeds up the reverse reaction (more $C$ and $D$ available to react back). Eventually the two rates become equal, concentrations stop changing, and the system has reached chemical equilibrium — reachable from either direction, whether you start with only $A$ and $B$, only $C$ and $D$, or any mixture of them. That the equilibrium is genuinely dynamic, not a reaction that has simply stopped, was proved decisively using deuterium. Haber's ammonia synthesis, $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, was run to equilibrium once with ordinary hydrogen and once with deuterium in its place. When the two equilibrium mixtures (one containing $\text{H}_2$, $\text{N}_2$, $\text{NH}_3$, the other $\text{D}_2$, $\text{N}_2$, $\text{ND}_3$) were combined and left to stand, mass-spectrometric analysis later found a full scramble of isotopic forms — $\text{NH}_3$, $\text{NH}_2\text{D}$, $\text{NHD}_2$, $\text{ND}_3$ and mixed $\text{H}_2$, $\text{HD}$, $\text{D}_2$ — even though the total ammonia concentration had not changed. If the reaction had truly stopped at equilibrium, no such mixing could ever occur; the scrambling is direct evidence that forward and reverse reactions never actually cease, they merely proceed at equal rates.

The law of chemical equilibrium and the equilibrium constant

In 1864 the Norwegian chemists Cato Guldberg and Peter Waage proposed, from experimental study of many reversible reactions, that for a general reaction at equilibrium the concentrations are related by a fixed ratio, raised to the stoichiometric coefficients. For the general reaction $aA + bB \rightleftharpoons cC + dD$:

where: square brackets denote equilibrium molar concentrations (mol/L), and the exponents $a$, $b$, $c$, $d$ are exactly the coefficients of the balanced equation — not arbitrary powers. This relation is the law of chemical equilibrium, also called the law of mass action (concentration was historically termed "active mass"). It was verified on the reaction $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ at $731\ \text{K}$: across six experiments starting from very different initial mixtures — four beginning with only $\text{H}_2$ and $\text{I}_2$, two beginning with only $\text{HI}$ — the ratio $[\text{HI}]^2/([\text{H}_2][\text{I}_2])$ converged to the same numerical value every single time, while simpler ratios like $[\text{HI}]/([\text{H}_2][\text{I}_2])$ did not. $K_c$ is independent of the initial concentrations used, is characteristic of the reaction at a given temperature only, and the equilibrium constant of the reverse reaction is simply its reciprocal, $K_c' = 1/K_c$. Multiplying a balanced equation throughout by a factor $n$ raises $K_c$ to the power $n$.

Kp, homogeneous and heterogeneous equilibria

For reactions in the gas phase it is often more convenient to work with partial pressures than molar concentrations, giving the equilibrium constant $K_p$. In a homogeneous equilibrium, all reactants and products share one phase — all gases, as in $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, or all in aqueous solution, as in ester hydrolysis. A heterogeneous equilibrium spans more than one phase, such as the thermal decomposition $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$. Because the molar concentration of a pure solid or pure liquid is a fixed constant (independent of how much of it is present), it is absorbed into the equilibrium constant rather than written explicitly — so for calcium carbonate's decomposition, $K_c' = [\text{CO}_2(g)]$ and $K_p = p_{\text{CO}_2}$ alone, even though solid $\text{CaCO}_3$ and $\text{CaO}$ must both still be physically present for the equilibrium to exist. The units of $K_c$ or $K_p$ depend on whether the exponents in the numerator and denominator balance; when the standard states (1 M for solutes, 1 bar for gases) are built into the definition, both become dimensionless, though their numerical values still differ because the standard states differ.

