Autoionization of water

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Intros
Lessons
  1. What is auto-ionization?
  2. Autoionization of water.
  3. The autoionization expression.
  4. Calculations using the expression.
  5. Why is neutral water pH 7?
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Examples
Lessons
  1. Understand the Kw expression for the ionic product of water.
    1. Write the expression for the ionic product of water, Kw, at 25oC. Include the equation for the dissociation of water.
    2. Explain why, at 45oC, a sample of water has a pH below 7 but it is still neutral.
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Topic Notes
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In this lesson, we will learn:

  • To recall the equation for the autoionization of water.
  • How to use the autoionization expression to find the concentration of ions.
  • How the autoionization expression changes with temperature and gives the pH value of pure water.

Notes:

  • Recall that when acids and bases react, a salt and water is formed. When strong bases and strong acids react, the reaction is more vigorous and gives off more heat energy because neutralization is an exothermic process.
    Removing the salt-forming ions which are spectator ions, this can be expressed in the equation:

    H+(aq) + OH- (aq) β‡Œ\rightleftharpoons H2O (aq)

  • The reverse of this reaction is known as the autoionization of water, symbol Kw (also known as the ionic product of water) – there is an equilibrium between neutral water and the dissociated H3O+ and OH- ions in solution. See the equation:

    2 H2O (l) β‡Œ\rightleftharpoons H3O+ (aq) + OH- (aq)

  • The ionic product Kw is a very small value (almost all the equilibrium mixture is still water, the reactant) but it is noticeable and has a constant value at 25oC:

    Kw = [H3O+ (aq)] [OH-(aq)] = 1.00 * 10 -14 at 25oC

    This ionic product can be indexed using a negative logarithm:

    pKw = -log(Kw) = 14 at 25oC

    This is used just like pH can be used to get the following equation:

    pKw = pH + pOH = 14 at 25oC

  • The Kw expression is just an equilibrium constant expression, with [H2O(l)] cancelled out. As [H3O+] increases, [OH-] decreases so their product stays equal to 1 * 10-14 constant value at 25oC.
    • This is chemically sound too; if an acid HX was added to neutral water, dissociation into protons, H+, and the conjugate base X- would occur.
    • These protons would quickly react to form H3O+ by protonating neutral H2O molecules and the very small amount of OH- ions, decreasing [OH-] further. As [H3O+] increases, a decrease in [OH-] would therefore be observed.
    • The addition of acid (or base) will neutralize the majority of the base (or acid) present in neutral water, causing the equilibrium to shift to the left and form more water. However, the initial amounts of [H3O+] and [OH-] are extremely small, orders of magnitude smaller than any standard solutions of acid/base to be added to neutral water. As such, the concentration of the added acid/base is effectively unchanged by initial [H3O+] or [OH-]. Assume [H3O+] = [HX] or [OH-] = [:B] where [HX] or [:B] is the concentration of a strong acid or base added to neutral water.

  • Remember that the Kw constant holds at 25oC only.
    This means in neutral water at 25oC, [H3O+] = 1 * 10-7 M and [OH-] = 1 * 10-7 M.
    As stated above though, neutralization is an exothermic process which means that the autoionization of water is an endothermic process. At higher temperatures, the equilibrium shifts right (Le Chatelier’s principle!), leading to greater Kw as the product of [H3O+] * [OH-] increases. Even though the water is still neutral, this greater dissociation at higher temperatures leads to higher Kw and lower pH, because pH is defined as –log [H3O+].