Equilibrium — NEET notes

Chemical Equilibrium

A reversible reaction reaches dynamic equilibrium when the rate of forward reaction equals the rate of backward reaction. Concentrations become constant (not equal) and all macroscopic properties stop changing, even though both reactions continue at the molecular level.

r_f = r_b (at equilibrium)

Reversible Reactions

A reversible reaction proceeds in both forward and backward directions simultaneously. Under given conditions, it never goes to completion — instead it reaches a state of balance.

Most chemical reactions are reversible — products can react to re-form reactants. We write them with a double arrow ( leftharpoons ). Example: N_2(g) + 3H_2(g) leftharpoons 2NH_3(g) . The forward reaction forms ammonia; the backward reaction decomposes it.

• Irreversible reactions go to completion (e.g., combustion of wood, neutralization of strong acid + strong base in dilute solution)
• Reversible reactions never go to completion — both reactants and products are always present
• The double arrow leftharpoons indicates reversibility; a single arrow arrow indicates irreversibility
• Reversibility depends on conditions — a reaction irreversible in one condition may be reversible in another

Dynamic Equilibrium

At dynamic equilibrium, the forward and backward reactions occur at equal rates. Concentrations remain constant but molecules are continuously interconverting.

When a reversible reaction starts, the forward rate is high (lots of reactants) and the backward rate is zero (no products yet). As products accumulate, the backward rate increases. Eventually, forward rate = backward rate. This is dynamic equilibrium.

At equilibrium: r_ forward = r_ backward

• Equilibrium is dynamic — reactions don't stop, both continue at equal rates
• Equilibrium is not static — molecules are continuously being formed and consumed
• Concentrations become constant (not equal!) at equilibrium
• Equilibrium can be attained from either direction — starting with reactants or products
• Equilibrium can only be achieved in a closed system

Characteristics of Equilibrium

Chemical equilibrium has definite characteristics: closed system, constant macroscopic properties, dynamic nature, approachable from either side, and dependence on conditions.

• Closed system required — no exchange of matter with surroundings (energy exchange is OK)
• Constant macroscopic properties — concentration, pressure, color, density all remain constant
• Dynamic nature — forward and backward rates are equal and non-zero
• Attainable from either direction — same equilibrium state whether you start with reactants or products
• Catalyst does NOT change equilibrium position — it only helps reach equilibrium faster

Equilibrium Constant

The equilibrium constant K quantifies the position of equilibrium. K_c uses molar concentrations, K_p uses partial pressures. They are related by K_p = K_c(RT)^(Delta n). The reaction quotient Q predicts the direction a reaction will shift to reach equilibrium.

K_p = K_c(RT)^ n

Law of Chemical Equilibrium

For a balanced reaction at equilibrium, the ratio of product concentrations to reactant concentrations (each raised to their stoichiometric coefficients) is constant at a given temperature.

For a general reaction aA + bB leftharpoons cC + dD , the equilibrium constant expression is:

K_c = [C]^c [D]^d [A]^a [B]^b

This is the Law of Chemical Equilibrium (also called the Law of Mass Action applied to equilibrium). K_c depends only on temperature — not on initial concentrations, pressure, or catalysts.

• Products in the numerator, reactants in the denominator
• Each concentration raised to the power of its stoichiometric coefficient
• K_c uses molar concentrations (mol/L)
• Pure solids and pure liquids are excluded from the expression (their activity = 1)
• K depends ONLY on temperature — not on concentration, pressure, volume, or catalyst

K_p Expression

For gaseous equilibria, K_p uses partial pressures instead of concentrations. K_p and K_c are related through the ideal gas equation.

