Ion Concentration and pH in Detail
Strong Acids      HClO4, HI, HBr, HCl, HNO3 & H2SO4
 
If we use the general symbol, X, to represent the anion of these acids, then the general equation becomes:
 
                      HX + H2O ---> H3O+(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.
 
                O2- and NH2- 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 [H3O+] essentially equals the original concentration of the solute.
 
eg.         a 1.0 x 10-3 mol/L (0.0010 mol/L) HCl solution
 
               HCl      +    H2O --------> H3O+(aq) + Cl-(aq)
  start  0.0010 M                               -------         -------

  after        0.0 M                            0.0010 M     0.0010 M
                                                   (1.0 x 10-3 M)

The pH will be 3, the pOH will be 11.
 

Weak Monoprotic Acids
The general equation for a weak acid is:
 
                        HX + H2O ------> H3O+ + X-
 
Applying the equilibrium law to this equation we get:
 

 
In dilute solutions of acids and bases the [H2O] is essentially a constant at 55.4 mol/L. (1000 mL of H2O = 1000 g of  H2O/18.02 g/mol = 55.4 moles). This, when multiplied with Keq gives us a new constant value which we call Ka.
 
      OR 
 
The Ka values for acids and bases stronger than H3O+ are not listed because of their high degree of dissociation, the denominator would approach zero and result in an infinitely large Ka.
 
Calculate
(a) the [H3O+],
(b) the pH and
(c) the % dissociation for a 0.100 mol/L solution of acetic acid at 25oC, Ka for CH3COOH is 1.8 x 10-5.
 
The equation for the dissociation is:

                 CH3COOH + H2O -------> H3O+(aq) + CH3 COO-(aq)
 

Let 'x' be the concentration of acetic acid that dissociates:

                CH3COOH + H2O -------> H3O+(aq) + CH3 COO-(aq)
   start          0.100 M                            -------            -------
  finish          0.100 - x                               x                      x
 


 
1.8 x 10-3    (x) (x)
                    0.100 - x

1.8 x 10-4 - 1.8 x 10-3 x = x2

Upon rearrangement x2 +1.8 x 10-3x -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 [H3O+] is 1.25 x 10-2 mol/L

b) The pH is 1.9 based upon -log[H3O+]

c) The % dissociation is 1.25 x 10-2 M * 100 = 12.5%
                                      0.100 M
 

            Go to the Acid-Base Ka, Kb, pKa, pKb Worksheet