Does ionisation of water increase with temperature?

Yes! Autoionisation of water is endothermic: H₂O ⇌ H⁺ + OH⁻, ΔH > 0. By Le Chatelier, increasing temperature shifts equilibrium forward → more ionisation → Kw increases. Statement I is CORRECT.

Does Kw increase or decrease with temperature?

Kw INCREASES with temperature. At 25°C: Kw = 1×10⁻¹⁴. At 37°C (body temperature): Kw ≈ 2.4×10⁻¹⁴. At 60°C: Kw ≈ 9.6×10⁻¹⁴. Statement II says it decreases — this is INCORRECT.

Is water still neutral at higher temperatures even if Kw increases?

Yes! Even though Kw increases, water is still neutral because [H⁺] = [OH⁻] always in pure water. pH changes: at 37°C, Kw = 2.4×10⁻¹⁴, so [H⁺] = √(2.4×10⁻¹⁴) = 1.55×10⁻⁷, pH = 6.81. But still neutral (pH = pOH). pH of neutrality ≠ 7 at temperatures other than 25°C.

What is the pH of pure water at body temperature 37°C?

Kw(37°C) ≈ 2.4×10⁻¹⁴. [H⁺] = √(2.4×10⁻¹⁴) = 1.55×10⁻⁷ M. pH = -log(1.55×10⁻⁷) ≈ 6.81. Water is still neutral at 37°C, but pH = 6.81 (not 7.0). Neutral does not mean pH=7; it means [H⁺]=[OH⁻].

What is pKw at different temperatures?

pKw = -log(Kw). At 0°C: Kw ≈ 1.2×10⁻¹⁵, pKw = 14.9. At 25°C: Kw = 10⁻¹⁴, pKw = 14.0. At 37°C: Kw ≈ 2.4×10⁻¹⁴, pKw ≈ 13.6. At 100°C: Kw ≈ 10⁻¹², pKw = 12. pKw decreases as T increases (Kw increases).

Statement I: The degree of ionisation of water increases with temperature.<br>Statement II: The ionic product of water (

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ChemistryElectrochemistry

Statement I: The degree of ionisation of water increases with temperature.
Statement II: The ionic product of water ($K_w$) decreases with increase in temperature.
Select the correct option:

Options

Statement I is correct and II is incorrect

Both I and II are correct

Statement I is incorrect and II is correct

Both I and II are incorrect

Correct Answer

Statement I is correct and II is incorrect

Solution

Statement I: Autoionisation of water is endothermic:

$$H_2O \rightleftharpoons H^+ + OH^-, \quad \Delta H > 0$$

Increasing T → forward shift → more ionisation. Statement I is TRUE ✓

Statement II: Since ionisation increases with T, $K_w = [H^+][OH^-]$ INCREASES with T.

At 25°C: $K_w = 10^{-14}$. At 60°C: $K_w \approx 10^{-13.6}$ (larger).

Statement II says $K_w$ decreases — FALSE ✗

$K_w$ increases with T (endothermic autoionisation).
Statement I: TRUE | Statement II: FALSE

Theory: Electrochemistry

1. Ionic Product of Water $K_w$

Water undergoes autoionisation (autoprotolysis): $H_2O + H_2O \rightleftharpoons H_3O^+ + OH^-$ or simply $H_2O \rightleftharpoons H^+ + OH^-$. Ionic product: $K_w = [H^+][OH^-]$. At 25°C: $K_w = 1.0 \times 10^{-14}$ (pKw = 14.0). This is an equilibrium constant so it depends on temperature. The autoionisation is endothermic ($\Delta H \approx +55.8$ kJ/mol). Le Chatelier: increase T → equilibrium shifts forward → $K_w$ increases. At 0°C: $K_w \approx 1.14\times10^{-15}$. At 25°C: $1.01\times10^{-14}$. At 37°C: $2.4\times10^{-14}$. At 60°C: $9.55\times10^{-14}$. At 100°C: $5.5\times10^{-13}$. pKw decreases from 14.9 (0°C) to 12.3 (100°C). Implication: pH of neutral water at 37°C is 6.81, not 7.0. Neutral means [H⁺]=[OH⁻], not necessarily pH=7.

2. pH Scale and Its Temperature Dependence

pH $= -\log[H_3O^+] = -\log[H^+]$. pOH $= -\log[OH^-]$. At any temperature: pH + pOH $= pK_w$. At 25°C: pH + pOH $= 14$. At 37°C: pH + pOH $= 13.6$. A solution with pH 7 at 37°C is actually slightly basic (since neutral pH = 6.81 at 37°C). pH values below neutral pH = acidic; above = basic. pH range: 0 to 14 (at 25°C). Can exceed this range for very concentrated strong acids/bases (e.g., 10 M HCl: pH = -1). pH $< 0$: superacids. pH $> 14$: superbasic solutions. Measurement: glass electrode (potential depends on [H⁺]), pH meter. Indicators change colour over 2 pH unit range. Litmus: red at pH < 5, blue at pH > 8. Phenolphthalein: colourless at pH < 8.3, pink at pH > 10.

