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Chemistry · Equilibrium

pH, Ionisation Constant, Buffer Solutions

This subtopic builds the quantitative core of ionic equilibrium from NCERT Class XI Chemistry, Unit 6 (sections 6.11–6.12): the self-ionisation of water and its ionic product $K_w$, the logarithmic pH scale, the ionisation constants $K_a$ and $K_b$ of weak acids and bases, Ostwald's dilution law, the bridge $K_a \cdot K_b = K_w$ for a conjugate pair, and the Henderson–Hasselbalch equation for buffers. It is a heavily examined region: NEET draws repeated calculation questions on buffer pH, salt hydrolysis and dilute-acid pH from precisely these formulae.

Ionisation of Water and Kw

Water is amphoteric: in pure water one molecule donates a proton while another accepts it, so water acts simultaneously as a Brønsted acid and a Brønsted base. This self-ionisation establishes the equilibrium

$$\ce{H2O(l) + H2O(l) <=> H3O^+(aq) + OH^-(aq)}$$

Because water is a pure liquid its concentration is constant and is absorbed into the equilibrium constant, giving the ionic product of water, $K_w = [\ce{H+}][\ce{OH^-}]$. Experimentally $[\ce{H+}] = 1.0 \times 10^{-7}\,\text{M}$ at 298 K, and since dissociation produces H⁺ and OH⁻ in equal numbers, $[\ce{OH^-}] = [\ce{H+}] = 1.0 \times 10^{-7}\,\text{M}$. Therefore

$$K_w = [\ce{H3O+}][\ce{OH^-}] = (1\times 10^{-7})^2 = 1\times 10^{-14}\ \text{M}^2 \quad (298\,\text{K})$$

$K_w$ is an equilibrium constant and is therefore temperature dependent. The relative magnitudes of $[\ce{H3O+}]$ and $[\ce{OH^-}]$ classify a solution: acidic when $[\ce{H3O+}] > [\ce{OH^-}]$, neutral when they are equal, and basic when $[\ce{H3O+}] < [\ce{OH^-}]$. Note that ionisation is slight — only about 2 molecules in $10^{9}$ are dissociated, so the equilibrium lies overwhelmingly toward undissociated water.

The pH Scale and pOH

Hydronium-ion concentrations span many orders of magnitude, so a logarithmic measure is used. The pH of a solution is the negative logarithm to base 10 of the hydrogen-ion activity; for dilute solutions (below 0.01 M) the activity equals the molarity in mol L⁻¹:

$$\text{pH} = -\log a_{\ce{H+}} = -\log\{[\ce{H+}]/\text{mol L}^{-1}\}$$

Pure water at 298 K has $[\ce{H+}] = 10^{-7}\,\text{M}$, giving $\text{pH} = -\log(10^{-7}) = 7$. Taking the negative logarithm of $K_w = [\ce{H3O+}][\ce{OH^-}] = 10^{-14}$ links pH and pOH:

$$\text{p}K_w = \text{pH} + \text{pOH} = 14 \quad (298\,\text{K})$$

Figure 1 The pH scale from 0 to 14 with acidic, neutral and basic regions 0 2 4 6 7 8 10 12 14 ACIDIC · pH < 7 NEUTRAL · pH = 7 BASIC · pH > 7

A change of one pH unit corresponds to a tenfold change in $[\ce{H+}]$; gastric juice (~1.2), human blood (~7.4) and 0.1 M NaOH (~13) span the practical range. Source: NCERT Table 6.5.

Because the scale is logarithmic, a unit change in pH is a factor-of-ten change in $[\ce{H+}]$, and a change of 2 units is a factor of 100. Although $K_w$ varies with temperature, the resulting variation in pH is so small that it is usually ignored in problem solving. A few representative pH values from NCERT Table 6.5 anchor intuition.

