Kw

Auto-Ionization of Water and Equilibrium Constant Derivation

  • Pure water continuously undergoes auto-ionization (self-ionization), establishing a dynamic equilibrium between liquid water molecules and aqueous hydrogen and hydroxide ions:   H2O(l)⇌H+(aq)+OH−(aq)\text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{OH}^-(aq)

  • The standard equilibrium expression (KcK_c) for this dissociation is formulated as:   Kc=[H+][OH−][H2O]K_c = \frac{[\text{H}^+][\text{OH}^-]}{[\text{H}_2\text{O}]}

  • In pure water and dilute aqueous solutions, the concentration of liquid water is effectively constant, and its thermodynamic activity is defined as 11.

    • Under International Baccalaureate (IB) chemistry specifications, students are required to know that the activity of liquid water equals 11; the underlying thermodynamic derivations for this convention are addressed at the university level.

  • Substituting an activity of 11 for liquid water simplifies the equilibrium expression to the ion product constant of water (KwK_w):   Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-]

  • Because hydrogen ions in aqueous solution exist as hydronium ions (H3O+\text{H}_3\text{O}^+), the expression is also written as:   Kw=[H3O+][OH−]K_w = [\text{H}_3\text{O}^+][\text{OH}^-]

Temperature Dependence of KwK_w

  • Like all equilibrium constants, KwK_w is temperature-dependent.

  • Consequently, the exact concentrations of [H+][\text{H}^+] and [OH−][\text{OH}^-] present in pure water depend directly on temperature.

  • The auto-ionization of water is an endothermic process, meaning KwK_w increases as temperature rises:

    • At 0∘C0^\circ\text{C}: Kw=1.14×10−15K_w = 1.14 \times 10^{-15}

    • At 10∘C10^\circ\text{C}: Kw=2.93×10−15K_w = 2.93 \times 10^{-15}

    • At 20∘C20^\circ\text{C}: Kw=6.81×10−15K_w = 6.81 \times 10^{-15}

    • At 21∘C21^\circ\text{C}: Kw=0.684×10−14K_w = 0.684 \times 10^{-14}

    • At 25∘C25^\circ\text{C} (Standard Ambient Temperature and Pressure / SATP): Kw=1.008×10−14K_w = 1.008 \times 10^{-14}

    • At 30∘C30^\circ\text{C}: Kw=1.471×10−14K_w = 1.471 \times 10^{-14}

    • At 40∘C40^\circ\text{C}: Kw=2.916×10−14K_w = 2.916 \times 10^{-14}

    • At 50∘C50^\circ\text{C}: Kw=5.476×10−14K_w = 5.476 \times 10^{-14}

    • At 100∘C100^\circ\text{C}: Kw=5.13×10−13K_w = 5.13 \times 10^{-13}

Quantitative Analysis and Calculated Values Across Temperatures

  • In pure, distilled, or deionized water, the stoichiometry of auto-ionization dictates that equal molar quantities of hydrogen and hydroxide ions are produced. Therefore, in pure water:   [H+]=[OH−][\text{H}^+] = [\text{OH}^-]

  • Quantitative Evaluation at 21∘C21^\circ\text{C}:

    • Kw=0.684×10−14K_w = 0.684 \times 10^{-14}

    • Setting x=[H+]=[OH−]x = [\text{H}^+] = [\text{OH}^-]:     0.684×10−14=x20.684 \times 10^{-14} = x^2

    • Solving for xx:     x=0.684×10−14=8.2704293×10−8 mol dm−3x = \sqrt{0.684 \times 10^{-14}} = 8.2704293 \times 10^{-8}\,\text{mol}\,\text{dm}^{-3}

    • Concentrations:     [H+]=[H3O+]=8.2704293×10−8 mol dm−3≈8.27×10−8 mol dm−3[\text{H}^+] = [\text{H}_3\text{O}^+] = 8.2704293 \times 10^{-8}\,\text{mol}\,\text{dm}^{-3} \approx 8.27 \times 10^{-8}\,\text{mol}\,\text{dm}^{-3}     [OH−]=8.2704293×10−8 mol dm−3≈8.27×10−8 mol dm−3[\text{OH}^-] = 8.2704293 \times 10^{-8}\,\text{mol}\,\text{dm}^{-3} \approx 8.27 \times 10^{-8}\,\text{mol}\,\text{dm}^{-3}

