The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however,

NCERT Exemplar · Q7

Q. The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H^+ ions and OH^- ions are equal. What will be the pH of pure water at 60°C?

(i) Equal to 7.0

(ii) Greater than 7.0

(iii) Less than 7.0

(iv) Equal to zero

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Concept understanding — Hydrogen Ion Concentration pH

Hydrogen Ion Concentration and pH

Imagine you have a glass of pure water. It looks simple, but inside, a tiny fraction of water molecules are constantly splitting apart and re-forming. This splitting creates two kinds of charged particles: a hydrogen ion ( $H^+$ ) and a hydroxide ion ( $OH^-$ ). In pure water, these two are perfectly balanced — there are exactly as many $H^+$ as $OH^-$ .

Now, if you add something like lemon juice (an acid), you increase the number of hydrogen ions. The balance tips: more $H^+$ than $OH^-$ . If you add baking soda (a base), you decrease $H^+$ or increase $OH^-$ , and the balance tips the other way.

The question is: how do we measure this imbalance in a simple, practical way? The numbers of $H^+$ ions are incredibly tiny — in pure water, only about 1 in every 10 million water molecules is split at any moment. Writing these numbers directly (like $0.0000001$ moles per litre) is clumsy. That's where pH comes in.

The Precise Definition

pH is a mathematical shortcut. It stands for "power of hydrogen" (from the French puissance d'hydrogène).

\text{pH} = -\log_{10}[H^+]

The logarithm base 10 does two things at once:

It compresses a huge range of numbers (from $10^{-14}$ to $10^{0}$ ) into a manageable scale of 0 to 14.
The negative sign flips the direction: higher $[H^+]$ gives a lower pH, and lower $[H^+]$ gives a higher pH.

What the Numbers Mean

$[H^+]$ (mol/L)	pH	Example
$10^{-1}$	1	Stomach acid
$10^{-3}$	3	Lemon juice
$10^{-7}$	7	Pure water (neutral)
$10^{-9}$	9	Baking soda solution
$10^{-13}$	13	Household bleach

Notice the pattern: each step of 1 in pH means a tenfold change in $[H^+]$ . A solution of pH 3 has 10 times more $H^+$ than pH 4, and 100 times more than pH 5.

The Key Insight

pH is not a measure of "how acidic" something is in a vague sense — it is a precise, logarithmic measure of the actual number of hydrogen ions present. The scale runs from 0 (most acidic, highest $[H^+]$ ) to 14 (most basic, lowest $[H^+]$ ), with 7 being neutral. …

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Heating water increases its degree of ionization, raising both $[\text{H}^+]$ and $[\text{OH}^-]$ equally. Since pH depends only on $[\text{H}^+]$ , and that concentration rises, the pH of pure water at 60°C will be less than 7.0.

Why pH changes with temperature

The pH scale measures hydrogen ion concentration: $\text{pH} = -\log_{10}[\text{H}^+]$ . At 25°C, pure water has $[\text{H}^+] = 10^{-7}\,\text{M}$ , giving pH = 7.0. But "neutral" does not mean "pH = 7 always." Neutral means $[\text{H}^+] = [\text{OH}^-]$ , which is true at any temperature for pure water.

Water's self-ionization is an equilibrium:

\text{H}_2\text{O} \rightleftharpoons \text{H}^+ + \text{OH}^-

with equilibrium constant $K_w = [\text{H}^+][\text{OH}^-]$ . At 25°C, $K_w = 1.0 \times 10^{-14}$ . This reaction is endothermic—it absorbs heat—so raising the temperature shifts the equilibrium to the right (Le Chatelier's principle), producing more ions. The value of $K_w$ increases.

Important

Neutral water always satisfies $[\text{H}^+] = [\text{OH}^-]$ , but the magnitude of these equal concentrations rises with temperature.