FluidApprox. pHFluidApprox. pH
Concentrated HCl~ −1.0Human saliva6.4
Gastric juice~1.2Human blood7.4
Lemon juice~2.2Egg white, sea water7.8
Soft drinks, vinegar~3.0Lime water10.5
Black coffee5.0Saturated NaOH~15

Ionisation Constants of Weak Acids

A weak acid HX is only partially ionised in water. With initial concentration $c$ and degree of ionisation $\alpha$, the equilibrium concentrations are $c(1-\alpha)$ for HX and $c\alpha$ each for H⁺ and X⁻:

$$\ce{HX(aq) + H2O(l) <=> H3O^+(aq) + X^-(aq)}$$

The acid ionisation constant follows directly from the equilibrium-constant expression, with water omitted because it is a pure liquid:

$$K_a = \frac{[\ce{H+}][\ce{X^-}]}{[\ce{HX}]} = \frac{c\alpha^2}{1-\alpha}$$

At a given temperature, a larger $K_a$ means a stronger acid. As with pH, a logarithmic measure is convenient: $\text{p}K_a = -\log K_a$, so a smaller $\text{p}K_a$ marks a stronger acid. The NCERT-listed constants below show the enormous range of acid strength among common weak acids.

Weak acidKa (298 K)pKa
Nitrous acid (HNO₂)4.5 × 10⁻⁴3.35
Hydrofluoric acid (HF)3.5 × 10⁻⁴3.46
Formic acid (HCOOH)1.8 × 10⁻⁴3.75
Benzoic acid (C₆H₅COOH)6.5 × 10⁻⁵4.19
Acetic acid (CH₃COOH)1.74 × 10⁻⁵4.76
Niacin (C₅H₄NCOOH)1.5 × 10⁻⁵4.82
Hypochlorous acid (HClO)3.0 × 10⁻⁸7.52
Hydrocyanic acid (HCN)4.9 × 10⁻¹⁰9.31
Phenol (C₆H₅OH)1.3 × 10⁻¹⁰9.89

Polyprotic acids ionise stepwise, each step weaker than the last because pulling a proton from an already-negative ion is harder. For phosphoric acid, $\ce{H3PO4}$ ionises in three stages with $K_{a_1} > K_{a_2} > K_{a_3}$, and the overall constant satisfies $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$ — a relation NEET tested directly in 2025.

NEET Trap

pH of a very dilute strong acid is not −log c

For a $10^{-8}\,\text{M}$ HCl solution, applying $\text{pH} = -\log(10^{-8}) = 8$ wrongly makes a dilute acid turn basic. At such dilution the H⁺ from water is comparable to that from the acid. Writing $[\ce{H3O+}] = 10^{-8} + x$ and imposing $K_w = (10^{-8}+x)(x) = 10^{-14}$ gives $[\ce{OH^-}] = x = 9.5\times 10^{-8}$, hence $\text{pOH} = 7.02$ and $\text{pH} = 6.98$.

Below about $10^{-6}\,\text{M}$, never ignore water's autoionisation; the answer stays just below 7 for acids and just above 7 for bases.

Degree of Ionisation and Ostwald's Law

The degree of ionisation $\alpha$ is the fraction of the dissolved electrolyte that has ionised. Starting from $K_a = c\alpha^2/(1-\alpha)$, when $\alpha$ is small enough that $(1-\alpha) \approx 1$ — true for most weak acids at ordinary concentrations — the expression simplifies and can be solved for $\alpha$:

$$K_a \approx c\alpha^2 \qquad \Rightarrow \qquad \alpha = \sqrt{\frac{K_a}{c}}$$

This is Ostwald's dilution law. It states that the degree of ionisation of a weak electrolyte depends on temperature (through $K_a$) and on concentration: at constant temperature $\alpha$ is inversely proportional to $\sqrt{c}$, so dilution increases ionisation. The hydrogen-ion concentration then follows as $[\ce{H+}] = c\alpha = \sqrt{K_a\,c}$, from which pH is obtained.

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Build the foundation first

The Brønsted–Lowry and Lewis frameworks behind $K_a$ and $K_b$ are developed in Ionic Equilibrium: Acids & Bases.

Weak Bases and the Ka·Kb Relation

A weak base MOH ionises partially, $\ce{MOH(aq) <=> M^+(aq) + OH^-(aq)}$, with a base ionisation constant $K_b = [\ce{M+}][\ce{OH^-}]/[\ce{MOH}] = c\alpha^2/(1-\alpha)$, and $\text{p}K_b = -\log K_b$. Many amines are weak bases; ammonia produces hydroxide in water:

$$\ce{NH3(aq) + H2O(l) <=> NH4^+(aq) + OH^-(aq)}$$

For a conjugate acid–base pair the two constants are tied together. Adding the acid ionisation of $\ce{NH4+}$ to the base ionisation of $\ce{NH3}$ yields the autoionisation of water, so the equilibrium constants multiply:

$$K_a \times K_b = [\ce{H3O+}][\ce{OH^-}] = K_w = 1.0\times 10^{-14}\ \text{M} \quad (298\,\text{K})$$

Taking negative logarithms gives the partner relation $\text{p}K_a + \text{p}K_b = \text{p}K_w = 14$. The immediate consequence is that a strong acid has a weak conjugate base and vice versa — knowing one constant fixes the other.

