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JEE Mains Chemistry · Equilibrium

pH of Acids and Bases

Finding pH for strong acids and bases, their mixtures and dilutions, and for weak acids and bases from Ka or Kb, including acids that give up two protons.

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Why this matters

Twenty PYQs, half of them numerical, and five from 2026. Thirteen deal with strong acids and bases, their mixtures and their dilution; seven use Ka or Kb of a weak acid or base, including acids that lose two protons.

Concept 1 of 2: Strong acids, bases and their mixtures

A strong acid or base ionises completely, so [H+][\mathrm{H^+}] or [OH−][\mathrm{OH^-}] comes straight from the concentration and the number of H or OH per formula. In a mixture, H+\mathrm{H^+} and OH−\mathrm{OH^-} cancel mole for mole. Whatever is left over, divided by the total volume, sets the pH.

Definition

  • pH=−log⁡[H+]\mathrm{pH}=-\log[\mathrm{H^+}], pOH=−log⁡[OH−]\mathrm{pOH}=-\log[\mathrm{OH^-}], pH+pOH=14\mathrm{pH+pOH}=14 at 25 °C.
  • Count per formula: H2SO4\mathrm{H_2SO_4} gives 2 H+\mathrm{H^+}; Ca(OH)2\mathrm{Ca(OH)_2} and Ba(OH)2\mathrm{Ba(OH)_2} give 2 OH−\mathrm{OH^-}.
  • Mixture: millimoles of H+\mathrm{H^+} minus millimoles of OH−\mathrm{OH^-}, divided by the total volume.
  • Diluting a strong acid n times raises its pH by log⁡n\log n. A change of [H+][\mathrm{H^+}] by a factor of 1000 moves the pH by 3.
  • A very dilute acid never crosses 7: for 10−810^{-8} M HCl, water's own 10−710^{-7} M dominates and the pH is about 6.98.
  • KwK_w rises with temperature, so hot pure water has pH below 7 but is still neutral, with [H+]=[OH−][\mathrm{H^+}]=[\mathrm{OH^-}].

pH of a strong acid-base mixture

pH=−log⁡nH+−nOH−Vtotal\mathrm{pH}=-\log\frac{n_{\mathrm{H^+}}-n_{\mathrm{OH^-}}}{V_{\text{total}}}

Worked example

40 mL of 0.1 M H2SO4\mathrm{H_2SO_4} is mixed with 60 mL of 0.1 M NaOH. Find the pH. (log⁡2=0.30\log2=0.30)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 29 July 2022 · Q124Moderate

Example 1 · Equilibrium · pH of Acids and Bases

200 mL200\text{ }mL of 0.01MHCl0.01MHCl is mixed with 400 mL400\text{ }mL of 0.01MH2SO40.01MH_{2}SO_{4}. The pHpH of the mixture is

Sulphuric acid counts twice

0.01 M H2SO4\mathrm{H_2SO_4} gives 0.02 M H+\mathrm{H^+}, and 0.01 M Ca(OH)2\mathrm{Ca(OH)_2} gives 0.02 M OH−\mathrm{OH^-}. Forgetting the 2 is the most common wrong option in mixture questions.

Dilution does not cross 7

Diluting HCl to 10−810^{-8} M does not give pH 8. An acid stays acidic; water's 10−710^{-7} M of H+\mathrm{H^+} must be added in.

Hot water is neutral at pH below 7

Heating raises KwK_w, so both [H+][\mathrm{H^+}] and [OH−][\mathrm{OH^-}] rise together. The pH falls, but the water is still neutral. "H+\mathrm{H^+} rises and OH−\mathrm{OH^-} falls" is wrong.
Drill 12 more on strong acids, bases and their mixtures

Concept 2 of 2: Weak acids and bases from Ka and Kb

A weak acid ionises only a little, so its concentration hardly changes. Then Ka≈Cα2K_a\approx C\alpha^2, and [H+]=Cα=KaC[\mathrm{H^+}]=C\alpha=\sqrt{K_aC}. A weak base works the same way with KbK_b and OH−\mathrm{OH^-}. An acid with two protons gives up the first far more easily, so the first step sets the pH.

