Aqueous Solutions of Strong Acids and Bases

Autoionization of Water

Pure water is a rather poor conductor of electricity because it contains very few ions to permit the conduction of electrical current.  Water does ionize into H3O+ and OH- ions to a very small extent.  A process we can represent by the following equilibrium:

H2O(l)  +  H2O(l)  <==> H3O+(aq)  +  OH-(aq) 

 We call this process the autoionization of water.  The equilibrium constant expression for the autoionization reaction is:

     [H3O+][OH-]
K = -----------
       [H2O]2
The concentration of water in aqueous solutions is typically very large (about 55M) and remains essentially constant for dilute solutions.  We therefore exclude the concentration of water from equilibrium constant expressions for aqueous solutions, just as we exclude the concentration of pure solids and liquids from equilibrium constant expressions for heterogeneous reactions.  Thus the equilibrium constant expression for the autoionization of water can be written as

Kw = K[H2O]2 = [H3O+][OH-]

This important equilibrium constant is called the ion-product constant for water.  At room temperature Kw has a value of 1.0 x 10 -14  Although the equilibrium between H3O+ and OH- (as well as other ionic equilibria) is affected somewhat by the presence of additional ions in solution, it is the customary practice to ignore these ionic effects except in very accurate work.  Thus our expression and value for Kw above is taken to be valid for aqueous solutions as well as for pure water and it can be used to calculate either  [H3O+] or [OH-] if one of these concentrations is known.  A solution for which   [H3O+] = [OH-] is said to be neutral.  In most solutions  [H3O+] and [OH-] are not equal.  As the concentration of one of these  ions increases, the concentration of the other must decrease, so that the ion product equals 1.0 x 10-14.  In acidic solutions [H3O+] exceeds [OH-].  In basic solutions, the reverse is true:   [OH-] exceeds  [H3O+]. 

Solutions of a Strong Acid or Base

Suppose we dissolve 1.0 mol of the strong acid, HBr in 1.0 L of aqueous solution, giving 1.0 M HBr.  We would like to know the concentration of H3O+ ion in this solution.  Since HBr is a strong acid it is 100% ionized and so there will be 1.0 mol H3O+ ion from the HBr.  The autoionization of water will contribute a small amount of H3O+ ion.  In pure water, the concentration of H3O+ from autoionization of water is 1.0 x 10-14 ;  in an acidic solution, the contribution of  H3O+ from water will be even smaller.  This is because of Le Chatelier's principle, when you increase the concentration  of  H3O+ in water by adding an acid, the self-ionization of water reverses until a new equilibrium is established.

This means that in a solution of a strong acid, you can usually ignore the autoionization of water as a source of  H3O+ ion.  This will be true unless you are working in an extremely dilute ( 10-6 M or so) solution.  Even though you can usually ignore the  autoionization of water in determining the H3O+  concentration in a solution of a strong acid, the equilibrium still exists and is responsible for the small concentration of OH- ion.  The following example illustrates this concept:

  Example 1                                 

Calculate the OH- ion concentration in 0.20 M HNO3.  Based on the discussion above we let the concentration of  H3O+ ion = 0.20 M, the same as the strong acid HNO3.  Next we substitute this value into the Kw expression and solve for [OH-].

 Kw = [H3O+][OH-] = 1.0 x 10-14 = (0.20)[OH-] 

[OH-] = 1.0 x 10-14 / 0.20 = 5.0 x 10-14 M

  Example 2

Calculate the H3O+ ion concentration in 0.20 M Ca(OH)2.  Since Ca(OH)2 is a strong base the concentration of OH- will be 0.40 M (since each mole of Ca(OH)2 produces 2 mol of OH-).  The concentration of H3O+ is obtained from:

 Kw = [H3O+][OH-] = 1.0 x 10-14 = [H3O+](0.40) 

[H3O+] = 1.0 x 10-14 / 0.40 = 2.5 x 10-14 M

 

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