For gaseous reactions, it's often more convenient to use partial pressures instead of concentrations. For aA(g) + bB(g) leftharpoons cC(g) + dD(g) :

K_p = (p_C)^c (p_D)^d (p_A)^a (p_B)^b

Since p = cRT (from ideal gas law, where c = molar concentration), we can relate K_p and K_c :

K_p = K_c(RT)^ n_g

where n_g = (c + d) - (a + b) = moles of gaseous products - moles of gaseous reactants

• R = 0.0821 L atm mol ^ -1 K ^ -1 (when pressure is in atm) or R = 8.314 J mol ^ -1 K ^ -1 (SI)
• If n_g = 0 → K_p = K_c (e.g., H _2 + I _2 leftharpoons 2HI, n_g = 2-2 = 0 )
• If n_g > 0 → K_p > K_c
• If n_g < 0 → K_p < K_c
• T must be in Kelvin — a common source of errors

Reaction Quotient Q

Q has the same expression as K but uses current (non-equilibrium) concentrations. Comparing Q with K tells you which direction the reaction will shift.

The reaction quotient Q has the exact same mathematical expression as K , but uses instantaneous concentrations (not equilibrium ones). It answers the question: 'Where is the reaction right now compared to equilibrium?'

Q_c = [C]^c [D]^d [A]^a [B]^b (at any instant)

• Q < K → Too many reactants, too few products → reaction shifts forward (toward products)
• Q > K → Too many products, too few reactants → reaction shifts backward (toward reactants)
• Q = K → System is at equilibrium → no net shift

Manipulating K

When you reverse, multiply, or add reactions, the equilibrium constant changes in predictable ways. These rules are essential for multi-step equilibrium problems.

The value of K depends on exactly how the equation is written. Three key rules govern how K transforms:

• Reverse the reaction → K' = 1 K
• Multiply coefficients by n → K' = K^n
• Divide coefficients by n → K' = K^ 1/n
• Add two reactions → K_ net = K_1 K_2
• Subtract a reaction → K_ net = K_1 K_2

Units of K

The units of K depend on Delta n. When Delta n = 0, K is dimensionless. NCERT treats K as having units, but thermodynamic K (using activities) is always dimensionless.

The units of K_c and K_p depend on n :

Units of K_c = (mol/L)^ n Units of K_p = (atm)^ n_g

• If n = 0 : K is unitless (e.g., H _2 + I _2 leftharpoons 2HI)
• If n > 0 : K_c has units of (mol/L) ^ n
• If n < 0 : K_c has units of (mol/L) ^ n = (L/mol) ^ | n|
• Thermodynamic K (using activities) is always dimensionless — but NCERT problems use K_c and K_p with units

Le Chatelier's Principle

If a system at equilibrium is disturbed (by changing concentration, pressure, or temperature), it shifts in the direction that partially counteracts the disturbance. Catalyst and inert gas (at constant volume) do NOT shift equilibrium. The Haber process is the classic industrial application.

If stress applied equilibrium shifts to reduce the stress

Le Chatelier's Principle

A system at equilibrium responds to an external stress by shifting in the direction that partially undoes the stress. It never completely eliminates the stress — only partially counteracts it.

Le Chatelier's Principle (1884): If a change in concentration, pressure, or temperature is imposed on a system at equilibrium, the equilibrium shifts in the direction that tends to undo the effect of the change.

• The shift is always partial — the system doesn't fully reverse the change
• The principle predicts the direction of shift, not the magnitude
• It applies to both chemical equilibria and physical equilibria
• Three main stresses: change in concentration, pressure/volume, or temperature
• A catalyst is NOT a stress — it does not shift equilibrium

Le Chatelier's Principle is the single most tested equilibrium topic in NEET. Expect 1-2 questions every year.

Effect of Concentration

Adding a reactant or removing a product shifts equilibrium forward. Adding a product or removing a reactant shifts it backward. The value of K remains unchanged.

For A + B leftharpoons C + D :

• Add reactant (A or B) → Q < K → shifts forward → more products formed
• Remove product (C or D) → Q < K → shifts forward → more products formed
• Add product (C or D) → Q > K → shifts backward → more reactants formed
• Remove reactant (A or B) → Q > K → shifts backward → more products consumed
• K does NOT change — only the position of equilibrium changes

Effect of Pressure & Volume

Increasing pressure (decreasing volume) shifts equilibrium toward the side with fewer moles of gas. If Delta n_g = 0, pressure change has no effect.

Pressure changes affect gaseous equilibria only. The key factor is n_g (difference in moles of gaseous products and reactants).

