5.3 Auto-ionization of Water
Water naturally auto-ionizes to produce equal concentrations of hydronium and hydroxide. The extent of this reaction is very small.
\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq)\]
\[K_{\mathrm{w}} = [\mathrm{H_3O^+}][\mathrm{OH^-}] = 1.0\times 10^{-14} ~~~(\mathrm{at}~25~^{\circ}\mathrm{C})\]
Note that water is amphiprotic in this reaction. One water donates a proton while another water accepts the proton.
The relationship of the strength of an acid to its conjugate base is related through Kw.
\[K_{\mathrm{a}} \times K_{\mathrm{b}} = K_{\mathrm{w}}\]
or in log form
\[\mathrm{p}K_{\mathrm{a}} + \mathrm{p}K_{\mathrm{b}} = \mathrm{p}K_{\mathrm{w}}\]
We can see this relationship clearly by writing out the acid- and base-ionization reactions and adding them together.
\[\begin{align*} \color{green}{\mathrm{HA}}(aq) + \mathrm{H_2O}(aq) &\rightleftharpoons \mathrm{H_3O^+}(aq) + \color{red}{\mathrm{A^-}}(aq) \\ \color{red}{\mathrm{A^-}}(aq) + \mathrm{H_2O}(aq) &\rightleftharpoons \mathrm{OH^–}(aq) + \color{green}{\mathrm{HA}}(aq)\\[1.5ex] \mathrm{2H_2O}(l) &\rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq) \end{align*}\]
Recall that adding two equations together has an equilibrium constant that is the result of multiplying the original equilibrium constants together.
\[\begin{align*} K_{\mathrm{w}} &= K_{\mathrm{a}} \times K_{\mathrm{b}} \\[1.5ex] &=\dfrac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\color{red}{\mathrm{A}^{-}}\right]}{[\color{green}{\mathrm{HA}}]} \times \dfrac{\left[\color{green}{\mathrm{HA}}\right]\left[\mathrm{OH}^{-}\right]}{[\color{red}{\mathrm{A^-}}]}\\[2.5ex] &= [\mathrm{H_3O^+}][\mathrm{OH^-}] \end{align*}\]
Increasing the temperature of water slightly shifts Kw to the right.
| T (°C) | Kw | pKw | pH | pOH |
|---|---|---|---|---|
0 |
1.14 × 10–15 |
14.94 |
7.47 |
7.47 |
10 |
2.93 × 10–15 |
14.53 |
7.27 |
7.26 |
20 |
6.81 × 10–15 |
14.17 |
7.08 |
7.08 |
25 |
1 × 10–14 |
14 |
7 |
7 |
30 |
1.47 × 10–14 |
13.83 |
6.92 |
6.92 |
40 |
2.92 × 10–14 |
13.54 |
6.77 |
6.77 |
50 |
5.48 × 10–14 |
13.26 |
6.63 |
6.63 |
100 |
5.13 × 10–13 |
12.29 |
6.14 |
6.14 |
Notice that pKw is equal to 14 at only one temperature… 25 °C. For all temperatures, pH is equal to pOH for pure water. This means that for any temperature, pure water is a neutral solution.
Endo or Exothermic?
Is the auto-ionization of water an endo or exothermic process? If water is heated, which way will the reaction go according to Le Chatelier’s Principle?
\[\mathrm{2H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH}^-(aq) \quad K_{\mathrm{w}} = [\mathrm{H_3O^+}][\mathrm{OH^-}]\]
According to the data, pKw decreases as temperature increases. Recall that Kw and pKw have an inverse relationship. If pKw decreases, Kw is increasing (with increasing temperature). Therefore, we can conclude that this process is endothermic.
Practice
What is the pH and pOH of pure water at 10 °C? Kw(H2O) = 2.93 × 10–15 (at 10 °C).
Solution
For the reaction
\[2\mathrm{H_2O}(l) \rightleftharpoons \mathrm{H_3O^+}(aq) + \mathrm{OH^-}(aq)\] Solve for x
\[\begin{align*} [\mathrm{H_3O^+}][\mathrm{OH^-}] &= K_{\mathrm{w}} \\[1.25ex] x^2 &= 2.93\times 10^{-15} \\[1.25ex] x &= 5.41\times 10^{-8}~M \\[1.25ex] [\mathrm{H_3O^+}] = [\mathrm{OH^-}] &= 5.41\times 10^{-8}~M \end{align*}\]
Find pH and pOH
\[\begin{align*} \mathrm{pH} &= -\log[\mathrm{H_3O^+}] \\[1.25ex] &= -\log(5.41\times 10^{-8}) \\[1.25ex] &= 7.27 \\[2ex] \mathrm{pOH} &= -\log[\mathrm{OH^-}] \\[1.25ex] &= -\log(5.41\times 10^{-8}) \\[1.25ex] &= 7.27 \\[2ex] \end{align*}\]
Practice
What is the Kb for the conjugate base of acetic acid at 50 °C?
Ka(CH3COOH) = 1.633 × 10–5 and Kw(H2O) = 5.48 × 10–14 (at 50 °C).
Solution
Kb(CH3COO–) = 3.36 × 10–9 at 50 °C.
\[\begin{align*} K_{\mathrm{a}} \times K_{\mathrm{b}} &= K_{\mathrm{w}} \\[1.5ex] K_{\mathrm{b}} &= \dfrac{K_{\mathrm{w}}}{K_{\mathrm{a}}} \\[1.5ex] &= \dfrac{5.48\times 10^{-14}}{1.633\times 10^{-5}} \\[1.5ex] &= 3.36\times 10{-9} \end{align*}\]
Math String: 5.48e-14 / 1.633e-5