Applications of the equilibrium constant: extent and direction

The magnitude of $K_c$ tells you how far a reaction proceeds before stopping, without saying anything about how fast it gets there. If $K_c > 10^3$, products dominate overwhelmingly at equilibrium (e.g. $\text{H}_2 + \text{Cl}_2 \rightleftharpoons 2\text{HCl}$ has $K_c = 4.0\times10^{31}$ at $300\ \text{K}$); if $K_c < 10^{-3}$, reactants dominate and the reaction barely proceeds (e.g. $\text{N}_2 + \text{O}_2 \rightleftharpoons 2\text{NO}$ has $K_c = 4.8\times10^{-31}$ at $298\ \text{K}$); and for $K_c$ between $10^{-3}$ and $10^3$, both reactants and products are present in appreciable amounts at equilibrium. To decide which way an out-of-equilibrium mixture will move, use the reaction quotient $Q_c$ — the same expression as $K_c$, but evaluated using concentrations at any arbitrary moment, not necessarily at equilibrium:

where: the subscript $t$ signals these are concentrations measured at some time $t$, which need not be the equilibrium values. Comparing $Q_c$ to $K_c$: if $Q_c < K_c$, the system has too little product relative to equilibrium, so the reaction proceeds forward (left to right); if $Q_c > K_c$, it proceeds in reverse (right to left); and if $Q_c = K_c$, the system is already at equilibrium and there is no net change. This single comparison is the working tool behind every "which direction will it shift" question in this chapter.

Le Chatelier's principle: the factors that shift equilibrium

Le Chatelier's principle states that when a system at equilibrium is subjected to a change in concentration, pressure, volume or temperature, the system shifts in whichever direction relieves or counteracts that change. It applies uniformly to physical and chemical equilibria and is the reasoning tool behind industrial process design.

Concentration. Adding a reactant (or removing a product) drives the reaction forward; removing a reactant (or adding a product) drives it in reverse. Adding $\text{H}_2$ to $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ makes $Q_c$ momentarily smaller than $K_c$, so the forward reaction runs until $Q_c = K_c$ again. Industry exploits this by continuously removing ammonia (liquefying it) from the Haber process, or $\text{CO}_2$ from a lime kiln, keeping $Q_c$ below $K_c$ to pull the reaction toward completion.

Pressure and volume. Compressing a gaseous mixture raises total pressure and shifts equilibrium toward the side with fewer moles of gas, since that relieves the pressure again; expanding the volume shifts it toward the side with more moles of gas. A reaction with equal gas moles on both sides is unaffected by pressure changes, and pressure has no effect on a heterogeneous equilibrium's solid or liquid phase, since their volumes barely change with pressure.

Inert gas addition. Adding an inert gas like argon at constant volume leaves every partial pressure and concentration of the reacting species unchanged, so equilibrium is undisturbed — none of the species the reaction quotient depends on has actually changed concentration.

Temperature. Unlike concentration and pressure, a temperature change genuinely alters the numerical value of $K$, not just the position of equilibrium. For an exothermic reaction ($\Delta H < 0$), raising temperature decreases $K$ and shifts equilibrium toward the reactants; for an endothermic reaction ($\Delta H > 0$), raising temperature increases $K$ and shifts it toward products. Ammonia synthesis is strongly exothermic ($\Delta H = -92.38\ \text{kJ/mol}$), so low temperature favours higher yield — but rate becomes impractically slow, which is why industrial Haber-process conditions compromise: an iron catalyst at around $500\,^{\circ}\text{C}$ and $200\ \text{atm}$, trading some yield for a workable rate.

Catalyst. A catalyst provides a new, lower-energy pathway for both forward and reverse reactions, speeding up each by exactly the same factor. It changes how fast equilibrium is reached, never the equilibrium composition or $K$ itself, and it never appears in the equilibrium constant expression. If a reaction's $K$ is exceedingly small, no catalyst can rescue the yield — it can only make a favourable reaction faster, not make an unfavourable one happen.

