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ChemistryPhysical ChemistryTop Priority (3 Qs in JEE)

Chemical & Ionic Equilibrium Formulas

K_p & K_c relations, pH calculations of weak acids/bases, buffer solutions, Henderson equation, and solubility product K_sp.

8 Formulas•486+ Questions in Bank

Core Formulas & Mathematical Identities

8
01

Kp and Kc Relation

Kp=Kc(RT)Δng,Δng=∑ngas (prod)−∑ngas (react)K_p = K_c (RT)^{\Delta n_g}, \quad \Delta n_g = \sum n_{\text{gas (prod)}} - \sum n_{\text{gas (react)}}

Connects partial pressure equilibrium constant K_p and molarity constant K_c.

Examiner Tip: If Δn_g = 0, K_p = K_c and equilibrium is independent of total pressure.
02

Ostwald's Dilution Law & pH of Weak Acid

α=KaC,[H+]=KaC=Cα,pH=12[pKa−log⁡10C]\alpha = \sqrt{\frac{K_a}{C}}, \quad [H^+] = \sqrt{K_a C} = C\alpha, \quad \text{pH} = \frac{1}{2}[\text{p}K_a - \log_{10} C]

Degree of dissociation and hydronium concentration for weak monoprotic acid.

Examiner Tip: Valid only when degree of dissociation α ≤ 0.05 (5%).
03

Henderson-Hasselbalch Equation (Buffer Solutions)

pH=pKa+log⁡10([Salt][Acid]),pOH=pKb+log⁡10([Salt][Base])\text{pH} = \text{p}K_a + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right), \quad \text{pOH} = \text{p}K_b + \log_{10}\left(\frac{[\text{Salt}]}{[\text{Base}]}\right)

Calculates pH of acidic and basic buffer solutions resistent to pH change.

Examiner Tip: Maximum buffer capacity occurs when [Salt] = [Acid], giving pH = pKa.
04

Salt Hydrolysis pH Formulas

Weak Acid + Strong Base: pH=7+12(pKa+log⁡10C),Weak Acid + Weak Base: pH=7+12(pKa−pKb)\text{Weak Acid + Strong Base: } \text{pH} = 7 + \frac{1}{2}(\text{p}K_a + \log_{10} C), \quad \text{Weak Acid + Weak Base: } \text{pH} = 7 + \frac{1}{2}(\text{p}K_a - \text{p}K_b)

Hydrolysis of salts producing alkaline or acidic aqueous solutions.

Examiner Tip: Salt of weak acid and weak base pH is independent of concentration C.
05

Solubility Product (K_sp) & Common Ion Effect

AxBy(s)⇌xAy++yBx−  ⟹  Ksp=xxyySx+y,S′=Ksp[Common Ion]yA_x B_y (s) \rightleftharpoons x A^{y+} + y B^{x-} \implies K_{\text{sp}} = x^x y^y S^{x+y}, \quad S' = \frac{K_{sp}}{[\text{Common Ion}]^y}

Equilibrium constant for sparingly soluble salts, and suppression of solubility in presence of shared ions.

Examiner Tip: Common ion effect drastically decreases solubility S without altering K_sp.
06

Strong Acid + Weak Base Salt Hydrolysis

pH=7−12(pKb+log⁡10C),Kh=KwKb,h=KwKbC\text{pH} = 7 - \frac{1}{2}(\text{p}K_b + \log_{10} C), \quad K_h = \frac{K_w}{K_b}, \quad h = \sqrt{\frac{K_w}{K_b C}}

Cationic hydrolysis producing an acidic solution (e.g. NH₄Cl).

Examiner Tip: Degree of hydrolysis h increases upon dilution.
07

Buffer Capacity & Effective Buffer Range

β=dBdpH=2.303CacidCsaltCacid+Csalt,Effective Range: pH=pKa±1\beta = \frac{d B}{d\text{pH}} = 2.303 \frac{C_{\text{acid}} C_{\text{salt}}}{C_{\text{acid}} + C_{\text{salt}}}, \quad \text{Effective Range: } \text{pH} = \text{p}K_a \pm 1

Quantitative measure of buffer resistance to pH change upon addition of strong acid/base.

Examiner Tip: Maximum buffer capacity occurs at pH = pKa (equimolar ratio of conjugate acid-base pair).
08

Van 't Hoff Isochore & Temperature Dependence

ln⁡(K2K1)=ΔH∘R(1T1−1T2),Inert gas at const. V:Δng no shift\ln\left(\frac{K_2}{K_1}\right) = \frac{\Delta H^\circ}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right), \quad \text{Inert gas at const. } V: \Delta n_g \text{ no shift}

Variation of equilibrium constant K with absolute temperature T governed by standard enthalpy of reaction ΔH°.

Examiner Tip: Endothermic reactions (ΔH° > 0) have higher K at higher temperatures; exothermic reactions have lower K.

Top Examiner Traps & Pitfalls

Frequently tested misconceptions in Chemical & Ionic Equilibrium

  • •For very dilute strong acids (e.g. 10⁻⁸ M HCl), pH is NOT 8! Must include water's [H⁺] (pH ≈ 6.96).
  • •Solubility product precipitation condition: precipitation occurs iff Ionic Product (Q_sp) > K_sp.