Equilibrium
Chemical equilibrium, Le Chatelier, acids & bases, pH, buffers and Ksp for NEET
Chemical Equilibrium
Equilibrium State, Kc, Kp and the Reaction QuotientTopic 1
Many reactions are reversible — they proceed in both directions. When such a reaction runs in a closed container, it reaches a state of equilibrium where the rate of the forward reaction equals the rate of the reverse reaction. Equilibrium is dynamic: molecules keep reacting both ways, but the concentrations of reactants and products no longer change. This dynamic-but-constant nature is the first idea NEET checks.
At equilibrium, the concentrations obey the law of mass action, giving the equilibrium constant. For a reaction $aA + bB \rightleftharpoons cC + dD$, $K_c = \dfrac{[C]^c [D]^d}{[A]^a [B]^b}$ in terms of molar concentrations. For gas-phase reactions an equivalent constant $K_p$ uses partial pressures. The two are related by $K_p = K_c (RT)^{\Delta n_g}$, where $\Delta n_g$ is the change in moles of gas — a formula NEET applies often.
The size of $K$ tells you how far a reaction goes: a large $K$ means products are favoured at equilibrium, a small $K$ means reactants dominate. Pure solids and pure liquids are left out of the expression (their 'concentration' is constant), a point that trips students in heterogeneous equilibria.
To predict the direction of a reaction not yet at equilibrium, compute the reaction quotient $Q$ using the same expression as $K$ but with current concentrations. If $Q < K$ the reaction moves forward (toward products); if $Q > K$ it moves backward; and if $Q = K$ it is already at equilibrium. This $Q$-versus-$K$ comparison is a guaranteed NEET tool.
| Quantity | Relation |
|---|---|
| $K_c$ | $\dfrac{[C]^c[D]^d}{[A]^a[B]^b}$ |
| $K_p$–$K_c$ | $K_p = K_c(RT)^{\Delta n_g}$ |
| $Q \lt K$ | forward reaction |
| $Q \gt K$ | reverse reaction |
For $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, find $\Delta n_g$ and write the $K_p$–$K_c$ relation.
Show solution
$\Delta n_g = 2 - (1+3) = -2$, so $K_p = K_c (RT)^{-2}$.
At a given moment $Q = 5$ while $K = 50$. Which way does the reaction proceed?
Show solution
$Q < K$, so the reaction proceeds in the forward direction (making more products) until $Q$ rises to equal $K$.
At equilibrium, the rates of the forward and reverse reactions are:
$K_p$ and $K_c$ are related by:
If $Q > K$, the reaction proceeds:
A large value of $K$ means at equilibrium:
In the equilibrium expression, pure solids and liquids are:
NEET tip: $K_p = K_c(RT)^{\Delta n_g}$ (count gas moles only). Compare $Q$ with $K$ for direction: $Q \lt K$ forward, $Q \gt K$ reverse. Omit pure solids/liquids from $K$.
Le Chatelier's PrincipleTopic 2
Once a reaction is at equilibrium, what happens if we disturb it? Le Chatelier's principle gives the answer: if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the equilibrium shifts in the direction that opposes (partly undoes) that change. This single principle predicts the outcome of most NEET equilibrium-disturbance questions.
Concentration: adding a reactant pushes the equilibrium toward products (to consume the added reactant); removing a product also pulls it forward. The reverse changes shift it backward. The value of $K$ stays the same — only the position moves.
Pressure (for gases): increasing the pressure (decreasing volume) shifts the equilibrium toward the side with fewer gas moles, to reduce the pressure. If both sides have equal gas moles, pressure has no effect. Adding an inert gas at constant volume does not shift the equilibrium (partial pressures are unchanged) — a classic NEET trap.
Temperature: this is the only factor that actually changes the value of $K$. Raising the temperature favours the endothermic direction (it absorbs the added heat), while lowering it favours the exothermic direction. A catalyst changes neither $K$ nor the equilibrium position — it only helps the system reach equilibrium faster by speeding both directions equally. These rules are applied directly in the industrial Haber process (high pressure, moderate temperature) and Contact process, common NEET contexts.
| Change | Equilibrium shifts |
|---|---|
| Add reactant | toward products (forward) |
| Increase pressure | toward fewer gas moles |
| Increase temperature | toward endothermic side (changes $K$) |
| Catalyst | no shift; reaches equilibrium faster |
For $\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ (exothermic), how does increasing pressure affect the yield of $\text{NH}_3$?