Equilibrium constant, Gibbs energy and spontaneity

The value of $K$ has nothing to do with reaction rate, but it is directly tied to the thermodynamics of the reaction through the Gibbs energy change. Recall from thermodynamics that $\Delta G = \Delta G^\circ + RT\ln Q$, where $\Delta G^\circ$ is the standard Gibbs energy change and $Q$ the reaction quotient. Since a reaction proceeds spontaneously in the direction that makes $\Delta G$ negative and stops exactly when $\Delta G = 0$, the point where $Q$ becomes $K$ is precisely the equilibrium point. This connection is worked out fully as a derivation below; the essential result is $\Delta G^\circ = -RT\ln K$, which means a large negative $\Delta G^\circ$ gives a large $K$ (products strongly favoured), a large positive $\Delta G^\circ$ gives a tiny $K$ (reactants strongly favoured), and $\Delta G^\circ = 0$ gives $K = 1$.

Ionic equilibrium: acids, bases and salts

Substances that conduct electricity in aqueous solution are electrolytes; those that don't (like sugar solution) are non-electrolytes. Strong electrolytes (like $\text{NaCl}$) ionize almost completely, while weak electrolytes (like acetic acid, under $5\%$ ionized) exist mostly as unionized molecules in a dynamic equilibrium with their ions — this is ionic equilibrium. Three progressively broader definitions describe acids and bases. The Arrhenius concept: an acid produces $\text{H}^+(aq)$ (really the hydrated hydronium ion $\text{H}_3\text{O}^+$) in water; a base produces $\text{OH}^-(aq)$. This works well for aqueous solutions but cannot explain the basicity of substances like ammonia, which have no hydroxyl group at all. The Bronsted-Lowry concept fixes that gap: an acid is a proton ($\text{H}^+$) donor, a base is a proton acceptor. When $\text{NH}_3$ dissolves in water, water donates a proton to ammonia:

Here water is the acid and ammonia the base going forward, but in the reverse reaction $\text{NH}_4^+$ donates a proton to $\text{OH}^-$, making $\text{NH}_4^+$ the acid and $\text{OH}^-$ the base. A pair of species differing by exactly one proton — like $\text{NH}_3$/$\text{NH}_4^+$ or $\text{H}_2\text{O}$/$\text{OH}^-$ — is a conjugate acid-base pair; a strong Bronsted acid always has a weak conjugate base, and vice versa. Water's ability to act as both an acid (donating a proton to a stronger base) and a base (accepting a proton from a stronger acid) is its amphoteric or dual character. The Lewis concept is broader still: an acid is any species that accepts an electron pair, a base any species that donates one. This captures acids with no proton at all, like $\text{BF}_3$, which reacts with $:\!\text{NH}_3$ purely by accepting its lone pair ($\text{BF}_3 + {:}\text{NH}_3 \rightarrow \text{BF}_3{:}\text{NH}_3$). Electron-deficient species like $\text{AlCl}_3$, $\text{Co}^{3+}$ and $\text{Mg}^{2+}$ are Lewis acids; species with a lone pair to donate, like $\text{H}_2\text{O}$, $\text{NH}_3$ and $\text{OH}^-$, are Lewis bases.

Ionization constants: Ka, Kb and the strength of weak acids and bases

For a weak acid $\text{HX}$ partially ionizing in water, $\text{HX}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{X}^-(aq)$, define $c$ as the initial concentration and $\alpha$ as the degree of ionization — the fraction of $\text{HX}$ that has ionized at equilibrium. The equilibrium concentrations become $c(1-\alpha)$ for $\text{HX}$ and $c\alpha$ for each ion, giving the ionization constant:

