Editing PH (section)

=== Weak acids and bases ===
A [[weak acid]] or the conjugate acid of a weak base can be treated using the same formalism.
* Acid HA: {{chem2|HA <-> H+ + A-}}
* Base A: {{chem2|HA+ <-> H+ + A}}
First, an acid dissociation constant is defined as follows. Electrical charges are omitted from subsequent equations for the sake of generality
: <math chem="">K_a = \frac \ce{[H] [A]}\ce{[HA]}</math>
and its value is assumed to have been determined by experiment. This being so, there are three unknown concentrations, [HA], [H<sup>+</sup>] and [A<sup>−</sup>] to determine by calculation. Two additional equations are needed. One way to provide them is to apply the law of [[mass conservation]] in terms of the two "reagents" H and A.
: <math chem="">C_\ce{A} = \ce{[A]}  +  \ce{[HA]}</math>
: <math chem="">C_\ce{H} = \ce{[H]}  +  \ce{[HA]}</math>

''C'' stands for [[analytical concentration]]. In some texts, one mass balance equation is replaced by an equation of charge balance. This is satisfactory for simple cases like this one, but is more difficult to apply to more complicated cases as those below. Together with the equation defining ''K''<sub>a</sub>, there are now three equations in three unknowns. When an acid is dissolved in water ''C''<sub>A</sub> = ''C''<sub>H</sub> = ''C''<sub>a</sub>, the concentration of the acid, so [A] = [H]. After some further algebraic manipulation an equation in the hydrogen ion concentration may be obtained.
: <math chem="">[\ce H]^2 + K_a[\ce H] - K_a C_a = 0</math>

Solution of this [[quadratic equation]] gives the hydrogen ion concentration and hence p[H] or, more loosely, pH. This procedure is illustrated in an [[ICE table]] which can also be used to calculate the pH when some additional (strong) acid or alkaline has been added to the system, that is, when ''C''<sub>A</sub> ≠ ''C''<sub>H</sub>.

For example, what is the pH of a 0.01&nbsp;M solution of [[benzoic acid]], p''K''<sub>a</sub> = 4.19?
* Step 1: <math>K_a = 10^{-4.19} = 6.46\times10^{-5}</math>
* Step 2: Set up the quadratic equation. <math chem="">[\ce{H}]^2 + 6.46\times 10^{-5}[\ce{H}] - 6.46\times 10^{-7} = 0 </math>
* Step 3: Solve the quadratic equation. <math chem="">[\ce{H+}] = 7.74\times 10^{-4};\quad \mathrm{pH} = 3.11</math>

For alkaline solutions, an additional term is added to the mass-balance equation for hydrogen. Since the addition of hydroxide reduces the hydrogen ion concentration, and the hydroxide ion concentration is constrained by the self-ionization equilibrium to be equal to <math chem="">\frac{K_w}\ce{[H+]}</math>, the resulting equation is:
: <math chem="">C_\ce{H} = \frac{[\ce H] + [\ce{HA}] -K_w}\ce{[H]}</math>