Module 8

Lesson 3.1  The Acid Ionization Constant, Ka



Key Concepts


Quantitatively, you can consider an equilibrium solution of a weak acid in water to behave much the same as the equilibrium systems you studied in Module 7. In other words, you can use the equilibrium law expression to describe the reaction and calculate an equilibrium constant. However, a special kind of equilibrium constant is used to describe the equilibrium position of a Bronsted-Lowry reaction in which a weak acid reacts with water. This equilibrium constant is known as Ka.   Ka indicates the extent to which an acid ionizes in solution.


Refer to the Ka values listed in the table of "Relative Strengths of Acids and Bases".


You will notice that, for the first six acids ("strong acids"),  the Ka values are listed simply as "very large".  Why is this?


As was the case for calculating Kc, Ka is calculated by substituting equilibrium concentrations into the equilibrium law expression.


For strong acids, the acid reacts quantitatively with water to form hydronium ions. Therefore, the equilibrium concentration of hydronium ions is equal to the initial concentration of the acid.

For example, consider a 0.10 mol/L solution of HCl:





We could have also used a single arrow here to indicate a quantitative reaction.


Substituting these equilibrium concentrations into the equilibrium law expression yields the following:





Note that the concentration of liquid water is omitted for the Ka expression. We assume that the value remains virtually constant.


At equilibrium, only negligible amounts of un-dissociated acid (HCl) remain in the reaction vessel. Because virtually all the acid has ionized, the numerator is very large relative to the denominator. For this reason, the value of Ka is very large. This is the same scenario for any of the six strong acids.


In contrast to the strong acids, the Ka values for weak acids are very small and vary from weak acid to weak acid. For a weak acid, only a small proportion of the initial acid concentration will be converted to hydronium ions. Therefore, the equilibrium concentration of hydronium ions is far less than the initial concentration of the weak acid.


Consider the reaction between acetic acid and water:






We can write the equilibrium law for this acid-base reaction as follows:




Note that the concentration of liquid water is omitted for the Ka expression. We assume that the value remains virtually constant.


For the reaction of a 0.10 mol/L CH3COOH solution, only 1.3 % of the CH3COOH molecules react (SATP conditions). In this case, the denominator is a relatively large number (compared to the numerator) because the denominator represents the equilibrium concentration of un-dissociated acetic acid molecules. For weak acids, Ka values are extremely small. The smaller the Ka, the weaker the acid.


To calculate Ka values and amount concentrations for weak acids, use ICE tables. Recall from Module 7 that ICE tables are an effective way to represent the changes that occur within a chemical system as it moves from its initial setup to equilibrium concentrations.


Utilize the "Rule of Thumb" (page 741 of textbook) when doing these calculations. The Rule of Thumb states that the initial concentration of the weak acid will remain unchanged at equilibrium if the initial concentration of the acid is at least 1000 times greater than its Ka value. As you can see in Sample Problem 16.3 on page 740 to 741, using this approximation can greatly simplify the math!


The instructional video below demonstrates, step by step, how to solve various types of problems using Ka and the equilibrium law expression.


Watch


Read pages 737 to 742 of the textbook.

Check Your Understanding


Complete Practice Questions  5 and 6 on page 743.


Page 743 Practice Question 5

  1. \( \mathrm { \%~ reaction = \dfrac{ [H_3O^+(aq)] }{ [propanoic ~ acid(aq)] } \times 100 = \dfrac{1.16 \times 10^{-3} \frac{mol}{L} }{ 0.100~ \frac{mol}{L} } \times 100 = 1.16\% } \)


  2. Amount Concentration [C2H5COOH(aq)]
    (mol/L)     		[C2H5COO-(aq)]
    (mol/L)     		[H3O+]
    (mol/L)
    Initial 0.100 0 0
    Change -1.16 x 10-3 +1.16 x 10-3 +1.16 x 10-3
    Equilibrium 0.099 1.16 x 10-3 1.16 x 10-3

    \( \mathrm { K_a = \dfrac{[C_2H_5COO^-(aq)] [H_3O^+(aq)] }{ [C_2H_5COOH(aq)] } } \)

    \( \mathrm { K_a = \dfrac{[1.16 \times 10^{-3} \frac{mol}{L}] [1.16 \times 10^{-3} \frac{mol}{L}] }{ [0.099 \frac{mol}{L}] } = 1.36 \times 10^{-5} } \)


  3. As with all equilibrium constants, the value is constant only at the temperature at which the data was collected.

Page 743 Practice Question 6

  1. \( \mathrm { [H_3O^+(aq)] = 10^{-2.43} = 3.7 \times 10^{-3} \frac{mol}{L} } \)

    \( \mathrm { \%~reaction = \dfrac{ [H_3O^+(aq)] }{ [\text{2-hydroxypropanoic acid(aq)}] } \times 100 } \)

    \( \mathrm { \%~reaction = \dfrac{ 3.7 \times 10^{-3} \frac{mol}{L} }{ 0.10 \frac{mol}{L} } \times 100 = 3.7\% } \)


  2. Amount [C2H4OHCOOH(aq)]
    (mol/L)         		[C2H4OHCOO-(aq)]
    (mol/L)         		[H3O+(aq)]
    (mol/L)
    Initial 0.10 0 0
    Change -3.7 x 10-3 +3.7 x 10-3 +3.7 x 10-3
    Equilibrium 0.10
    = 0.10 - (3.7 x 10-3) =
    0.096     		3.7 x 10-3 3.7 x 10-3

    \( \mathrm { K_a = \dfrac{[C_2H_4OHCOO^-(aq)] [H_3O^+(aq)] }{ [C_2H_4OHCOOH(aq)] } } \)

    \( \mathrm { K_a = \dfrac{[3.7 \times 10^{-3} \frac{mol}{L}] [3.7 \times 10^{-3} \frac{mol}{L}] }{ [0.10 \frac{mol}{L}] }= 1.4 \times 10^{-4} } \)


  3. The addition of the hydroxyl group increases the Ka by a factor of 10. Therefore, 2-hydroxypropanoic acid ionizes ten times more than propanoic acid.