Step-by-step reasoning

At 25°C: $K_w = 1.0 \times 10^{-14}$ . For pure water, $[\text{H}^+] = [\text{OH}^-] = \sqrt{K_w} = 10^{-7}\,\text{M}$ , so pH = 7.0.
At 60°C: The ionization constant increases. A typical value is $K_w \approx 9.6 \times 10^{-14}$ (the exact number varies slightly by source, but the trend is what matters).
Equal concentrations maintained: Pure water still has $[\text{H}^+] = [\text{OH}^-]$ , so each equals $\sqrt{K_w}$ .
Calculate the new hydrogen ion concentration:

$[\text{H}^+] = \sqrt{9.6 \times 10^{-14}} \approx 3.1 \times 10^{-7}\,\text{M}$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
$Q. The mass of sodium hydroxide that must be dissolved in 1.0 L solution to get pH of 12 is (A) 0.2 g (B) 0.3 g (C) 0.4 g (D) 0.8 g$
›Reveal solutionSolution
pH 12 means pOH = 2, so [OH⁻] = 0.01 M. For NaOH (strong base), this requires 0.01 mol in 1.0 L, which is 0.40 g. The correct option is (C).

Concept & Intuition

pH is a measure of hydrogen ion concentration, but in a strong base like NaOH, the hydroxide ion concentration directly determines the pH via the water dissociation constant. Since NaOH dissociates completely, the mass needed is simply the molar mass times the required moles of OH⁻. The key is to convert pH to pOH first, because pOH gives [OH⁻] directly.

Step-by-step solution

Relate pH and pOH

At 25 °C, pH + pOH = 14.

Given pH = 12, so pOH = 14 − 12 = 2.

Find [OH⁻] from pOH

pOH = −log₁₀[OH⁻], so [OH⁻] = 10⁻² M = 0.01 mol/L.

Determine moles of NaOH needed

NaOH is a strong base: NaOH → Na⁺ + OH⁻ completely.

Thus, [NaOH] = [OH⁻] = 0.01 M.

For 1.0 L solution: moles of NaOH = 0.01 mol/L × 1.0 L = 0.01 mol.

Convert moles to mass

Molar mass of NaOH = 23 (Na) + 16 (O) + 1 (H) = 40 g/mol. …
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TG EAPCET 2026Set ap-2026-05-04-AN1 markMCQ
$pK_a$
›Reveal solutionSolution
This is a buffer problem where NaOH reacts with acetic acid, shifting the buffer ratio. The final pH is 4.91, corresponding to option (A).

The key concept here is the Henderson–Hasselbalch equation for a buffer solution. A buffer containing a weak acid (acetic acid, CH₃COOH) and its conjugate base (acetate, CH₃COO⁻) resists pH change when small amounts of strong base are added. The equation is:
$\text{pH} = \text{p}K_a + \log \frac{[\text{conjugate base}]}{[\text{acid}]}$
When NaOH is added, it reacts completely with the acid:
$\text{CH}_3\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{O}$
So the moles of acid decrease and the moles of acetate increase by the same amount. The trick is to track these changes precisely.

Find moles of NaOH added

Molar mass of NaOH = 40 g/mol.

Moles of NaOH = $\frac{8}{40} = 0.2$ mol.

Initial moles in the buffer

We have 1 mole of acetic acid (HA) and 1 mole of sodium acetate (A⁻) in 1.0 L solution.

So initially: $n_{\text{HA}} = 1.0$ , $n_{\text{A}^-} = 1.0$ .

Reaction with NaOH

The 0.2 mol OH⁻ will consume 0.2 mol of HA and produce 0.2 mol of A⁻.

After reaction:

$n_{\text{HA}} = 1.0 - 0.2 = 0.8 \text{ mol}$ $n_{\text{A}^-} = 1.0 + 0.2 = 1.2 \text{ mol}$

Concentrations in the same volume Volume remains 1.0 L, so concentrations are numerically equal to moles:

$[\text{HA}] = 0.8 \text{ M}, \quad [\text{A}^-] = 1.2 \text{ M}$

Apply Henderson–Hasselbalch …
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TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$pK_a$
›Reveal solutionSolution
This is a buffer solution of acetic acid and sodium acetate; after accounting for dilution, the ratio of conjugate base to acid is 1:1, so the pH equals the pKa, which is 4.76.

