# 6.1: 6.1 Dissociation of Water and the pH Scale


An understanding of acid and base chemistry principles, as well as the ability to solve quantitative problems, requires first learning how water behaves and relates to these concepts. We know the formula for water is H2O(l), and that it is a permanent dipole due to the electronegativity of the oxygen atom. The oxygen draws the two pairs of shared electrons closer to it and slightly away from the hydrogen atoms, creating a slightly negative charge around the oxygen atom.

Because the electrons are nearer the oxygen atom, a hydrogen atom can actually ‘pivot’ away from the rest of the molecule and reorient itself in solution, either towards the oxygen atom in another water molecule, associate with another anion, or combine with a hydroxide ion to form a new water molecule. While these reactions are transient and continuous, they are significant enough to be of relevance. We write this reaction, simplified, as:

$\ce{ H2O(l) \rightleftharpoons H^{+} (aq) + OH^{-} (aq)} \nonumber$

For most of our discussion of acid base chemistry, we will drop the phase designations and generally work with aqueous phase ions:

$\ce{ H2O \rightleftharpoons H^{+} + OH^{-}} \nonumber$

In this reaction, one hydrogen ion is “dis-associating” with the rest of the molecule. So, this and other acid reactions like it are called dissociation reactions. For water it is also sometimes called hydrolysis. (‘hydro’ and lysis’) The hydrogen ion, H+ is called a proton. The OH- ion is hydroxide.

Just like other reactions we studied in thermodynamics, this reaction is reversible and has an equilibrium that is sensitive to temperature, etc. So, acid base chemistry is really just, you guessed it, more thermodynamics.

Using the ideal, dilute solution assumption (meaning we can use concentration instead of activity), we can write the expression for the equilibrium constant for this reaction as:

$K_{eq}=\dfrac{[\ce{H^{+}}][\ce{OH^{-}}]}{[\ce{H2O}]} \nonumber$

Because this Keq is for the dissociation of water, we call it Kw. And, because the activity of water is ~1.0, we can re-write the equation:

Using our equation ΔG = -RTlnKeq, we can calculate the value of Kw,298K = 1×10-14.

$K_w=[\ce{H^{+}}][\ce{OH^{-}}] \nonumber$

Now, this equilibrium of water is usually responding to changes in system pH that are caused by other species – strong or weak acids and bases. So, we wouldn’t have a case with pure water where the protons from water had to exactly match the hydroxide ions from water. In fact this is rarely the case. Instead we have solutions that are acidic and basic all the time. There will be other stuff in solution that will take care of conservation of mass and electroneutrality.

Kw is sensitive to temperature, as are all equilibrium constants. We already know how to adjust this value for temperature.

In the term pH, the H refers to the concentrations of protons, H+. The p is a mathematical shorthand, referring to the negative base-10 logarithm. So,

pH = -log[H+]

Although not as common, you could just as easily (and legitimately) us the term ‘pOH’:

pOH = -log[OH-]

Because Kw is an equilibrium constant (or, an equilibrium coefficient, if you consider that it is not truly constant but can change with temperature), we can always relate pH and pOH, or [H+(aq)] and [OH-(aq)]. Let’s do an example.

Example $$\PageIndex{1}$$

What is pH when [H+] = [OH-] ?

Solution

Kw = [H+] [OH-] = 1×10-14,       and [H+] = [OH-] = x,             so x2 = 1×10-14

x = [H+] = 1×10-7                           -log10[H+] = -log10(1×10-7)  = pH                          pH = 7

Neutral pH is when [OH-] = [H+] = 1×10-7. Acidic solutions have [H+] > 1×10-7, or pH < 7. Basic, or alkaline, solutions have [H+] < 1×10-7, or pH >

Exercise $$\PageIndex{1}$$

For these other solutions, fill in the missing values.

 Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 pH = pH = pH = 7 pH = pH = pOH = pOH = 10 pOH = pOH = pOH = [H+] = 1x10-1 M [H+] = [H+] = [H+] = 1x10-9 M [H+] = [OH-] = [OH-] = [OH-] = [OH-] = [OH-] = 1x10-11 M

The table is repeated below, with missing values filled in.

 Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 pH = 1 pH = 4 pH = 7 pH = 9 pH = 3 pOH = 13 pOH = 10 pOH = 7 pOH = 5 pOH = 11 [H+] = 1x10-1 M [H+] = 1x10-4 M [H+] = 1x10-7 M [H+] = 1x10-9 M [H+] = 1x10-3 M [OH-] =  1x10-13 M [OH-] = 1x10-10 M [OH-] = 1x10-7 M [OH-] = 1x10-5 M [OH-] = 1x10-11 M

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Another useful relationship follows from these:

pH + pOH          =            -log[H+] + -log[OH-]      =            -log10-14            =               14