• Increase pressure (or decrease volume) → shifts toward fewer moles of gas
• Decrease pressure (or increase volume) → shifts toward more moles of gas
• If n_g = 0 (same moles on both sides) → no effect of pressure change
• This applies only to changes in total pressure by compression/expansion — NOT by adding inert gas at constant volume

Effect of Temperature

Temperature is the only factor that changes the value of K. Increasing temperature favors the endothermic direction; decreasing temperature favors the exothermic direction.

Temperature is special — it's the only stress that actually changes the value of K . Treat heat as a reactant (endothermic) or product (exothermic) and apply Le Chatelier's.

• Exothermic reaction ( H < 0 ): heat is a 'product'. Increasing T → shifts backward → K decreases
• Endothermic reaction ( H > 0 ): heat is a 'reactant'. Increasing T → shifts forward → K increases
• Decreasing temperature always favors the exothermic direction
• This is the ONLY factor that changes the numerical value of K

Effect of Catalyst

A catalyst increases the rates of both forward and backward reactions equally. It helps reach equilibrium faster but does not change the equilibrium position or K.

A catalyst provides an alternative reaction pathway with lower activation energy. It speeds up BOTH the forward and backward reactions by the same factor.

• Catalyst does NOT change K
• Catalyst does NOT shift the equilibrium position
• Catalyst does NOT change G or H
• Catalyst only reduces the time to reach equilibrium
• Catalyst lowers activation energy for both forward AND backward reactions equally

Haber Process

The industrial synthesis of ammonia is the textbook application of Le Chatelier's Principle. Every condition is optimized to maximize NH₃ yield while maintaining practical reaction rates.

N_2(g) + 3H_2(g) leftharpoons 2NH_3(g) H = -92 kJ/mol

• High pressure (200 atm): n_g = 2-4 = -2 , so high P favors product side (fewer gas moles). Yield increases
• Moderate temperature (~700 K): Reaction is exothermic, so low T increases K . But too low = too slow. 700 K is a compromise
• Iron catalyst (Fe + Mo promoter): Speeds up the reaction at 700 K. Doesn't change K but makes the compromise temperature workable
• Remove NH₃ continuously: Removing product shifts equilibrium forward — more NH _3 produced

The Haber process is NCERT's favorite example. If NEET asks 'what conditions favor NH₃ production?' — answer: high pressure, moderate temperature, catalyst, continuous removal of NH₃.

Ionic Equilibrium & Acids-Bases

Ionic equilibrium deals with equilibria involving ions in aqueous solutions. Three definitions of acids and bases — Arrhenius, Bronsted-Lowry, Lewis — progressively broaden the concept. Strong electrolytes dissociate completely; weak electrolytes establish equilibrium characterized by K_a or K_b.

K_a = [H^+][A^-] [HA]

Arrhenius Theory

The simplest acid-base definition: acids produce H⁺ ions and bases produce OH⁻ ions in aqueous solution. Limited because it only works in water.

Arrhenius (1887): An acid is a substance that produces H^+ (or H_3O^+ ) in aqueous solution. A base is a substance that produces OH^- in aqueous solution.

• HCl → H ^+ + Cl ^- (Arrhenius acid)
• NaOH → Na ^+ + OH ^- (Arrhenius base)
• Limitation: Only works in aqueous solutions. Cannot explain why NH _3 is basic (no OH ^- in its formula), or why HCl + NH _3 react in the gas phase
• Cannot explain basicity of Na _2 CO _3 , NaHCO _3 (no OH ^- in formula)
• H ^+ doesn't exist freely in water — it's hydrated as H_3O^+ (hydronium ion)

Bronsted-Lowry Theory

A broader definition: acids donate protons (H⁺) and bases accept protons. This introduces conjugate acid-base pairs and works in any solvent, not just water.

