Ka : The acid ionization constant


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Important note: all constants refered to: Kc, Kw, Ka, and Kb are temperature-dependant. All discussions are assumed to be at 25 °C, i.e. standard temperature.

Basic Information

1) Weak acids are less than 100% ionized in solution.
2) Acetic acid (formula = HC2H3O2) is the most common weak acid example used by instructors.
3) Another way to write acetic acid's formula is CH3COOH.
4) A common abbreviation for acetic acid is HAc, where Ac¯ refers to the acetate polyatomic ion.

The following equation describes the reaction between acetic acid and water:

HAc + H2O <==> H3O+ + Ac¯

Note that it is an equilibrium condition.

The equilibrium constant for this reaction is written as follows:

Kc = ( [H3O+] [Ac¯] ) / ( [HAc] [H2O] )

However, in pure liquid water, [H2O] is a constant value. To demonstrate this, consider 1000 mL of water with a density of 1.00 g/mL. This 1.00 liter (1000 mL) would weigh 1000 grams. This mass divided by the molecular weight of water (18.0152 g/mol) gives 55.5 moles. The "molarity" of this water would then be 55.5 mol / 1.00 liter or 55.5 M.

The solutions studied in introductory chemistry are so dilute that the "concentration" of water is unaffected. So 55.5 molar can be considered to be a constant if the solution is dilute enough.

Moving [H2O] to the other side gives:

Kc [H2O] = ( [H3O+] [Ac¯] ) / [HAc]

Since the term Kc [H2O] is a constant, let it be symbolized by Ka, giving:

Ka = ( [H3O+] [Ac¯] ) / [HAc]

This constant, Ka, is called the acid ionization constant. It can be determined by experiment and each acid has its own unique value. For example, acetic acid's value is 1.77 x 10¯5.

From the chemical equation above, it can be seen that H3O+ and Ac¯ concentrations are in the molar ratio of one-to-one. This will have an important consequence as we move into solving weak acid poblems.


Go to Solving Ka Problems: Part One

Go to Solving Ka Problems: Part Two

Go to Solving Ka Problems: Part Three

Go to a listing of many types of acid base problems and their solutions

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