Chemistry·Revision Notes

Ionic Equilibrium in Solution — Revision Notes

NEET UG
Version 1Updated 22 Mar 2026

⚡ 30-Second Revision

  • ElectrolytesStrong (α1\alpha \approx 1), Weak (0<α<10 < \alpha < 1).
  • Ostwald's Dilution Lawα=Ka/C\alpha = \sqrt{K_a/C} (for weak acid).
  • Ionic Product of WaterKw=[H+][OH]=1.0×1014K_w = [H^+][OH^-] = 1.0 \times 10^{-14} at 25circC25^circ C.
  • pH ScalepH=log[H+]pH = -log[H^+], pOH=log[OH]pOH = -log[OH^-], pH+pOH=14pH + pOH = 14.
  • Acid/Base StrengthKaK_a (acid), KbK_b (base). Larger KaK_a/KbK_b = stronger. pKa=logKapK_a = -logK_a, pKb=logKbpK_b = -logK_b. Smaller pKapK_a/pKbpK_b = stronger.
  • Conjugate PairKa×Kb=KwK_a \times K_b = K_w or pKa+pKb=pKwpK_a + pK_b = pK_w.
  • Salt Hydrolysis

- SA+SB: Neutral - SA+WB: Acidic (Kh=Kw/KbK_h = K_w/K_b) - WA+SB: Basic (Kh=Kw/KaK_h = K_w/K_a) - WA+WB: pH depends on KaK_a vs KbK_b.

  • Buffer SolutionsWeak acid/conjugate base OR Weak base/conjugate acid.
  • Henderson-HasselbalchpH=pKa+log10[Salt][Acid]pH = pK_a + log_{10}\frac{[Salt]}{[Acid]} (acidic buffer).
  • Solubility ProductFor AxByA_xB_y, Ksp=[Ay+]x[Bx]yK_{sp} = [A^{y+}]^x[B^{x-}]^y.
  • Common Ion EffectDecreases solubility of sparingly soluble salt.
  • PrecipitationOccurs if Qsp>KspQ_{sp} > K_{sp}.

2-Minute Revision

Ionic equilibrium describes the dynamic balance of ions and undissociated molecules in electrolyte solutions. Strong electrolytes dissociate completely, while weak electrolytes establish equilibrium, quantified by the degree of dissociation (alphaalpha) and governed by Ostwald's Dilution Law.

Water's autoionization gives rise to the ionic product (Kw=1014K_w = 10^{-14} at 25circC25^circ C) and the pH scale (pH=log[H+]pH = -log[H^+]), where pH+pOH=14pH+pOH=14. Acid and base strengths are measured by KaK_a and KbK_b (or pKapK_a and pKbpK_b), with the crucial relationship Ka×Kb=KwK_a \times K_b = K_w for conjugate pairs.

Salts can hydrolyze, affecting solution pH depending on the strengths of their parent acid and base. Buffer solutions, mixtures of weak acid/base and their conjugates, resist pH changes, with their pH calculated using the Henderson-Hasselbalch equation.

Finally, sparingly soluble salts establish solubility equilibrium, characterized by the solubility product (KspK_{sp}), which is used to calculate molar solubility (SS) and predict precipitation. The common ion effect reduces the solubility of such salts.

5-Minute Revision

Ionic equilibrium is the study of reactions involving ions in solution, specifically the dynamic balance between dissociated ions and undissociated molecules. This is particularly relevant for weak electrolytes (weak acids, weak bases), which only partially ionize. The **degree of dissociation (alphaalpha) quantifies this extent, and Ostwald's Dilution Law** states that α\alpha increases with dilution.

Water's autoionization (H2OH++OHH_2O \rightleftharpoons H^+ + OH^-) is fundamental, yielding the **ionic product of water (Kw=[H+][OH]K_w = [H^+][OH^-])**, which is 1.0×10141.0 \times 10^{-14} at 25circC25^circ C. This forms the basis of the pH scale (pH=log[H+]pH = -log[H^+]), where pH+pOH=14pH+pOH=14. Solutions are acidic (pH<7pH<7), neutral (pH=7pH=7), or basic (pH>7pH>7) at 25circC25^circ C.