Bronsted-Lowry (1923): An acid is a proton ( H^+ ) donor. A base is a proton acceptor. Every acid-base reaction is a proton transfer.

acid_1 HA + base_2 B leftharpoons acid_2 BH^+ + base_1 A^-

• Every acid has a conjugate base (acid minus one proton): HA → A ^-
• Every base has a conjugate acid (base plus one proton): B → BH ^+
• Conjugate pair = species that differ by one proton. E.g., HCl/Cl ^- , NH _3 /NH _4^+ , H _2 O/OH ^-
• Water is amphoteric — acts as acid (donates H ^+ to NH _3 ) or base (accepts H ^+ from HCl)
• Strong acid → weak conjugate base. Weak acid → strong conjugate base

Lewis Theory

The broadest definition: a Lewis acid accepts an electron pair, a Lewis base donates an electron pair. This extends acid-base chemistry beyond proton transfer to include BF₃, AlCl₃, metal ions, etc.

Lewis (1923): A Lewis acid is an electron pair acceptor. A Lewis base is an electron pair donor. The acid-base reaction forms a coordinate (dative) bond.

• Lewis acids: Species with empty orbitals or electron-deficient centers — BF _3 , AlCl _3 , FeCl _3 , H ^+ , metal cations (Cu ^ 2+ , Fe ^ 3+ )
• Lewis bases: Species with lone pairs — NH _3 , H _2 O, OH ^- , Cl ^- , ROH, amines
• Every Bronsted acid is a Lewis acid, but NOT vice versa (BF _3 is Lewis acid but doesn't donate H ^+ )
• Every Bronsted base is a Lewis base (they all have lone pairs to accept H ^+ )
• Lewis theory is the most general — it encompasses both Arrhenius and Bronsted-Lowry

Strong vs Weak Acids/Bases

Strong acids/bases dissociate completely in water (α ≈ 1). Weak acids/bases reach an ionization equilibrium (α << 1) characterized by K_a or K_b. Ostwald's dilution law connects K, α, and concentration.

Strong acids (HCl, HBr, HI, HNO _3 , H _2 SO _4 , HClO _4 ) and strong bases (NaOH, KOH, Ca(OH) _2 , Ba(OH) _2 ) dissociate completely in dilute aqueous solution. Weak acids and bases establish an ionization equilibrium.

CH_3COOH leftharpoons CH_3COO^- + H^+ K_a = [CH_3COO^-][H^+] [CH_3COOH]

NH_3 + H_2O leftharpoons NH_4^+ + OH^- K_b = [NH_4^+][OH^-] [NH_3]

• Degree of ionization ( ) = fraction of total acid/base that ionizes. Strong: 1 . Weak: 1
• Ostwald's Dilution Law: K_a = C ^2 1- C ^2 (when 1 ), so = K_a/C
• Dilution increases (more ionization) but decreases [H^+] — don't confuse the two!
• Larger K_a → stronger weak acid. Larger K_b → stronger weak base
• For a conjugate pair: K_a K_b = K_w = 10^ -14 at 25°C

pH Scale

The pH scale is a logarithmic measure of H⁺ concentration: pH = -log[H⁺]. Water auto-ionizes with K_w = [H⁺][OH⁻] = 10⁻¹⁴ at 25°C. pH + pOH = 14. For weak acids/bases, pH calculations require K_a/K_b and Ostwald's dilution law.

pH = - [H^+]

Auto-ionization of Water

Water molecules spontaneously transfer protons to each other, establishing an equilibrium that defines K_w, the ionic product of water. At 25°C, K_w = 10⁻¹⁴.

H_2O(l) + H_2O(l) leftharpoons H_3O^+(aq) + OH^-(aq)

K_w = [H^+][OH^-] = 10^ -14 at 25°C (298 K)

In pure water: [H^+] = [OH^-] = K_w = 10^ -7 M. This means even 'pure' water has a tiny but measurable ion concentration.

• K_w is called the ionic product of water (or auto-protolysis constant)
• K_w depends on temperature: at 25°C, K_w = 10^ -14 ; at higher T , K_w increases (auto-ionization is endothermic)
• In pure water at 25°C: [H^+] = [OH^-] = 10^ -7 M → pH = 7 (neutral)
• At temperatures above 25°C, neutral pH < 7 (because K_w > 10^ -14 , so [H^+] > 10^ -7 )
• K_w = K_a K_b for any conjugate acid-base pair

pH & pOH

pH and pOH convert unwieldy H⁺ and OH⁻ concentrations into simple numbers. The relationship pH + pOH = 14 (at 25°C) is a direct consequence of K_w = 10⁻¹⁴.