We have a mixture of a weak acid (acetic acid) and its conjugate base (acetate from sodium acetate). That’s a classic buffer. The pH of a buffer is given by the Henderson–Hasselbalch equation:
$\text{pH} = \text{p}K_a + \log \frac{[\text{conjugate base}]}{[\text{acid}]}$
The key here is that we are mixing two solutions and then diluting to a final volume. The ratio of concentrations in the final solution is the same as the ratio of moles (since both are in the same final volume). So we don’t need the final concentrations — just the moles of acid and base.

Let’s work through it step by step.

Find moles of acetic acid (HA)

Volume = 10 mL = 0.010 L, concentration = 1.0 M

Moles of HA = $0.010 \times 1.0 = 0.010$ mol.

Find moles of acetate (A⁻)

Volume = 20 mL = 0.020 L, concentration = 0.5 M

Moles of A⁻ = $0.020 \times 0.5 = 0.010$ mol.

Dilution effect

The mixture is diluted to 100 mL. Both species are in the same final volume, so the ratio of their concentrations is exactly the ratio of their moles:
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TG EAPCET 2021Set ap-2021-08-09-AN1 markMCQ
$0.2\,\text{M}\ \mathrm{CH_3COOH}$
›Reveal solutionSolution
This is a buffer solution of a weak acid and its conjugate base, so the Henderson–Hasselbalch equation applies directly. The pH is $4.74 + \log(0.1/0.2) = 4.74 - 0.30 = 4.44$ , which rounds to 4.43, option (B).

The key concept here is the buffer solution. A mixture of a weak acid (acetic acid, $\mathrm{CH_3COOH}$ ) and its salt with a strong base (sodium acetate, $\mathrm{CH_3COONa}$ ) resists drastic pH changes. The salt fully dissociates, providing a reservoir of the conjugate base ( $\mathrm{CH_3COO^-}$ ). The weak acid remains mostly undissociated. The pH of such a mixture is governed by the equilibrium:
$\mathrm{CH_3COOH \rightleftharpoons H^+ + CH_3COO^-}$
Because the acid is weak and the conjugate base is present in significant amount from the salt, we can use the Henderson–Hasselbalch equation, which is derived directly from the acid dissociation constant expression:
$\mathrm{pH = pK_a + \log \frac{[salt]}{[acid]}}$
This works because the equilibrium concentration of the acid is approximately its initial concentration (very little dissociates), and the equilibrium concentration of the conjugate base is approximately the initial concentration from the salt (the small amount from acid dissociation is negligible).

Let’s apply it step by step.

Identify the given values.

$[\mathrm{CH_3COOH}] = 0.2\ \mathrm{M}$ (weak acid)

$[\mathrm{CH_3COONa}] = 0.1\ \mathrm{M}$ (salt, gives $[\mathrm{CH_3COO^-}] = 0.1\ \mathrm{M}$ )

$K_a = 1.8 \times 10^{-5}$ , so $\mathrm{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74$

Plug into the Henderson–Hasselbalch equation.

$\mathrm{pH} = 4.74 + \log \left( \frac{0.1}{0.2} \right)$

Simplify the ratio.

$\frac{0.1}{0.2} = 0.5$
So:
$\mathrm{pH} = 4.74 + \log(0.5)$

Evaluate the logarithm. $\log(0.5) = \log(1/2) = -\log 2 \approx -0.3010$ …
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TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
$(1+\alpha) \times$
›Reveal solutionSolution

$[\mathrm{H^+}]=10^{-2}$ gives $\alpha=0.1$ , so the effective concentration is $(1+\alpha)C=0.11\,\mathrm{M}$ and $\pi=0.11\,RT$ — option (B).

Given: a $0.1\ \mathrm{M}$ monobasic acid with $\mathrm{pH}=2$ .

Degree of dissociation.

$[\mathrm{H^+}]=10^{-\mathrm{pH}}=10^{-2}=0.01\ \mathrm{M},\qquad \alpha=\frac{[\mathrm{H^+}]}{C}=\frac{0.01}{0.1}=0.1.$ …
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Q.The pH of neutral water at 25°C is 7.0. As the temperature increases, ionisation of water increases, however, the concentration of H^+ ions and OH^- ions are equal. What will be the pH of pure water at 60°C?

Why pH changes with temperature

Step-by-step reasoning

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