Ion Concentration and pH in Detail 
Strong Acids
HClO_{4}, HI, HBr, HCl, HNO_{3} & H_{2}SO_{4}

If we use the general symbol, X, to represent the anion of
these
acids, then the general equation becomes:

HX + H_{2}O > H_{3}O^{+}(aq) + X^{}(aq)

Dissociation is 100% complete.

Strong Bases Any alkali metal with
hydroxide
is considered to be a strong base and capable of undergoing complete
dissociation.

O^{2} and NH_{2}^{} also react with water to
give OH^{} to completion.

It is easy to calculate the [H^{+}] and the pH of
a strong
acid or base. Since a strong acid is completely dissociated in dilute
solution
the [H_{3}O^{+}] essentially equals the original
concentration
of the solute.

eg. a 1.0
x 10^{3}
mol/L (0.0010 mol/L) HCl solution

HCl + H_{2}O
>
H_{3}O^{+}(aq) + Cl^{}(aq)
start 0.0010 M   after 0.0
M
0.0010 M 0.0010 M
The pH will be 3, the pOH will be 11.

Weak Monoprotic Acids 
The general equation for a weak acid is:

HX + H_{2}O > H_{3}O^{+} + X^{}

Applying the equilibrium law to this equation we get:


In dilute solutions of acids and bases the [H_{2}O]
is essentially
a constant at 55.4 mol/L. (1000 mL of H_{2}O = 1000 g of
H_{2}O/18.02 g/mol = 55.4 moles). This, when multiplied with K_{eq}
gives us a new constant value which we call K_{a}.

OR

The K_{a }values for acids and bases stronger
than H_{3}O^{+}
are not listed because of their high degree of dissociation, the
denominator
would approach zero and result in an infinitely large K_{a}.

Calculate
(a) the [H_{3}O^{+}], (b) the pH and (c) the % dissociation for a 0.100 mol/L solution of acetic acid at 25^{o}C, K_{a }for CH_{3}COOH is 1.8 x 10^{5}. 
The equation for the dissociation is:
CH_{3}COOH + H_{2}O > H_{3}O^{+}(aq)
+ CH_{3
}COO^{}(aq)

Let 'x' be the concentration of acetic acid that
dissociates:
CH_{3}COOH + H_{2}O > H_{3}O^{+}(aq)
+ CH_{3
}COO^{}(aq)


1.8 x 10^{3 }= (x) (x)
0.100  x 1.8 x 10^{4 } 1.8 x 10^{3 }x = x^{2} Upon rearrangement x^{2 }+1.8 x 10^{3}x
1.8 x 10^{4
}=
0

Use the quadratic equation to solve for 'x'
a = 1 b = 1.8 x 10^{3 }c = 1.8 x 10^{4} Therefore the [H_{3}O^{+}] is 1.25 x 10^{2} mol/L b) The pH is 1.9 based upon log[H_{3}O^{+}] c) The % dissociation is 1.25 x 10^{2 }M
*
100 = 12.5%