pH = - [H^+] pOH = - [OH^-]

pH + pOH = pK_w = 14 (at 25°C)

• Acidic solution: pH < 7 , [H^+] > 10^ -7 M
• Neutral solution: pH = 7 , [H^+] = 10^ -7 M (at 25°C)
• Basic solution: pH > 7 , [H^+] < 10^ -7 M
• pH scale typically ranges from 0 to 14, but can go beyond (e.g., 10 M HCl has pH -1 )
• Each unit change in pH = 10-fold change in [H^+] (logarithmic scale)

pH of Strong Acids & Bases

For strong acids and bases, dissociation is complete. pH is calculated directly from the concentration without needing K_a or K_b.

Strong acids and bases dissociate completely. So [H^+] equals the acid concentration (for monobasic acids), and [OH^-] equals the base concentration (for monoacidic bases).

• Strong monobasic acid (HCl, HNO _3 ): [H^+] = C → pH = - C
• Strong dibasic acid (H _2 SO _4 ): [H^+] = 2C → pH = - (2C)
• Strong monoacidic base (NaOH, KOH): [OH^-] = C → pOH = - C → pH = 14 - pOH
• Strong diacidic base (Ca(OH) _2 , Ba(OH) _2 ): [OH^-] = 2C → pOH = - (2C)

pH of Weak Acids & Bases

For weak acids/bases, only a fraction ionizes. Use K_a (or K_b) and concentration to find [H⁺] (or [OH⁻]), then compute pH.

For a weak acid HA with concentration C and ionization constant K_a :

[H^+] = K_a C (when 1 , i.e., K_a C)

pH = 1 2 (pK_a - C) = 1 2 pK_a - 1 2 C

For a weak base B with concentration C and ionization constant K_b :

[OH^-] = K_b C pOH = 1 2 (pK_b - C)

pH = 14 - pOH = 14 - 1 2 (pK_b - C)

• The approximation [H^+] = K_a C is valid when < 0.05 (less than 5% ionization)
• If is not negligible, solve the full quadratic: K_a = x^2 C-x where x = [H^+]
• pK_a = - K_a and pK_b = - K_b
• Relationship: pK_a + pK_b = pK_w = 14 (at 25°C)
• Lower pK_a → stronger acid. Lower pK_b → stronger base

Buffer Solutions

Buffer solutions resist changes in pH upon addition of small amounts of acid or base. Acidic buffers (weak acid + conjugate base salt) and basic buffers (weak base + conjugate acid salt) are described by the Henderson-Hasselbalch equation. Buffer capacity depends on total concentration and the ratio of components.

pH = pK_a + [Salt] [Acid]

Buffer Solutions

A buffer solution maintains nearly constant pH when small amounts of acid or base are added. It consists of a weak acid/base and its conjugate salt.

A buffer solution is a solution that resists significant pH change upon addition of small amounts of strong acid or strong base. It works by having a reservoir of both a weak acid (to neutralize added base) and its conjugate base (to neutralize added acid).

• Acidic buffer: Weak acid + salt of its conjugate base (e.g., CH _3 COOH + CH _3 COONa). pH < 7
• Basic buffer: Weak base + salt of its conjugate acid (e.g., NH _4 OH + NH _4 Cl). pH > 7
• Buffer does NOT prevent pH change entirely — it minimizes the change
• Buffer fails if the added acid/base exceeds the buffer's capacity (one component gets used up)

Henderson-Hasselbalch Equation

This equation directly gives the pH of a buffer solution from pK_a (or pK_b) and the ratio of salt to acid (or salt to base). It's the single most useful equation for buffer problems.

For an acidic buffer (weak acid HA + salt NaA):

pH = pK_a + [A^-] [HA] = pK_a + [Salt] [Acid]

For a basic buffer (weak base B + salt BHCl):

pOH = pK_b + [BH^+] [B] = pK_b + [Salt] [Base]

• When [Salt] = [Acid]: (1) = 0 → pH = p K_a (buffer pH equals p K_a of the acid)
• When [Salt] > [Acid]: > 0 → pH > p K_a
• When [Salt] < [Acid]: < 0 → pH < p K_a
• Buffer works best when [Salt]/[Acid] is between 0.1 and 10 (pH within 1 of p K_a )
• To prepare buffer at a target pH, choose an acid with p K_a close to the target pH

Buffer Capacity

Buffer capacity is the amount of acid or base a buffer can neutralize before its pH changes significantly. It depends on total concentration and the ratio of components.