where: $K_a = [\text{H}^+][\text{X}^-]/[\text{HX}]$ equivalently, and a larger $K_a$ means a stronger acid. When $\alpha$ is small (weak acid, not too dilute), the approximation $1 - \alpha \approx 1$ simplifies this to $K_a \approx c\alpha^2$, useful for quick estimates but always worth checking against the exact quadratic when $\alpha$ turns out not to be small. An identical treatment gives the base ionization constant $K_b = c\alpha^2/(1-\alpha)$ for a weak base $\text{MOH} \rightleftharpoons \text{M}^+ + \text{OH}^-$. Since $K_a$ and $K_b$ values span many orders of magnitude, they are reported logarithmically: $\text{p}K_a = -\log K_a$ and $\text{p}K_b = -\log K_b$ — smaller $\text{p}K_a$ means a stronger acid. Acid strength depends on two competing factors: the strength of the H–A bond (weaker bond, easier to break, stronger acid) and its polarity (larger electronegativity difference, easier heterolytic cleavage, stronger acid). Down a group of the periodic table, bond strength dominates as atomic size increases, so acid strength rises: $\text{HF} \ll \text{HCl} \ll \text{HBr} \ll \text{HI}$. Across a period, bond polarity dominates as electronegativity rises, so acid strength again rises: $\text{CH}_4 < \text{NH}_3 < \text{H}_2\text{O} < \text{HF}$. Polyprotic acids like $\text{H}_2\text{X}$ ionize in successive steps, each with its own constant, $K_{a1}$, $K_{a2}$, and so on, and always $K_{a1} \gg K_{a2} \gg K_{a3}$ — it is much harder to pull a positively charged proton away from an already negatively charged ion like $\text{HX}^-$ than from the neutral parent $\text{H}_2\text{X}$, because of the extra electrostatic attraction holding that proton on.

Water, the pH scale and pKw

Pure water ionizes very slightly, one molecule acting as acid and donating a proton to another acting as base:

Since $[\text{H}_2\text{O}]$ is essentially constant, it is folded into a new constant, the ionic product of water:

where: experimentally, $[\text{H}^+] = [\text{OH}^-] = 1.0\times10^{-7}\ \text{M}$ at $298\ \text{K}$, so $K_w = 1.0\times10^{-14}\ \text{M}^2$ at that temperature (it rises with temperature, since water's self-ionization is endothermic, though this shift is usually small enough to ignore). A solution is acidic if $[\text{H}^+] > [\text{OH}^-]$, basic if $[\text{H}^+] < [\text{OH}^-]$, and neutral if they are equal. Because hydrogen-ion concentrations range over many orders of magnitude, chemists use the logarithmic pH scale:

where: pH $7$ is neutral (pure water at $298\ \text{K}$), pH below $7$ is acidic, and pH above $7$ is basic. Taking $-\log$ of the $K_w$ expression gives $\text{pK}_w = \text{pH} + \text{pOH} = 14$ at $298\ \text{K}$ — a relation that holds for every aqueous solution at that temperature, acidic, neutral or basic alike, and lets you find pOH the instant you know pH, or vice versa.

Common ion effect and hydrolysis of salts

Adding an ion already present in an ionic equilibrium suppresses further ionization — the common ion effect, a direct application of Le Chatelier's principle. Adding sodium acetate to acetic acid raises $[\text{Ac}^-]$, which pushes $\text{HAc} \rightleftharpoons \text{H}^+ + \text{Ac}^-$ back toward $\text{HAc}$, lowering $[\text{H}^+]$ and raising pH; conversely, adding $\text{H}^+$ from an external strong acid drives the equilibrium toward more undissociated $\text{HAc}$. Salts formed from acid-base neutralization are not automatically neutral. A salt of a strong acid and strong base ($\text{NaCl}$) simply hydrates without hydrolysing, giving pH $7$. A salt of a weak acid and strong base (e.g. $\text{CH}_3\text{COONa}$) hydrolyses at the anion: $\text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq)$, releasing $\text{OH}^-$ and giving pH $> 7$. A salt of a strong acid and weak base (e.g. $\text{NH}_4\text{Cl}$) hydrolyses at the cation, releasing $\text{H}^+$ and giving pH $< 7$. A salt of a weak acid and weak base (e.g. $\text{CH}_3\text{COONH}_4$) hydrolyses at both ions simultaneously, and its pH depends only on the relative strengths of the parent acid and base, independent of concentration:

where: pH exceeds $7$ if the parent acid is weaker than the parent base ($\text{p}K_a > \text{p}K_b$), and is below $7$ in the opposite case.