NEET Trap

Ka·Kb = Kw is for the conjugate pair only

The relation $K_a \cdot K_b = K_w$ holds between an acid and its own conjugate base — for example $\ce{CH3COOH}$ and $\ce{CH3COO^-}$, or $\ce{NH4+}$ and $\ce{NH3}$. It does not connect the $K_a$ of one acid with the $K_b$ of an unrelated base. For a salt of a weak acid and a weak base (e.g. dimethylammonium acetate), the solution pH is $\text{pH} = 7 + \tfrac{1}{2}(\text{p}K_a - \text{p}K_b)$.

If $\text{p}K_a > \text{p}K_b$ the salt is slightly basic (pH > 7); if $\text{p}K_a < \text{p}K_b$ it is slightly acidic.

Buffer Solutions and the Henderson Equation

Many body fluids such as blood and urine maintain a definite pH, and deviation signals malfunction. Solutions that resist a change in pH on dilution or on adding small amounts of acid or alkali are called buffer solutions. An acidic buffer is made from a weak acid and its salt with a strong base; a basic buffer from a weak base and its salt with a strong acid.

For a weak acid HA in equilibrium with its conjugate base $\ce{A^-}$, rearranging $K_a = [\ce{H+}][\ce{A^-}]/[\ce{HA}]$ and taking logarithms gives the Henderson–Hasselbalch equation. Because the acid is weak, $[\ce{HA}] \approx$ concentration of acid added and $[\ce{A^-}] \approx$ concentration of salt added:

$$\text{pH} = \text{p}K_a + \log\frac{[\text{Salt}]}{[\text{Acid}]}$$

When $[\text{Salt}] = [\text{Acid}]$ the log term vanishes and $\text{pH} = \text{p}K_a$. Hence a buffer works best when the required pH is close to the $\text{p}K_a$ of the chosen acid: acetic acid / sodium acetate buffers around pH 4.76, and ammonium hydroxide / ammonium chloride around pH 9.25. The analogous basic-buffer result is $\text{pH} = \text{p}K_a + \log([\text{Base}]/[\text{Conjugate acid}])$, obtained via $\text{pH} + \text{pOH} = 14$.

Figure 2 Buffer action: added acid is neutralised by conjugate base, added alkali by the weak acid BUFFER HA & A⁻ pH ≈ pKa Add H⁺ A⁻ + H⁺ → HA Add OH⁻ HA + OH⁻ → A⁻ + H₂O

Added acid is consumed by the conjugate base A⁻ and added alkali by the weak acid HA, so $[\text{Salt}]/[\text{Acid}]$ barely shifts and pH holds steady. The pH of a buffer is unaffected by dilution because that ratio is unchanged. Source: NCERT §6.12.

Worked Examples

Example 1 · Strong acid

Calculate the pH of a $3.8 \times 10^{-3}\,\text{M}$ solution (the hydrogen-ion concentration measured in a soft drink).

A strong acid is fully ionised, so $[\ce{H+}] = 3.8\times 10^{-3}\,\text{M}$ directly. Then $\text{pH} = -\log(3.8\times 10^{-3}) = -\{\log 3.8 + \log 10^{-3}\} = -\{0.58 - 3.0\} = 2.42$. The solution is acidic, as expected. (NCERT Problem 6.16.)

Example 2 · Weak acid via Ka

Calculate the pH and percent dissociation of $0.08\,\text{M}$ hypochlorous acid, $\ce{HOCl}$, given $K_a = 2.5\times 10^{-5}$.

For $\ce{HOCl + H2O <=> H3O+ + ClO^-}$ with $[\ce{H3O+}] = x$, $K_a = x^2/(0.08 - x)$. Since $x \ll 0.08$, take $0.08 - x \approx 0.08$, so $x^2 = (2.5\times 10^{-5})(0.08) = 2.0\times 10^{-6}$, giving $x = 1.41\times 10^{-3}\,\text{M}$. Hence $\text{pH} = -\log(1.41\times 10^{-3}) = 2.85$ and percent dissociation $= (1.41\times 10^{-3}/0.08)\times 100 = 1.76\%$. (NCERT Problem 6.20.)