3. Strong and Weak Acids

Strong acids (complete dissociation): HCl, HBr, HI, HNO₃, H₂SO₄ (1st ionisation), HClO₄. pH = -log(C) for monoprotic strong acid. 0.01 M HCl: pH = 2. Weak acids: partially dissociated. Ka = [H⁺][A⁻]/[HA]. pH $= \frac{1}{2}(pK_a - \log C)$. For 0.1 M CH₃COOH (Ka = 1.8×10⁻⁵): $[H^+] = \sqrt{K_a C} = \sqrt{1.8\times10^{-6}} = 1.34\times10^{-3}$ M. pH = 2.87. Degree of dissociation $\alpha = [H^+]/C = 1.34\times10^{-3}/0.1 = 1.34\%$. Ostwald dilution law: $K_a = C\alpha^2/(1-\alpha)$. For small $\alpha$: $K_a \approx C\alpha^2$, $\alpha = \sqrt{K_a/C}$ (decreases with increasing C). Strong acid + weak base: acidic salt (NH₄Cl). Weak acid + strong base: basic salt (CH₃COONa). Weak acid + weak base: pH ≈ 7 + (pKa - pKb)/2.

4. Buffer Solutions

Buffer: solution that resists pH change on addition of small amounts of acid or base. Acidic buffer: weak acid + salt of weak acid with strong base. Example: CH₃COOH/CH₃COONa. pH $= pK_a + \log\dfrac{[salt]}{[acid]}$ (Henderson-Hasselbalch). Basic buffer: weak base + salt of weak base with strong acid. NH₃/NH₄Cl. pOH $= pK_b + \log\dfrac{[salt]}{[base]}$. Maximum buffer capacity: when [acid] = [salt], pH = pKa. Buffer range: $pK_a \pm 1$. Biological buffers: blood uses bicarbonate buffer (pH 7.4): $H_2CO_3/HCO_3^-$, pKa = 6.1. Blood pH 7.4 is maintained partly by lung (CO₂ exhalation) and kidney (H⁺ excretion). Other biological: phosphate buffer (pH 6-8), protein buffers (amino acid side chains act as acids/bases), haemoglobin buffer.

5. Acid-Base Titrations and Indicators

Neutralisation: $H^+ + OH^- \to H_2O$, $\Delta H = -57.1$ kJ/mol. Titration curves: Strong acid + strong base: equivalence point pH = 7. Sharp break near equivalence point. Any indicator with pH range near 7. Weak acid + strong base: equivalence point pH > 7 (basic). Use phenolphthalein (range 8.3-10.0). Weak base + strong acid: equivalence point pH < 7 (acidic). Use methyl orange (range 3.1-4.4). Weak acid + weak base: gradual curve, no sharp break. Difficult to titrate accurately. Indicators change colour at specific pH ranges. Good indicator: pKin should be near equivalence point pH. At transition point: [HIn] = [In⁻]. pH of transition = pKin. Colour change span: about 2 pH units (pH = pKin ± 1).

6. Salt Hydrolysis

Salts of strong acid + strong base: do not hydrolyse. pH = 7. Example: NaCl, KNO₃. Salts of strong acid + weak base: cationic hydrolysis (acidic). NH₄Cl: NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺. pH < 7. $K_h = K_w/K_b$. For NH₄Cl: pH = 7 - ½(pKb - log C) = 7 - ½(4.74 - log C). Salts of weak acid + strong base: anionic hydrolysis (basic). CH₃COONa: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. pH > 7. $K_h = K_w/K_a$. pH = 7 + ½(pKa + log C). Salts of weak acid + weak base: both hydrolyse. pH ≈ 7 + ½(pKa - pKb) approximately. CH₃COONH₄ (ammonium acetate): pKa ≈ pKb ≈ 4.74, so pH ≈ 7.

7. Nernst Equation

$E_{cell} = E^\circ_{cell} - \dfrac{RT}{nF}\ln Q = E^\circ_{cell} - \dfrac{0.0592}{n}\log Q$ (at 298 K). $n$ = number of electrons transferred. $Q$ = reaction quotient (concentrations/pressures). At equilibrium: $E_{cell} = 0$ and $Q = K$: $E^\circ_{cell} = \dfrac{0.0592}{n}\log K$. Daniel cell: $Zn/Zn^{2+}(1M) || Cu^{2+}(1M)/Cu$. $E^\circ = +1.10$ V. If $[Zn^{2+}] = 0.1$ M, $[Cu^{2+}] = 0.01$ M: $Q = [Zn^{2+}]/[Cu^{2+}] = 10$. $E = 1.10 - (0.0592/2)\log(10) = 1.10 - 0.0296 = 1.0704$ V. Applications: concentration cells (different concentrations of same electrolyte), pH measurement (glass electrode uses Nernst equation for H⁺).