Buffer solutions

A buffer solution resists a change in pH when small amounts of acid or base are added, or on dilution. The two standard recipes are a weak acid with a salt of its conjugate base (an acidic buffer, such as acetic acid plus sodium acetate, buffering near pH $4.76$), or a weak base with a salt of its conjugate acid (a basic buffer, such as ammonia plus ammonium chloride, buffering near pH $9.25$). Blood itself is buffered by dissolved carbonic acid and bicarbonate, holding pH within a very narrow physiological range. The governing relation, the Henderson-Hasselbalch equation, is derived in full below; its acidic-buffer form is $\text{pH} = \text{p}K_a + \log([\text{salt}]/[\text{acid}])$ — when salt and acid concentrations are equal, the log term vanishes and pH simply equals $\text{p}K_a$, which is exactly why you choose an acid whose $\text{p}K_a$ sits close to the target pH when designing a buffer.

Solubility equilibria: the solubility product

For a sparingly soluble ionic salt $\text{M}_x\text{X}_y(s) \rightleftharpoons x\text{M}^{p+}(aq) + y\text{X}^{q-}(aq)$ in equilibrium with its saturated solution, the concentration of the undissolved solid is constant and is folded into a new constant, the solubility product:

where: if the salt's molar solubility is $S$, then $[\text{M}^{p+}] = xS$ and $[\text{X}^{q-}] = yS$, giving $K_{sp} = x^x y^y S^{(x+y)}$ — for example, $\text{BaSO}_4$ (a 1:1 salt) gives $K_{sp} = S^2$, while $\text{A}_2\text{X}_3$ gives $K_{sp} = 108S^5$. Comparing $K_{sp}$ with the equivalent reaction quotient $Q_{sp}$ predicts whether a salt will precipitate ($Q_{sp} > K_{sp}$) or continue dissolving ($Q_{sp} < K_{sp}$). Just as with weak acids, a common ion suppresses solubility: adding excess $\text{OH}^-$ from $\text{NaOH}$ to a solution in contact with $\text{Ni(OH)}_2(s)$ drastically reduces how much $\text{Ni(OH)}_2$ dissolves, since the solid must satisfy the fixed product $K_{sp} = [\text{Ni}^{2+}][\text{OH}^-]^2$ with an artificially inflated $[\text{OH}^-]$ already present. This is exploited in gravimetric analysis to force a target ion to precipitate as completely as possible, and in reverse, salts of weak acids like phosphates become more soluble at lower pH, because protonating the anion pulls the equilibrium toward dissolving more solid.

Board level - Equilibrium constant for the formation of ammonia (NCERT Problem 6.1)

At $500\ \text{K}$, an equilibrium mixture of $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ has $[\text{N}_2] = 1.5\times10^{-2}\ \text{M}$, $[\text{H}_2] = 3.0\times10^{-2}\ \text{M}$ and $[\text{NH}_3] = 1.2\times10^{-2}\ \text{M}$. Calculate $K_c$.

Step 1. Write the equilibrium expression from the balanced equation, using each species' stoichiometric coefficient as the exponent: $K_c = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$.

Step 2. Compute the numerator: $[\text{NH}_3]^2 = (1.2\times10^{-2})^2 = 1.44\times10^{-4}$.

Step 3. Compute the denominator: $[\text{H}_2]^3 = (3.0\times10^{-2})^3 = 2.7\times10^{-5}$, so $[\text{N}_2][\text{H}_2]^3 = (1.5\times10^{-2})(2.7\times10^{-5}) = 4.05\times10^{-7}$.