Show solution
The product side has fewer gas moles (2 vs 4). Increasing pressure shifts equilibrium toward $\text{NH}_3$, increasing its yield — the basis of high pressure in the Haber process.
Does adding a catalyst increase the equilibrium yield of a reaction?
Show solution
No. A catalyst speeds up both forward and reverse reactions equally, so equilibrium is reached faster but the position and yield (and $K$) are unchanged.
Le Chatelier's principle says a disturbed equilibrium shifts to:
Increasing pressure shifts equilibrium toward the side with:
The only factor that changes the value of $K$ is:
Adding an inert gas at constant volume:
A catalyst affects the equilibrium position by:
NEET trap: only temperature changes $K$; a catalyst and an inert gas (constant V) do NOT shift equilibrium. Pressure increase to fewer-gas-moles side.
Ionic Equilibrium
Acids, Bases and pHTopic 3
Acids and bases are described by three theories of increasing scope. The Arrhenius view: an acid gives $\text{H}^+$ and a base gives $\text{OH}^-$ in water. The broader Bronsted-Lowry view: an acid is a proton donor and a base a proton acceptor, which introduces conjugate acid-base pairs that differ by one $\text{H}^+$ (e.g. $\text{HCl}/\text{Cl}^-$). The most general Lewis view: an acid is an electron-pair acceptor and a base an electron-pair donor. NEET regularly asks you to identify conjugate pairs and Lewis acids/bases.
Acids and bases differ in strength. Strong acids/bases ionise completely in water; weak ones ionise only partly, described by a dissociation constant $K_a$ (acids) or $K_b$ (bases). A larger $K_a$ means a stronger acid. For a conjugate pair, $K_a \times K_b = K_w$, linking the strength of an acid to that of its conjugate base.
Water itself ionises slightly: $\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-$, with the ionic product of water $K_w = [\text{H}^+][\text{OH}^-] = 1.0\times10^{-14}$ at $25^{\circ}\text{C}$. In pure (neutral) water $[\text{H}^+] = [\text{OH}^-] = 10^{-7}\ \text{M}$.
Acidity is reported on the pH scale: $\text{pH} = -\log[\text{H}^+]$, with $\text{pOH} = -\log[\text{OH}^-]$ and $\text{pH} + \text{pOH} = 14$ at $25^{\circ}\text{C}$. So pH $< 7$ is acidic, $= 7$ neutral, and $> 7$ basic. A change of one pH unit means a tenfold change in $[\text{H}^+]$ — a point NEET likes to test numerically. Being fluent with $\text{pH} = -\log[\text{H}^+]$ and $K_w$ handles most acid-base questions.
| Concept | Relation |
|---|---|
| Bronsted acid/base | proton donor / acceptor |
| Ionic product of water | $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$ |
| pH | $-\log[\text{H}^+]$; $\text{pH}+\text{pOH}=14$ |
| Conjugate pair | $K_a \times K_b = K_w$ |
Find the pH of a $0.001\ \text{M}$ solution of a strong acid (HCl).
Show solution
HCl is strong, so $[\text{H}^+] = 0.001 = 10^{-3}\ \text{M}$. $\text{pH} = -\log(10^{-3}) = 3$.
Identify the conjugate base of $\text{HCO}_3^-$.
Show solution
Removing one $\text{H}^+$ from $\text{HCO}_3^-$ gives $\text{CO}_3^{2-}$, which is its conjugate base. (Adding $\text{H}^+$ would give its conjugate acid, $\text{H}_2\text{CO}_3$.)
A Bronsted-Lowry base is a:
The ionic product of water at $25^{\circ}\text{C}$ is:
The pH of a neutral solution at $25^{\circ}\text{C}$ is:
A Lewis acid is an:
For a conjugate acid-base pair, $K_a \times K_b$ equals:
NEET tip: $\text{pH} = -\log[\text{H}^+]$, $\text{pH}+\text{pOH}=14$, $K_w=10^{-14}$. Conjugate pairs differ by one $\text{H}^+$ and $K_aK_b=K_w$. A 1-unit pH change = 10× $[\text{H}^+]$.