Buffer capacity ( ) measures the effectiveness of a buffer — how many moles of strong acid or base can be added per liter before pH changes by one unit.

• Buffer capacity is maximum when [Acid] = [Salt] (i.e., pH = p K_a )
• Higher total concentration = higher buffer capacity (1 M buffer resists more than 0.01 M buffer)
• Buffer capacity decreases as the ratio [Salt]/[Acid] deviates from 1
• Effective buffer range: p K_a 1 (pH range where buffer works well)
• Buffer is overwhelmed when all of one component is consumed

Solubility Equilibrium

Sparingly soluble salts establish a solubility equilibrium characterized by K_sp (solubility product). Molar solubility can be calculated from K_sp and vice versa. The common ion effect reduces solubility. Precipitation occurs when Q (ionic product) exceeds K_sp.

K_ sp = [A^+]^m[B^-]^n for A_mB_n

K_ sp Expression

K_sp is the equilibrium constant for the dissolution of a sparingly soluble salt. It represents the maximum product of ion concentrations that can exist in solution.

For a sparingly soluble salt A_mB_n dissolving in water:

A_mB_n(s) leftharpoons mA^ n+ (aq) + nB^ m- (aq)

K_ sp = [A^ n+ ]^m [B^ m- ]^n

The solid doesn't appear in the expression (activity = 1). K_ sp depends only on temperature.

• AgCl(s) leftharpoons Ag ^+ + Cl ^- : K_ sp = [Ag^+][Cl^-]
• PbCl _2 (s) leftharpoons Pb ^ 2+ + 2Cl ^- : K_ sp = [Pb^ 2+ ][Cl^-]^2
• Ca _3 (PO _4 ) _2 (s) leftharpoons 3Ca ^ 2+ + 2PO _4^ 3- : K_ sp = [Ca^ 2+ ]^3[PO_4^ 3- ]^2
• Larger K_ sp → more soluble (for salts with the same ion ratio)
• K_ sp is valid ONLY for sparingly soluble salts (not for NaCl, KNO _3 etc.)

Molar Solubility

Molar solubility (s) is the number of moles of salt that dissolve per liter to form a saturated solution. It's calculated from K_sp using stoichiometry.

If the molar solubility of A_mB_n is s mol/L, then [A^ n+ ] = ms and [B^ m- ] = ns .

K_ sp = (ms)^m (ns)^n = m^m n^n s^ (m+n)

• AB type (AgCl, BaSO _4 ): K_ sp = s^2 → s = K_ sp
• AB _2 type (PbCl _2 , CaF _2 ): K_ sp = 4s^3 → s = ( K_ sp 4 )^ 1/3
• A _2 B type (Ag _2 CrO _4 ): K_ sp = 4s^3 → s = ( K_ sp 4 )^ 1/3
• A _2 B _3 type (Bi _2 S _3 ): K_ sp = 108s^5 → s = ( K_ sp 108 )^ 1/5
• Solubility in g/L = s M (where M = molar mass)

Common Ion Effect

Adding a common ion to a solution of a sparingly soluble salt decreases its solubility. This is a direct application of Le Chatelier's Principle to solubility equilibrium.

The common ion effect: Adding an ion that is already present in the equilibrium shifts the dissolution equilibrium backward, decreasing the solubility of the salt.

• Common ion effect decreases solubility of the sparingly soluble salt
• K_ sp remains unchanged — only the solubility ( s ) changes
• In the presence of a common ion at concentration C : set up the K_ sp expression with the extra ion concentration included
• For AgCl in 0.1 M NaCl: K_ sp = [Ag^+](0.1 + s) [Ag^+] 0.1 (since s 0.1 )

Precipitation Condition

Compare the ionic product Q with K_sp: if Q > K_sp, precipitation occurs. If Q < K_sp, more salt can dissolve. If Q = K_sp, the solution is exactly saturated.