Step 4. Divide: $K_c = \dfrac{1.44\times10^{-4}}{4.05\times10^{-7}} \approx 3.56\times10^{2}$. Answer: $K_c \approx 3.6\times10^{2}\ \text{M}^{-2}$, comfortably inside the $10^{-3}$ to $10^{3}$ range, so both reactants and appreciable ammonia coexist at this equilibrium — consistent with ammonia synthesis needing a catalyst and pressure to push the yield further, rather than a reaction that is essentially complete on its own.

Competitive level (JEE / NEET) - Equilibrium composition of dissociating PCl5 (NCERT Problem 6.9)

$3.00\ \text{mol}$ of $\text{PCl}_5$ is placed in a $1\ \text{L}$ closed vessel and allowed to reach equilibrium at $380\ \text{K}$, where $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$ has $K_c = 1.80$. Find the equilibrium composition of the mixture.

Step 1. Set up an ICE (initial, change, equilibrium) table. Let $x$ mol/L of $\text{PCl}_5$ dissociate: initial concentrations are $[\text{PCl}_5]_0 = 3.0$, $[\text{PCl}_3]_0 = [\text{Cl}_2]_0 = 0$; at equilibrium, $[\text{PCl}_5] = 3.0 - x$, $[\text{PCl}_3] = [\text{Cl}_2] = x$.

Step 2. Substitute into the equilibrium expression: $K_c = \dfrac{[\text{PCl}_3][\text{Cl}_2]}{[\text{PCl}_5]} = \dfrac{x^2}{3.0-x} = 1.80$.

Step 3. Clear the denominator and collect into standard quadratic form: $x^2 = 1.80(3.0-x) = 5.4 - 1.8x$, so $x^2 + 1.8x - 5.4 = 0$. Using the quadratic formula with $a=1$, $b=1.8$, $c=-5.4$: $x = \dfrac{-1.8 \pm \sqrt{1.8^2 - 4(1)(-5.4)}}{2} = \dfrac{-1.8 \pm \sqrt{24.84}}{2} = \dfrac{-1.8 \pm 4.98}{2}$.

Step 4. Take only the physically meaningful root — concentration cannot be negative, and $x$ cannot exceed the starting $3.0\ \text{M}$ — so reject the negative root and keep $x = \dfrac{3.18}{2} = 1.59$. Answer: $[\text{PCl}_5] = 1.41\ \text{M}$, $[\text{PCl}_3] = [\text{Cl}_2] = 1.59\ \text{M}$. Verify by back-substitution: $(1.59)^2/1.41 = 2.53/1.41 \approx 1.79 \approx K_c$, confirming the answer. Timing note: under three minutes once the ICE table habit is automatic. The one place JEE/NEET problems try to trap you is the quadratic's two roots — always sanity-check against the initial concentration (here, $x < 3.0$) and reject any root that gives a negative concentration for any species; do not simply pick "the positive one" if both roots happen to be positive but only one is chemically valid. When $K_c$ is very small compared to the initial concentration, skip the quadratic entirely and approximate $3.0 - x \approx 3.0$ to save time, but always state that you have made the approximation.

Board level - pH and degree of dissociation of hydrofluoric acid (NCERT Problem 6.18)

The ionization constant of $\text{HF}$ is $K_a = 3.2\times10^{-4}$. Calculate the degree of dissociation and the pH of a $0.02\ \text{M}$ $\text{HF}$ solution.

Step 1. Set up the ICE table in terms of degree of ionization $\alpha$: for $\text{HF} + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{F}^-$, initial $[\text{HF}] = 0.02$, and at equilibrium $[\text{HF}] = 0.02(1-\alpha)$, $[\text{H}_3\text{O}^+] = [\text{F}^-] = 0.02\alpha$.

