Ka, Kw, Kb - The Acid, Water, and Base Dissociation Constant for AP Chemistry (page 2)

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By — McGraw-Hill Professional
Updated on Feb 9, 2011


Because the concentration of the hydronium ion, H3O+, can vary tremendously in solutions of acids and bases, a scale to easily represent the acidity of a solution was developed. It is called the pH scale and is related to the [H3O+]:

    pH = –log [H3O+] or –log [H+] using the shorthand notation

Remember that in pure water Kw = [H3O+][OH] = 1.0 × 10–14. Since both the hydronium ion and hydroxide ions are formed in equal amounts, the Kw expression can be expressed as:

Solving for [H3O+] gives us [H3O+] = 1.0 × 10–7. If you then calculate the pH of pure water:

The pH of pure water is 7.00. On the pH scale this is called neutral. A solution whose [H3O+] is greater than in pure water will have a pH less than 7.00 and is called acidic. A solution whose [H3O+] is less than in pure water will have a pH greater than 7.00 and is called basic. Figure 15.2 shows the pH scale and the pH values of some common substances.

The pOH of a solution can also be calculated. It is defined as pOH = –log[OH]. The pH and the pOH are related:

    pH+pOH= pKw =14.00 at 25°C

Acid–Base Equilibrium

In any of the problems above in which [H+] or [OH] was calculated, you can now calculate the pH or pOH of the solution.

You can estimate the pH of a solution by looking at its [H+]. For example, if a solution has an [H+] = 1 × 10–5, its pH would be 5. This value was determined from the value of the exponent in the [H+].

Kb—The Base Dissociation Constant

Weak bases (B), when placed into water, also establish an equilibrium system much like weak acids:

The equilibrium constant expression is called the weak base dissociation constant, Kb, and has the form:

The same reasoning that was used in dealing with weak acids is also true here: [HB+] = [OH]; [HB] ≈ Minitially; the numerator can be represented as [OH]2; and knowing the initial molarity and Kb of the weak base, the [OH] can easily be calculated. And if the initial molarity and [OH] are known, Kb can be calculated.

For example, a 0.500 M solution of ammonia has a pH of 11.48. What is the Kb of ammonia?

The Ka and Kb of conjugate acid–base pairs are related through the Kw expression:

    Ka × Kb = Kw

This equation shows an inverse relationship between Ka and Kb for any conjugate acid–base pair.

This relationship may be used in problems such as: Determine the pH of a solution made by adding 0.400 mol of strontium acetate to sufficient water to produce 2.000 L of solution.


    The initial molarity is 0.400 mol/2.000 L = 0.200 M.

    When a salt is added to water dissolution will occur:

The resultant solution, since strontium acetate is soluble, has 0.200 M Sr2+ and 0.400 M C2H3O2

Ions such as SrSr2+, which come from strong acids or strong bases, may be ignored in this type of problem. Ions such as C2H3O2, from weak acids or bases, will undergo hydrolysis. The acetate ion is the conjugate BASE of acetic acid (Ka = 1.8 × 10–5). Since acetate is not a strong base this will be a Kb problem, and OH will be produced. The equilibrium is:

Determining Kb from Ka (using Kw = KaKb = 1.0 × 10–14 gives:

with x = 1.5 × 10–5 = [OH], and pH = 9.180

Practice problems for these concepts can be found at:

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