Direct answer

Acids and bases, pH, weak acid dissociation, salt hydrolysis, buffers, solubility product and common ion effect.

Acids and bases, pH, weak acid dissociation, salt hydrolysis, buffers, solubility product and common ion effect.

Key ideas

Ionic Equilibrium (pH, Buffers, Ksp): what you must understand

  • pH + pOH = 14 at 25 °C
  • Buffers resist pH change; Henderson equation gives their pH
  • Common ion lowers solubility
  • Precipitation happens when ionic product > Ksp
Formula sheet

Must-know formulas for Ionic Equilibrium (pH, Buffers, Ksp)

  • Weak acid: [H⁺] = √(Ka C)
  • Henderson: pH = pKa + log([salt]/[acid])
  • Ksp for AB₂: 4s³
  • Salt of weak acid + strong base: pH = 7 + ½(pKa + log C)
Print

One-page formula sheet

Printable Ionic Equilibrium (pH, Buffers, Ksp) formula sheet: formulas, key ideas and traps on one page. Save it as a PDF for offline revision.

Avoid

Common mistakes students make

  • Using the strong-acid formula for weak acids
  • Forgetting dilution when two solutions are mixed
  • Wrong Ksp expression for the formula
For the exam

How this topic is asked in JEE

Appears frequently. Buffer and Ksp numericals appear regularly. Learn the standard pH formulas as a table.

Prerequisites: Chemical Equilibrium.

NCERT map

Where Ionic Equilibrium (pH, Buffers, Ksp) is in NCERT

Chapter names follow the current rationalised NCERT textbooks. Read these sections before solving PYQs.

ClassBookChapterSections JEE asks from
Class 11Chemistry Part IEquilibriumIonization of Acids and Bases; Buffer Solutions; Solubility Equilibria of Sparingly Soluble Salts; Common Ion Effect; Ionization of Water and pH Scale
JEE Advanced level

A JEE Advanced-level example on Ionic Equilibrium (pH, Buffers, Ksp)

Question. A buffer solution is prepared by mixing 100 mL of 0.2 M CH₃COOH with 100 mL of 0.1 M NaOH. Solid AgCH₃COO is added to this buffer until saturation. Given Ka of CH₃COOH = 2 × 10⁻⁵ and Ksp of AgCH₃COO = 2 × 10⁻³, find the concentration of Ag⁺ ions in the saturated solution.

Show solution
  1. Moles of CH₃COOH = 0.02 mol, moles of NaOH = 0.01 mol. NaOH neutralizes equal moles of acid, so after reaction: CH₃COOH = 0.01 mol, CH₃COO⁻ = 0.01 mol in 200 mL total volume.
  2. Concentration of acetate ion in buffer: [CH₃COO⁻] = 0.01 mol / 0.2 L = 0.05 M. This acts as the common ion.
  3. When AgCH₃COO dissolves: AgCH₃COO ⇌ Ag⁺ + CH₃COO⁻. At saturation, Ksp = [Ag⁺][CH₃COO⁻].
  4. Apply common ion effect. The buffer maintains [CH₃COO⁻] ≈ 0.05 M since Ag⁺ dissolved is small compared to buffer capacity.
  5. [Ag⁺] = Ksp / [CH₃COO⁻] = (2 × 10⁻³) / (0.05) = 4 × 10⁻² M

Answer: [Ag⁺] = 4 × 10⁻² M (or 0.04 M)

This example was solved twice independently and both solutions agree.

Free lectures

Best free lectures for Ionic Equilibrium (pH, Buffers, Ksp)

Full-chapter free lectures picked from YouTube search results by views, from established JEE channels (mostly in Hinglish). Use the subtopic list to jump straight to the part you need (timestamps come from each video's own chapter list). Watch one, then solve PYQs the same day.

FAQs

Frequently asked questions

Is Ionic Equilibrium (pH, Buffers, Ksp) important for JEE Main 2027?

Appears frequently. It is part of the official 'Equilibrium' unit in the JEE Main 2027 syllabus.

What are the most important formulas in Ionic Equilibrium (pH, Buffers, Ksp)?

Weak acid: [H⁺] = √(Ka C); Henderson: pH = pKa + log([salt]/[acid]); Ksp for AB₂: 4s³.

How should I study Ionic Equilibrium (pH, Buffers, Ksp) for JEE?

Understand the key ideas from NCERT or one full lecture, learn the formula sheet, then solve 25–30 past JEE questions on the topic. Buffer and Ksp numericals appear regularly. Learn the standard pH formulas as a table.

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