Example 3 · Buffer via Henderson

A solution contains $0.2\,\text{M}\ \ce{NH4Cl}$ and $0.1\,\text{M}\ \ce{NH3}$, with $\text{p}K_b$ of ammonia $= 4.75$. Find the pH.

This is a basic buffer. With $K_b = 10^{-4.75} = 1.77\times 10^{-5}$ and $\ce{NH3 + H2O <=> NH4+ + OH^-}$, $K_b = (0.20)(x)/(0.10)$ after neglecting $x$. So $[\ce{OH^-}] = x = 0.88\times 10^{-5}\,\text{M}$, then $[\ce{H+}] = K_w/[\ce{OH^-}] = 1.12\times 10^{-9}$, giving $\text{pH} = -\log(1.12\times 10^{-9}) = 8.95$. Equivalently, using $\text{pH} = \text{p}K_a + \log([\text{base}]/[\text{salt}])$ with $\text{p}K_a = 9.25$ returns the same value. (NCERT Problem 6.22.)

Quick Recap

The formula sheet for this subtopic

  • Ionic product of water: $K_w = [\ce{H+}][\ce{OH^-}] = 1.0\times 10^{-14}$ at 298 K (temperature dependent).
  • pH definition and partner: $\text{pH} = -\log[\ce{H+}]$; $\text{pH} + \text{pOH} = 14$.
  • Weak acid: $K_a = c\alpha^2/(1-\alpha)$; $\text{p}K_a = -\log K_a$ — smaller pKa = stronger acid.
  • Ostwald's dilution law: $\alpha = \sqrt{K_a/c}$, so $\alpha$ increases on dilution; $[\ce{H+}] = \sqrt{K_a c}$.
  • Conjugate pair: $K_a\cdot K_b = K_w$ and $\text{p}K_a + \text{p}K_b = 14$.
  • Acidic buffer: $\text{pH} = \text{p}K_a + \log([\text{Salt}]/[\text{Acid}])$; pH = pKa when salt and acid are equimolar; pH is unchanged by dilution.
  • Very dilute strong acid (≤ $10^{-6}\,\text{M}$): include water's H⁺, so pH stays just below 7.

NEET PYQ Snapshot — pH, Ionisation Constant, Buffer Solutions

Real NEET previous-year questions drawn from this subtopic, with years and worked solutions.

NEET 2022

The pH of the solution containing 50 mL each of 0.10 M sodium acetate and 0.01 M acetic acid is [Given pKa of CH₃COOH = 4.57]

  • (1) 3.57
  • (2) 4.57
  • (3) 2.57
  • (4) 5.57
Answer: (4) 5.57

A mixture of a weak acid and its conjugate base is an acidic buffer. $\text{pH} = \text{p}K_a + \log([\text{Salt}]/[\text{Acid}]) = 4.57 + \log(0.1/0.01) = 4.57 + 1 = 5.57$. (Equal volumes cancel in the ratio.)

NEET 2021

The pKb of dimethylamine and pKa of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethylammonium acetate solution is:

  • (1) 6.25
  • (2) 8.50
  • (3) 5.50
  • (4) 7.75
Answer: (4) 7.75

Dimethylammonium acetate is the salt of a weak acid and a weak base, so $\text{pH} = 7 + \tfrac{1}{2}(\text{p}K_a - \text{p}K_b) = 7 + \tfrac{1}{2}(4.77 - 3.27) = 7.75$.

NEET 2018

Solutions were prepared by mixing NaOH and HCl: (a) 60 mL M/10 HCl + 40 mL M/10 NaOH; (b) 55 mL M/10 HCl + 45 mL M/10 NaOH; (c) 75 mL M/5 HCl + 25 mL M/5 NaOH; (d) 100 mL M/10 HCl + 100 mL M/10 NaOH. The pH of which one will be equal to 1?

  • (1) b
  • (2) a
  • (3) d
  • (4) c
Answer: (4) c

In (c), milliequivalents of HCl $= (1/5)\times 75 = 15$ and of NaOH $= (1/5)\times 25 = 5$; net acid = 10 meq in 100 mL, so normality $= 10^{-1}$ and $\text{pH} = -\log(10^{-1}) = 1$. The other mixtures give pH 1.7, 2 or neutral.