Buffers, Hydrolysis and Solubility ProductTopic 4
A buffer solution resists changes in pH when small amounts of acid or base are added. It is made from a weak acid and its salt (acidic buffer, e.g. $\text{CH}_3\text{COOH} + \text{CH}_3\text{COONa}$) or a weak base and its salt (basic buffer). Buffers are vital biologically — blood is buffered near pH $7.4$, a fact NEET likes to reference.
The pH of an acidic buffer is given by the Henderson-Hasselbalch equation: $\text{pH} = \text{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$. When the salt and acid concentrations are equal, $\text{pH} = \text{p}K_a$ and the buffer has its maximum capacity. This equation is a guaranteed NEET calculation.
When a salt dissolves in water, its ions may react with water — salt hydrolysis — affecting the pH. A salt of a strong acid and strong base (e.g. NaCl) gives a neutral solution; a salt of a weak acid and strong base (e.g. $\text{CH}_3\text{COONa}$) is basic; and a salt of a strong acid and weak base (e.g. $\text{NH}_4\text{Cl}$) is acidic. Predicting the nature of a salt solution from its parent acid and base is a frequent NEET question.
For sparingly soluble salts, equilibrium between the solid and its dissolved ions is described by the solubility product $K_{sp}$. A precipitate forms when the ionic product exceeds $K_{sp}$; the solution is just saturated when they are equal. Adding a common ion shifts this equilibrium back, decreasing solubility — the common-ion effect, used to control precipitation and in qualitative salt analysis. Mastering the Henderson equation, salt-nature rules, and the $K_{sp}$/common-ion idea covers the bulk of ionic-equilibrium questions.
| Concept | Key point |
|---|---|
| Buffer (Henderson) | $\text{pH} = \text{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$ |
| Salt of weak acid + strong base | basic solution |
| Precipitation | ionic product $> K_{sp}$ |
| Common-ion effect | decreases solubility |
A buffer has equal concentrations of acetic acid and sodium acetate, with $\text{p}K_a = 4.74$. Find its pH.
Show solution
$\text{pH} = \text{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]} = 4.74 + \log(1) = 4.74$ (equal concentrations to maximum buffer capacity).
Is an aqueous solution of $\text{NH}_4\text{Cl}$ acidic, basic or neutral?
Show solution
$\text{NH}_4\text{Cl}$ is the salt of a strong acid (HCl) and a weak base ($\text{NH}_3$). The $\text{NH}_4^+$ hydrolyses to give $\text{H}^+$, so the solution is acidic (pH < 7).
An acidic buffer is made of a weak acid and its:
The Henderson equation is:
A solution of $\text{CH}_3\text{COONa}$ (salt of weak acid + strong base) is:
A precipitate forms when the ionic product is:
The common-ion effect on solubility:
NEET tip: Buffer pH from Henderson ($\text{pH}=\text{p}K_a$ when salt = acid). Salt nature: WA+SB to basic, SA+WB to acidic, SA+SB to neutral. Precipitate when ionic product $> K_{sp}$; common ion lowers solubility.
Quick Revision — Equilibrium
- Equilibrium is dynamic — forward and reverse rates equal; concentrations constant.
- Equilibrium constant: $K_c$ (concentrations), $K_p$ (partial pressures), with $K_p = K_c (RT)^{\Delta n_g}$.
- Reaction quotient $Q$: $Q \lt K$ forward, $Q \gt K$ reverse, $Q=K$ equilibrium.
- Le Chatelier: a system at equilibrium shifts to oppose any change in concentration, pressure or temperature (a catalyst changes neither $K$ nor the position).
- Acids/bases: Arrhenius, Bronsted-Lowry (proton donor/acceptor, conjugate pairs), Lewis (electron-pair acceptor/donor).
- pH $= -\log[\text{H}^+]$; $K_w = [\text{H}^+][\text{OH}^-] = 10^{-14}$ at $25^{\circ}\text{C}$; neutral pH $7$.
- Buffer: resists pH change; $\text{pH} = \text{p}K_a + \log\dfrac{[\text{salt}]}{[\text{acid}]}$ (Henderson).
- Solubility product $K_{sp}$: precipitation when ionic product $> K_{sp}$; common-ion effect lowers solubility.
Frequently Asked Questions
Ready to test yourself?
Attempt the full timed mock tests — Main & Advanced level.
Start Mock Test 1 →