Step 2. Substitute into $K_a = \dfrac{(0.02\alpha)^2}{0.02(1-\alpha)} = \dfrac{0.02\alpha^2}{1-\alpha} = 3.2\times10^{-4}$. Since $K_a$ is not negligible compared with $c = 0.02$, the small-$\alpha$ shortcut is unsafe here, so keep the full quadratic: dividing through by $0.02$ gives $\alpha^2 + 0.016\alpha - 0.016 = 0$.

Step 3. Solve using the quadratic formula: $\alpha = \dfrac{-0.016 \pm \sqrt{(0.016)^2 + 4(0.016)}}{2} = \dfrac{-0.016 \pm \sqrt{0.064256}}{2} = \dfrac{-0.016 \pm 0.2535}{2}$. Rejecting the negative root, $\alpha = \dfrac{0.2375}{2} \approx 0.12$.

Step 4. Compute the equilibrium concentrations and pH: $[\text{H}_3\text{O}^+] = [\text{F}^-] = 0.02 \times 0.12 = 2.4\times10^{-3}\ \text{M}$, and $[\text{HF}] = 0.02 \times 0.88 = 1.76\times10^{-2}\ \text{M}$. Then $\text{pH} = -\log(2.4\times10^{-3}) = 2.62$. Answer: $\alpha \approx 0.12$ ($12\%$ dissociated), and $\text{pH} = 2.62$.

Competitive level (JEE / NEET) - Common-ion suppression of Ni(OH)2 solubility (NCERT Problem 6.28)

The solubility product of $\text{Ni(OH)}_2$ is $K_{sp} = 2.0\times10^{-15}$. Calculate its molar solubility (a) in pure water, and (b) in $0.10\ \text{M}$ $\text{NaOH}$, and compare the two.

Step 1. In pure water, let $S$ be the molar solubility of $\text{Ni(OH)}_2(s) \rightleftharpoons \text{Ni}^{2+}(aq) + 2\text{OH}^-(aq)$, so $[\text{Ni}^{2+}] = S$ and $[\text{OH}^-] = 2S$. Then $K_{sp} = (S)(2S)^2 = 4S^3 = 2.0\times10^{-15}$, giving $S^3 = 5.0\times10^{-16}$ and $S = (5.0\times10^{-16})^{1/3} \approx 7.9\times10^{-6}\ \text{M}$.

Step 2. In $0.10\ \text{M}$ $\text{NaOH}$, the solution already carries a large "common ion" concentration of $\text{OH}^-$ before any $\text{Ni(OH)}_2$ dissolves. Let the new solubility be $S'$: $[\text{Ni}^{2+}] = S'$ and total $[\text{OH}^-] = (0.10 + 2S')$.

Step 3. Because $K_{sp}$ is so small, $2S'$ will turn out to be utterly negligible next to $0.10$, so approximate $(0.10 + 2S') \approx 0.10$ and solve: $K_{sp} = S'(0.10)^2 = 2.0\times10^{-15}$, giving $S' = \dfrac{2.0\times10^{-15}}{0.01} = 2.0\times10^{-13}\ \text{M}$.

Step 4. Compare the two solubilities: $\dfrac{S}{S'} = \dfrac{7.9\times10^{-6}}{2.0\times10^{-13}} \approx 4\times10^{7}$. Answer: solubility drops from about $7.9\times10^{-6}\ \text{M}$ in pure water to about $2.0\times10^{-13}\ \text{M}$ in $0.10\ \text{M}$ NaOH — a roughly forty-million-fold suppression by the common ion. Timing note: under three minutes if you spot immediately that $2S' \ll 0.10$ makes the cubic collapse into a simple square. The trap is forgetting to add the pre-existing $0.10\ \text{M}$ $\text{OH}^-$ from NaOH to whatever $\text{OH}^-$ the dissolving salt itself contributes — treating the two sources of $\text{OH}^-$ as independent, rather than summing them in one $K_{sp}$ expression, is the single most common error on common-ion solubility questions.