NEET 2025

Phosphoric acid ionises in three steps with ionisation constants $K_{a_1}, K_{a_2}, K_{a_3}$, while K is the overall ionisation constant. Which statements are true? A. $\log K = \log K_{a_1} + \log K_{a_2} + \log K_{a_3}$   B. H₃PO₄ is a stronger acid than H₂PO₄⁻ and HPO₄²⁻   C. $K_{a_1} > K_{a_2} > K_{a_3}$   D. $K_{a_1} = K_{a_3} + 2K_{a_2}$

  • (1) A, B and C only
  • (2) A and B only
  • (3) A and C only
  • (4) B, C and D only
Answer: (1) A, B and C only

Each successive proton is harder to remove from a more negative ion, so $K_{a_1} > K_{a_2} > K_{a_3}$ (C) and $\ce{H3PO4}$ is the strongest of the three species (B). The overall constant is the product of the stepwise constants, so its logarithm is the sum (A). Statement D has no basis.

NEET 2017 · Concept

For a $0.02\,\text{M}$ solution of HF with $K_a = 3.2\times 10^{-4}$, estimate the degree of dissociation $\alpha$ and the pH using Ostwald's dilution law.

Answer: α ≈ 0.12, pH ≈ 2.62

Using $\alpha = \sqrt{K_a/c} = \sqrt{3.2\times 10^{-4}/0.02} = 0.126$; the more exact quadratic $\alpha^2 + 1.6\times 10^{-2}\alpha - 1.6\times 10^{-2} = 0$ gives $\alpha = 0.12$. Then $[\ce{H+}] = c\alpha = 2.4\times 10^{-3}\,\text{M}$ and $\text{pH} = -\log(2.4\times 10^{-3}) = 2.62$. (Concept item modelled on NCERT Problem 6.18.)

FAQs — pH, Ionisation Constant, Buffer Solutions

The recurring conceptual snags around pH, Ka·Kb and buffers.

Why does the pH of a 10⁻⁸ M HCl solution come out close to 7 and not 8?

At such high dilution the H⁺ contributed by the autoionisation of water is no longer negligible compared with the acid. The total [H⁺] = 10⁻⁸ + x, where x is the contribution from water, fixed by Kw = (10⁻⁸ + x)(x) = 10⁻¹⁴. Solving the quadratic gives [OH⁻] = x = 9.5 × 10⁻⁸ M, so pOH = 7.02 and pH = 6.98. A blindly applied pH = −log(10⁻⁸) = 8 would absurdly make a dilute acid basic, which is the trap.

What is the relation between Ka and Kb for a conjugate acid–base pair?

For a conjugate acid–base pair, Ka × Kb = Kw = 1.0 × 10⁻¹⁴ at 298 K. Taking negative logarithms gives pKa + pKb = pKw = 14. Therefore a strong acid has a weak conjugate base and vice versa; knowing one constant immediately gives the other.

How is Ostwald's dilution law used to find the degree of ionisation?

For a weak electrolyte of initial concentration c, Ka = cα²/(1 − α). When α is small (1 − α ≈ 1) this simplifies to Ka = cα², giving α = √(Ka/c). This is Ostwald's dilution law: the degree of ionisation is inversely proportional to the square root of concentration, so α rises on dilution.

What is the Henderson–Hasselbalch equation for an acidic buffer?

For a buffer of a weak acid and its salt, pH = pKa + log([Salt]/[Acid]). When [Salt] = [Acid] the log term is zero and pH = pKa, so the buffer is most effective near pH equal to the pKa of the acid.

Does diluting a buffer change its pH?

No. In the Henderson–Hasselbalch equation only the ratio [Salt]/[Acid] appears, and dilution scales both concentrations by the same factor. The ratio is unchanged, so the pH of a buffer is essentially unaffected by moderate dilution.

How is the pH of a salt of a weak acid and a weak base calculated?

For a salt of a weak acid and a weak base (such as dimethylammonium acetate), pH = 7 + ½(pKa − pKb), where pKa is that of the weak acid and pKb that of the weak base. The result is independent of concentration.

Related Topics
Equilibrium Back to the full chapter hub and subtopic map. Ionic Equilibrium: Acids & Bases Arrhenius, Brønsted–Lowry and Lewis concepts behind Ka and Kb. Hydrolysis of Salts and pH Why salts of weak acids or bases are not pH-neutral. Solubility Product & Common-Ion Effect Ksp, sparingly soluble salts and the common-ion shift.