Acids and bases are defined by Arrhenius (H+/OH- producers), Brønsted-Lowry (proton donors/acceptors), and Lewis (electron pair acceptors/donors). Their strengths are quantified by **acid dissociation constant (KaK_a) and base dissociation constant (KbK_b)**. A larger KaK_a means a stronger acid, and a larger KbK_b means a stronger base. For a conjugate acid-base pair, a critical relationship exists: Ka×Kb=KwK_a \times K_b = K_w, or in logarithmic form, pKa+pKb=pKw=14pK_a + pK_b = pK_w = 14.

Salt hydrolysis occurs when ions from a salt react with water, altering the solution's pH. Salts of strong acid-strong base are neutral. Salts of strong acid-weak base are acidic (e.g., NH4ClNH_4Cl). Salts of weak acid-strong base are basic (e.g., CH3COONaCH_3COONa). Salts of weak acid-weak base have a pH determined by the relative strengths of the acid and base (pH=7+12pKa12pKbpH = 7 + \frac{1}{2}pK_a - \frac{1}{2}pK_b).

Buffer solutions resist pH changes upon addition of small amounts of acid or base. They consist of a weak acid and its conjugate base (acidic buffer) or a weak base and its conjugate acid (basic buffer). Their pH can be calculated using the Henderson-Hasselbalch equation: pH=pKa+log10[Salt][Acid]pH = pK_a + log_{10}\frac{[Salt]}{[Acid]} for acidic buffers.

Solubility equilibrium deals with sparingly soluble salts. For a salt AxByA_xB_y, the **solubility product constant (KspK_{sp})** is Ksp=[Ay+]x[Bx]yK_{sp} = [A^{y+}]^x[B^{x-}]^y. This constant relates to the **molar solubility (SS)** of the salt (e.g., Ksp=S2K_{sp}=S^2 for ABAB, Ksp=4S3K_{sp}=4S^3 for AB2AB_2). The common ion effect reduces the solubility of a sparingly soluble salt when a common ion is added. Precipitation occurs if the **ionic product (QspQ_{sp})** exceeds KspK_{sp}.

Worked Example: Calculate the pH of a 0.01,M0.01,\text{M} NH4OHNH_4OH solution if Kb=1.8×105K_b = 1.8 \times 10^{-5}.

    1
  1. NH4OHNH4++OHNH_4OH \rightleftharpoons NH_4^+ + OH^-
  2. 2
  3. Initial: 0.010.01, 00, 00.
  4. 3
  5. Equilibrium: 0.01x0.01-x, xx, xx.
  6. 4
  7. Kb=x20.01xx20.01=1.8×105K_b = \frac{x^2}{0.01-x} \approx \frac{x^2}{0.01} = 1.8 \times 10^{-5}.
  8. 5
  9. x2=1.8×107    x=[OH]=1.8×107=4.24×104,Mx^2 = 1.8 \times 10^{-7} \implies x = [OH^-] = \sqrt{1.8 \times 10^{-7}} = 4.24 \times 10^{-4},\text{M}.
  10. 6
  11. pOH=log(4.24×104)=4log(4.24)=40.627=3.373pOH = -log(4.24 \times 10^{-4}) = 4 - log(4.24) = 4 - 0.627 = 3.373.
  12. 7
  13. pH=14pOH=143.373=10.627pH = 14 - pOH = 14 - 3.373 = 10.627.

Prelims Revision Notes

    1
  1. ElectrolytesSubstances producing ions in solution. Strong (complete dissociation, e.g., HCl, NaOH, NaCl). Weak (partial dissociation, equilibrium, e.g., CH3COOHCH_3COOH, NH4OHNH_4OH).
  2. 2
  3. Degree of Dissociation ($alpha$)Fraction dissociated. For weak electrolytes, α=K/C\alpha = \sqrt{K/C} (Ostwald's Dilution Law). α\alpha increases with dilution.
  4. 3
  5. Ionic Product of Water ($K_w$)H2OH++OHH_2O \rightleftharpoons H^+ + OH^-. Kw=[H+][OH]K_w = [H^+][OH^-]. At 25circC25^circ C, Kw=1.0×1014K_w = 1.0 \times 10^{-14}. KwK_w increases with temperature.
  6. 4
  7. pH ScalepH=log[H+]pH = -log[H^+], pOH=log[OH]pOH = -log[OH^-]. pH+pOH=14pH + pOH = 14 at 25circC25^circ C. Acidic (pH<7pH<7), Neutral (pH=7pH=7), Basic (pH>7pH>7).
  8. 5
  9. Acid-Base Theories

* Arrhenius: Acid (H+H^+ producer), Base (OHOH^- producer). * Brønsted-Lowry: Acid (proton donor), Base (proton acceptor). Conjugate acid-base pairs (e.g., HCl/ClHCl/Cl^-). * Lewis: Acid (electron pair acceptor), Base (electron pair donor).