    • pH Calculation:     pH=−log⁡([H+])\text{pH} = -\log([\text{H}^+])     pH=−log⁡(8.2704293×10−8)=7.08247195...≈7.08\text{pH} = -\log(8.2704293 \times 10^{-8}) = 7.08247195... \approx 7.08

    • Result: Pure water at 21∘C21^\circ\text{C} has a neutral pH of 7.087.08

  • Quantitative Evaluation at 100∘C100^\circ\text{C}:

    • Kw=5.13×10−13K_w = 5.13 \times 10^{-13}

    • Concentration Calculation:     [H+]=[OH−]=5.13×10−13=7.16240183×10−7 mol dm−3[\text{H}^+] = [\text{OH}^-] = \sqrt{5.13 \times 10^{-13}} = 7.16240183 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}

    • Adjusted for significant figures:     [H+]=7.16×10−7 mol dm−3[\text{H}^+] = 7.16 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}     [OH−]=7.16×10−7 mol dm−3[\text{OH}^-] = 7.16 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}

    • pH Calculation:     pH=−log⁡(7.16240183×10−7)=6.14\text{pH} = -\log(7.16240183 \times 10^{-7}) = 6.14

  • Quantitative Evaluation at 30∘C30^\circ\text{C}:

    • Kw=1.471×10−14K_w = 1.471 \times 10^{-14}

    • Equilibrium setup:     1.471×10−14=x21.471 \times 10^{-14} = x^2     x=1.471×10−14=1.213×10−7 mol dm−3x = \sqrt{1.471 \times 10^{-14}} = 1.213 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}

    • Concentrations:     [H+]=1.213×10−7 mol dm−3[\text{H}^+] = 1.213 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}     [OH−]=1.213×10−7 mol dm−3[\text{OH}^-] = 1.213 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}

    • Logarithmic indices:     pH=−log⁡(1.213×10−7)=6.92\text{pH} = -\log(1.213 \times 10^{-7}) = 6.92     pOH=−log⁡(1.213×10−7)=6.92\text{pOH} = -\log(1.213 \times 10^{-7}) = 6.92

  • Quantitative Evaluation at 10∘C10^\circ\text{C}:

    • Kw=2.93×10−15K_w = 2.93 \times 10^{-15}

    • Concentration:     [H+]=2.93×10−15=5.37×10−8 mol dm−3[\text{H}^+] = \sqrt{2.93 \times 10^{-15}} = 5.37 \times 10^{-8}\,\text{mol}\,\text{dm}^{-3}

    • pH Calculation:     pH=−log⁡(5.37×10−8)=7.27\text{pH} = -\log(5.37 \times 10^{-8}) = 7.27

  • Quantitative Evaluation at 50∘C50^\circ\text{C}:

    • Kw=5.476×10−14K_w = 5.476 \times 10^{-14}

    • Concentrations:     [H+]=5.476×10−14=2.340×10−7 mol dm−3[\text{H}^+] = \sqrt{5.476 \times 10^{-14}} = 2.340 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}     [OH−]=2.340×10−7 mol dm−3[\text{OH}^-] = 2.340 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3}

    • pH Calculation:     pH=−log⁡(2.340×10−7)=6.631\text{pH} = -\log(2.340 \times 10^{-7}) = 6.631

Defining Neutrality across Temperature Ranges

  • A solution is defined as neutral if and only if [H+]=[OH−][\text{H}^+] = [\text{OH}^-]. Pure, distilled, or deionized water is always neutral at any temperature.

  • The numerical pH corresponding to neutrality changes as temperature changes:

    • The neutral pH point is 7.007.00 only at 25∘C25^\circ\text{C}.

    • As temperature decreases, the neutral pH value increases.

    • As temperature increases, the neutral pH value decreases.

  • Neutral pH values at specific temperatures:

    • At 0∘C0^\circ\text{C}: Neutral pH =7.47= 7.47

    • At 10∘C10^\circ\text{C}: Neutral pH =7.27= 7.27

    • At 20∘C20^\circ\text{C}: Neutral pH =7.08= 7.08

    • At 25∘C25^\circ\text{C}: Neutral pH =7.00= 7.00

    • At 30∘C30^\circ\text{C}: Neutral pH =6.92= 6.92

    • At 40∘C40^\circ\text{C}: Neutral pH =6.77= 6.77

    • At 50∘C50^\circ\text{C}: Neutral pH =6.63= 6.63

    • At 100∘C100^\circ\text{C}: Neutral pH =6.14= 6.14

  • Application Examples for Solution Classification at Non-Standard Temperatures:

    • At 10∘C10^\circ\text{C} (Neutral threshold pH =7.27= 7.27):

    • A solution with a pH of 7.27.2 is acidic (because 7.2<7.277.2 < 7.27).