Definition

  • Weak acid HA at concentration C: Ka=Cα21−α≈Cα2K_a=\frac{C\alpha^2}{1-\alpha}\approx C\alpha^2, so α=Ka/C\alpha=\sqrt{K_a/C}.
  • [H+]=KaC[\mathrm{H^+}]=\sqrt{K_aC}, pH=12(pKa−log⁡C)\mathrm{pH}=\tfrac12(\mathrm{p}K_a-\log C).
  • Weak base: [OH−]=KbC[\mathrm{OH^-}]=\sqrt{K_bC}, pOH=12(pKb−log⁡C)\mathrm{pOH}=\tfrac12(\mathrm{p}K_b-\log C).
  • Diprotic acid H2X\mathrm{H_2X}: Ka1K_{a1} sets [H+][\mathrm{H^+}], and [X2−]≈Ka2[\mathrm{X^{2-}}]\approx K_{a2}, whatever C is.
  • In a strong acid, the extra H+\mathrm{H^+} suppresses ionisation: [HA−]=Ka1[H2A][H+][\mathrm{HA^-}]=\frac{K_{a1}[\mathrm{H_2A}]}{[\mathrm{H^+}]}.
  • The overall step H2A⇌2H++A2−\mathrm{H_2A\rightleftharpoons 2H^++A^{2-}} has K=Ka1Ka2K=K_{a1}K_{a2}.

Weak acid

[H+]=KaC,pH=12(pKa−log⁡C)[\mathrm{H^+}]=\sqrt{K_aC},\qquad \mathrm{pH}=\tfrac12\left(\mathrm{p}K_a-\log C\right)

Worked example

Find the pH and the degree of ionisation of a 0.04 M weak acid with Ka=1.0×10−6K_a=1.0\times10^{-6}. (log⁡2=0.30\log2=0.30)
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 28 Jan 2026 Shift 1 · Q50Moderate

Example 2 · Equilibrium · pH of Acids and Bases

Consider the dissociation equilibrium of the following weak acid HA⇌H+(aq)+A−(aq)HA \rightleftharpoons H^{+}(aq) + A^{-}(aq) If the pKa of the acid is 4 , then the pH of 10 mM HA solution is ____\_\_\_\_ . (Nearest integer) [Given : The degree of dissociation can be neglected with respect to unity]

Take the square root

[H+][\mathrm{H^+}] is KaC\sqrt{K_aC}, not KaCK_aC, and pH is half of pKa−log⁡C\mathrm{p}K_a-\log C. Forgetting the half gives a pH twice too large.

In strong acid, divide by the strong acid's H+\mathrm{H^+}

A weak acid dissolved in 0.1 M HCl barely ionises. With [H2A]=[H+]=0.1[\mathrm{H_2A}]=[\mathrm{H^+}]=0.1 M, [HA−]=Ka1[\mathrm{HA^-}]=K_{a1}. The option 0.1 M treats the weak acid as fully ionised.
Drill 6 more on weak acids and bases from ka and kb

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

  • Strong acids, bases and their mixtures

    pH of a strong acid-base mixture

    pH=−log⁡nH+−nOH−Vtotal\mathrm{pH}=-\log\frac{n_{\mathrm{H^+}}-n_{\mathrm{OH^-}}}{V_{\text{total}}}
  • Weak acids and bases from Ka and Kb

    Weak acid

    [H+]=KaC,pH=12(pKa−log⁡C)[\mathrm{H^+}]=\sqrt{K_aC},\qquad \mathrm{pH}=\tfrac12\left(\mathrm{p}K_a-\log C\right)

Watch out for (5)

Test yourself on Equilibrium

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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