    1
  1. Acid/Base Dissociation ConstantsKa=[H+][A][HA]K_a = \frac{[H^+][A^-]}{[HA]}, Kb=[B+][OH][BOH]K_b = \frac{[B^+][OH^-]}{[BOH]}. pKa=logKapK_a = -logK_a, pKb=logKbpK_b = -logK_b. Stronger acid = larger KaK_a (smaller pKapK_a). Stronger base = larger KbK_b (smaller pKbpK_b).
  2. 2
  3. Conjugate Pair RelationshipFor a conjugate acid-base pair, Ka×Kb=KwK_a \times K_b = K_w or pKa+pKb=pKw=14pK_a + pK_b = pK_w = 14.
  4. 3
  5. Salt HydrolysisReaction of salt ions with water to produce acidity or basicity.

* SA+SB (e.g., NaCl): No hydrolysis, pH=7pH=7. * **SA+WB (e.g., NH4ClNH_4Cl)**: Cation hydrolysis (NH4++H2ONH3+H3O+NH_4^+ + H_2O \rightleftharpoons NH_3 + H_3O^+), acidic solution, pH<7pH < 7. Kh=Kw/KbK_h = K_w/K_b. * **WA+SB (e.g., CH3COONaCH_3COONa)**: Anion hydrolysis (CH3COO+H2OCH3COOH+OHCH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-), basic solution, pH>7pH > 7. Kh=Kw/KaK_h = K_w/K_a. * **WA+WB (e.g., CH3COONH4CH_3COONH_4)**: Both hydrolyze. pH=7+12pKa12pKbpH = 7 + \frac{1}{2}pK_a - \frac{1}{2}pK_b.

    1
  1. Buffer SolutionsResist pH change.

* Acidic Buffer: Weak acid + its conjugate base (e.g., CH3COOH/CH3COONaCH_3COOH/CH_3COONa). pH=pKa+log10[Salt][Acid]pH = pK_a + log_{10}\frac{[Salt]}{[Acid]}. * Basic Buffer: Weak base + its conjugate acid (e.g., NH4OH/NH4ClNH_4OH/NH_4Cl). pOH=pKb+log10[Salt][Base]pOH = pK_b + log_{10}\frac{[Salt]}{[Base]}.

    1
  1. Solubility Product ($K_{sp}$)For sparingly soluble salt AxBy(s)xAy+(aq)+yBx(aq)A_xB_y(s) \rightleftharpoons xA^{y+}(aq) + yB^{x-}(aq), Ksp=[Ay+]x[Bx]yK_{sp} = [A^{y+}]^x[B^{x-}]^y.

* **Molar Solubility (SS)**: For ABAB, Ksp=S2K_{sp}=S^2. For AB2AB_2, Ksp=4S3K_{sp}=4S^3. For A2B3A_2B_3, Ksp=108S5K_{sp}=108S^5.

    1
  1. Common Ion EffectDecreases solubility of a sparingly soluble salt by adding a common ion.
  2. 2
  3. Precipitation ConditionCompare Ionic Product (QspQ_{sp}) with KspK_{sp}. If Qsp>KspQ_{sp} > K_{sp}, precipitation occurs.

Vyyuha Quick Recall

Please Help All Boys Solve Buffer Solubility Equations:

  • PH: pH=log[H+]pH = -log[H^+]
  • Henderson-Hasselbalch: pH=pKa+log[Salt][Acid]pH = pK_a + log\frac{[Salt]}{[Acid]}
  • Acid-Base Conjugates: Ka×Kb=KwK_a \times K_b = K_w
  • Buffer Action: Resists pH changes
  • Salt Hydrolysis: Determines pH of salt solutions
  • Basic/Acidic: pH<7pH<7 (acidic), pH>7pH>7 (basic)
  • Solubility Product: KspK_{sp} for sparingly soluble salts
  • Equilibrium: Dynamic balance in weak electrolytes
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