    • Solutions are basic at 10∘C10^\circ\text{C} only if their pH exceeds 7.277.27

    • At 30∘C30^\circ\text{C} (Neutral threshold pH =6.92= 6.92):

    • A solution with a pH of 7.07.0 is basic (because 7.0>6.927.0 > 6.92).

    • Solutions are acidic at 30∘C30^\circ\text{C} only if their pH is below 6.926.92

    • At 50∘C50^\circ\text{C} (Neutral threshold pH =6.63= 6.63):

    • Solution with pH=8.00\text{pH} = 8.00 is basic (8.00>6.638.00 > 6.63)

    • Solution with pH=7.10\text{pH} = 7.10 is basic (7.10>6.637.10 > 6.63)

    • Solution with pH=7.00\text{pH} = 7.00 is basic (7.00>6.637.00 > 6.63)

    • Solution with pH=6.90\text{pH} = 6.90 is basic (6.90>6.636.90 > 6.63)

    • Solution with pH=6.00\text{pH} = 6.00 is acidic (6.00<6.636.00 < 6.63)

Calculating Concentration and pOH in Aqueous Solutions

  • The equation Kw=[H+][OH−]K_w = [\text{H}^+][\text{OH}^-] allows for the direct calculation of [H+][\text{H}^+] or [OH−][\text{OH}^-] in any aqueous solution provided KwK_w at that temperature and one ion concentration are known.

  • Sample Problem 1: Fresh Milk at 25∘C25^\circ\text{C}:

    • Given: Fresh milk has a hydrogen ion concentration [H+]=2.00×10−7 mol dm−3[\text{H}^+] = 2.00 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3} at 25∘C25^\circ\text{C}.

    • Goal: Calculate [OH−][\text{OH}^-] and classify as acidic or basic.

    • Execution: Because milk is predominantly water, Kw=1.008×10−14K_w = 1.008 \times 10^{-14} at 25∘C25^\circ\text{C}.     1.008×10−14=[H+][OH−]1.008 \times 10^{-14} = [\text{H}^+][\text{OH}^-]     1.008×10−14=(2.00×10−7)[OH−]1.008 \times 10^{-14} = (2.00 \times 10^{-7})[\text{OH}^-]     [OH−]=1.008×10−142.00×10−7=5.04×10−8 mol dm−3[\text{OH}^-] = \frac{1.008 \times 10^{-14}}{2.00 \times 10^{-7}} = 5.04 \times 10^{-8}\,\text{mol}\,\text{dm}^{-3}

    • Classification: Since [H+]=2.00×10−7 mol dm−3[\text{H}^+] = 2.00 \times 10^{-7}\,\text{mol}\,\text{dm}^{-3} is greater than [OH−]=5.04×10−8 mol dm−3[\text{OH}^-] = 5.04 \times 10^{-8}\,\text{mol}\,\text{dm}^{-3}, fresh milk is acidic.

  • Sample Problem 2: Coca-Cola at 10∘C10^\circ\text{C}:

    • Given: Coca-Cola has a hydrogen ion concentration [H+]=0.0031 mol dm−3[\text{H}^+] = 0.0031\,\text{mol}\,\text{dm}^{-3} at 10∘C10^\circ\text{C}.

    • Goal: Calculate [OH−][\text{OH}^-] and the pOH of the solution.

    • Execution: Using Kw=2.93×10−15K_w = 2.93 \times 10^{-15} at 10∘C10^\circ\text{C}:     2.93×10−15=(0.0031)[OH−]2.93 \times 10^{-15} = (0.0031)[\text{OH}^-]     [OH−]=2.93×10−150.0031=9.45×10−13 mol dm−3[\text{OH}^-] = \frac{2.93 \times 10^{-15}}{0.0031} = 9.45 \times 10^{-13}\,\text{mol}\,\text{dm}^{-3}

    • pOH Calculation:     pOH=−log⁡([OH−])\text{pOH} = -\log([\text{OH}^-])     pOH=−log⁡(9.45×10−13)=12.02\text{pOH} = -\log(9.45 \times 10^{-13}) = 12.02