8. Electrolysis — Faraday's Laws

First law: mass deposited $m = ZQ = ZIt$. Second law: masses deposited by same charge are proportional to equivalent weights. Combined: $m = \dfrac{MIt}{nF}$ where $M$ = molar mass, $n$ = n-factor, $F = 96485$ C/mol. For Cu²⁺ (n=2): depositing 1 mol Cu requires $2F = 192970$ C. At 10 A for 1 hour ($= 36000$ C): moles Cu $= 36000/(2 \times 96485) = 0.1865$ mol $= 11.86$ g. Electrolytic series determines what is deposited at cathode (reduction occurs): more positive reduction potential ion deposits first. In CuSO₄ with Cu electrodes: Cu²⁺ reduced at cathode, Cu oxidised at anode → Cu transfers from anode to cathode (electrorefining principle). In CuSO₄ with Pt electrodes: Cu deposits at cathode, O₂ evolved at anode.

Frequently Asked Questions

1. Why does Kw increase with temperature and what does this mean for pH? ⌄

Autoionisation: $H_2O \rightleftharpoons H^+ + OH^-$ is endothermic ($\Delta H \approx +56$ kJ/mol). Increasing temperature provides more energy, enabling more water molecules to ionise. By van't Hoff equation: $d\ln K_w/dT = \Delta H/(RT^2) > 0$, so Kw increases with T. This means at higher temperatures, both [H⁺] and [OH⁻] are larger in pure water. pH is lower (since [H⁺] higher). But water remains neutral because [H⁺] = [OH⁻] still. The confusion arises because "neutral" is often mistakenly equated with "pH 7". Correct: neutral means [H⁺] = [OH⁻], regardless of pH value.

2. What is the significance of Kw = 10⁻¹⁴ at 25°C? ⌄

This value means: in pure water at 25°C, [H⁺] = [OH⁻] = 10⁻⁷ M. pH = pOH = 7. The scale of 0-14 for pH arises from this: diluting strong acid to [H⁺] = 1 M gives pH = 0 (lower limit practical). Adding strong base to [OH⁻] = 1 M gives pOH = 0, pH = 14 (upper limit). The product [H⁺][OH⁻] = 10⁻¹⁴ at 25°C constrains solutions: in acidic solution [H⁺] > 10⁻⁷, so [OH⁻] < 10⁻⁷. In basic: [OH⁻] > 10⁻⁷, [H⁺] < 10⁻⁷. This constraining relationship is why we can always find [OH⁻] from pH (or vice versa) at 25°C.

3. How does pH of blood maintain at 7.4? ⌄

Blood pH 7.4 (slightly basic) is maintained by multiple buffer systems: (1) Bicarbonate buffer ($H_2CO_3/HCO_3^-$, pKa = 6.1): accounts for ~80% of blood buffering capacity. CO₂(aq) + H₂O ⇌ H₂CO₃ ⇌ H⁺ + HCO₃⁻. Lung regulates: [CO₂] (exhalation removes CO₂, reduces acidity). Kidney regulates: [HCO₃⁻] (reabsorption or excretion). (2) Protein buffers: haemoglobin (~10%), plasma proteins (~5%). (3) Phosphate buffer (~5%). (4) Bone carbonate (slow, reserve). Acidosis: pH < 7.35. Alkalosis: pH > 7.45. Even 0.1 pH unit change (from 7.4 to 7.3) can be life-threatening (alters protein folding, enzyme function, O₂ binding to haemoglobin via Bohr effect).

4. What is the Henderson-Hasselbalch equation and when does it apply? ⌄

Henderson-Hasselbalch: pH $= pK_a + \log([A^-]/[HA])$. Derived from: $K_a = [H^+][A^-]/[HA]$. Taking $-\log$: $pH = pK_a - \log([HA]/[A^-]) = pK_a + \log([A^-]/[HA])$. Valid when: (1) Weak acid and its conjugate base both present. (2) Both concentrations are significant (not extreme dilution). (3) $K_a$ is not too large (not a strong acid) or too small (if Ka/C < 0.01, approximation valid). (4) [A⁻]/[HA] ratio between 0.1 and 10 (within buffer range of pKa ± 1). Limitations: does not account for activity coefficients (dilute solutions assumed), doesn't work well for very dilute buffers or for polyprotic acids near multiple pKa values.

5. Why are some salts acidic and some basic in solution? ⌄

When a salt dissolves, the ions may react with water (hydrolysis). The reaction depends on the relative strength of the parent acid and base. If acid (HA) is strong: A⁻ is a weak conjugate base → does not hydrolyse significantly. If acid is weak: A⁻ is relatively strong conjugate base → A⁻ + H₂O ⇌ HA + OH⁻ (hydrolysis) → solution basic. Similarly for cation (from base): if base (BOH) is strong: B⁺ is weak conjugate acid → no hydrolysis. If base is weak: B⁺ + H₂O ⇌ BOH + H⁺ → solution acidic. Rule: strong acid + strong base → neutral (NaCl). Weak acid + strong base → basic (CH₃COONa). Strong acid + weak base → acidic (NH₄Cl). Weak acid + weak base → depends on relative Ka and Kb.

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