• Writing initial concentrations into $K_c$ instead of equilibrium concentrations. The equilibrium expression only ever uses concentrations after the system has stopped changing — always build an ICE table rather than substituting the numbers given in the problem statement directly.
• Including pure solids or liquids in the equilibrium expression. For heterogeneous equilibria like $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$, only $[\text{CO}_2]$ (or $p_{\text{CO}_2}$) appears; the solids must be physically present, but their concentration never appears in the expression.
• Confusing $K_p$ and $K_c$, or misapplying $\Delta n$. $\Delta n$ is moles of gaseous products minus moles of gaseous reactants in the balanced equation as written — recompute it if you scale the equation, and remember $K_p = K_c$ only when $\Delta n = 0$.
• Believing a catalyst shifts equilibrium composition. A catalyst changes the rate at which equilibrium is reached, never the position of equilibrium or the value of $K$.
• Mixing up the direction of the temperature effect. For an exothermic reaction, raising temperature decreases $K$ and shifts equilibrium backward; for an endothermic reaction, raising temperature increases $K$. Get $\Delta H$'s sign wrong and the direction flips.
• Using the small-$\alpha$ or small-$x$ approximation blindly. The shortcut $1-\alpha \approx 1$ or $c-x \approx c$ is only valid when $K$ is very small relative to the initial concentration; always check the size of the correction term afterward, and fall back to the full quadratic when $K_a$ or $K_b$ is not small enough.
• Forgetting that $[\text{H}^+]$ from water's own autoionization matters at very low concentrations. For strong acid or base solutions more dilute than about $10^{-6}\ \text{M}$, water's own $10^{-7}\ \text{M}$ contribution can no longer be ignored, and the pH will not simply equal $-\log(\text{acid concentration})$.
• Treating $\text{p}K_a + \text{p}K_b = 14$ as universal rather than pair-specific. It applies only to a genuine conjugate acid-base pair (like $\text{NH}_4^+$/$\text{NH}_3$) at $298\ \text{K}$, not to any two unrelated acids and bases you happen to be given.
• Assuming a weak acid automatically means a weak conjugate base. It is the opposite: the weaker the acid, the stronger its conjugate base — a direct consequence of $K_a \times K_b = K_w$ being fixed.

• Write the correct $K_c$ or $K_p$ expression for any reaction, including heterogeneous ones, correctly omitting pure solids and liquids.
• Convert between $K_p$ and $K_c$ using $K_p = K_c(RT)^{\Delta n}$, computing $\Delta n$ correctly from the balanced equation.
• Use the reaction quotient $Q$ to predict whether a given mixture will react forward, in reverse, or is already at equilibrium.
• Apply Le Chatelier's principle to predict the effect of concentration, pressure/volume, inert gas addition, temperature and catalysts on both the position of equilibrium and the value of $K$.
• Connect $\Delta G^\circ$ and $K$ using $\Delta G^\circ = -RT\ln K$, and read off spontaneity from the sign of $\Delta G^\circ$.
• Classify a species as an Arrhenius, Bronsted-Lowry or Lewis acid or base, and identify conjugate acid-base pairs.
• Calculate $K_a$, $K_b$, degree of ionization, pH and pOH for weak acids and weak bases, including using the ICE-table method.
• Use $K_w$, $\text{pH} + \text{pOH} = 14$ and $K_a \times K_b = K_w$ to move fluidly between any of pH, pOH, $K_a$ and $K_b$.
• Predict whether a salt's aqueous solution will be acidic, basic or neutral from the strengths of its parent acid and base, and calculate the pH of a buffer using the Henderson-Hasselbalch equation.
• Calculate solubility from $K_{sp}$ and explain how the common ion effect suppresses the solubility of a sparingly soluble salt.

Ask the tutor: if you want to work through more Le Chatelier "predict the shift" scenarios or practice buffer and $K_{sp}$ numericals step by step, just ask.