The ionic product (IP or Q ) has the same form as K_ sp but uses actual (non-equilibrium) ion concentrations. Comparing Q with K_ sp tells you whether precipitation will occur.

• Q < K_ sp : Solution is unsaturated → no precipitate, more salt can dissolve
• Q = K_ sp : Solution is saturated → at equilibrium, no net change
• Q > K_ sp : Solution is supersaturated → precipitate forms until Q = K_ sp

Hydrolysis of Salts

When salts dissolve in water, the ions from the weak component undergo hydrolysis (react with water), making the solution acidic or basic. Salt of strong acid + weak base → acidic. Salt of weak acid + strong base → basic. Salt of weak acid + weak base → depends on K_a vs K_b.

pH = 7 + 1 2 pK_a - 1 2 pK_b (WA + WB salt)

Salt Hydrolysis

Hydrolysis is the reverse of neutralization. The ion from the weak component reacts with water, shifting the [H⁺]/[OH⁻] balance and making the solution acidic or basic.

Hydrolysis = the reaction of a salt's ions with water. Only the ion from the weak component undergoes hydrolysis. The ion from the strong component is a spectator.

• Strong acid + Strong base salt (NaCl, KNO _3 ): No hydrolysis → pH = 7 (neutral)
• Strong acid + Weak base salt (NH _4 Cl, FeCl _3 ): Cation hydrolyzes → pH < 7 (acidic)
• Weak acid + Strong base salt (CH _3 COONa, Na _2 CO _3 ): Anion hydrolyzes → pH > 7 (basic)
• Weak acid + Weak base salt (CH _3 COONH _4 ): Both ions hydrolyze → pH depends on relative K_a and K_b

Hydrolysis is really just Le Chatelier applied to the auto-ionization of water. Adding CH _3 COO ^- removes H ^+ (by forming CH _3 COOH) → water ionizes more → [OH ^- ] increases → solution becomes basic.

Strong Acid + Weak Base Salt

In salts like NH₄Cl, the cation (NH₄⁺) is the conjugate acid of a weak base. It donates H⁺ to water, making the solution acidic.

NH_4^+(aq) + H_2O(l) leftharpoons NH_3(aq) + H_3O^+(aq)

K_h = K_w K_b pH = 7 - 1 2 pK_b - 1 2 C

• The cation (from weak base) hydrolyzes, producing H _3 O ^+ → acidic solution
• The anion (from strong acid, e.g., Cl ^- ) does NOT hydrolyze
• K_h = K_w / K_b where K_b is the ionization constant of the parent weak base
• Degree of hydrolysis h = K_h/C = K_w/(K_b C)
• Higher C → lower h (less hydrolysis) but lower pH (more acidic)

Weak Acid + Strong Base Salt

In salts like CH₃COONa, the anion (CH₃COO⁻) is the conjugate base of a weak acid. It accepts H⁺ from water, making the solution basic.

CH_3COO^-(aq) + H_2O(l) leftharpoons CH_3COOH(aq) + OH^-(aq)

K_h = K_w K_a pH = 7 + 1 2 pK_a - 1 2 C

• The anion (from weak acid) hydrolyzes, producing OH ^- → basic solution
• The cation (from strong base, e.g., Na ^+ ) does NOT hydrolyze
• K_h = K_w / K_a where K_a is the ionization constant of the parent weak acid
• Degree of hydrolysis h = K_h/C = K_w/(K_a C)
• Higher C → lower h but higher pH (more basic)

Weak Acid + Weak Base Salt

In salts like CH₃COONH₄, both the cation and anion hydrolyze. The pH depends on the relative strengths of the parent acid and base (K_a vs K_b).

pH = 7 + 1 2 pK_a - 1 2 pK_b

• Both cation AND anion hydrolyze simultaneously
• If K_a > K_b (p K_a < p K_b ): pH < 7 → acidic solution
• If K_a < K_b (p K_a > p K_b ): pH > 7 → basic solution
• If K_a = K_b : pH = 7 → neutral solution
• pH is independent of concentration C — unique to WA + WB salts