Ka and Kb are known as ionization constants. A strong acid or base undergoes complete ionization in an aqueous solution. However, weak acids and bases dissociate to a small extent in an aqueous solution. The dissociation of an acid release H+ ions, while the dissociation of a base liberates OH– ions in water.
The extent of ionization of a weak acid or a base can be determined by calculating their Ka and Kb values, respectively.
The acid dissociation constant (Ka) and base dissociation constant (Kb) are interrelated parameters.
In this article, you will learn how to calculate Ka from Kb with solved examples. But before that, let us introduce you to what is Ka and Kb. So, continue reading!
The ionization equilibrium for the dissociation of a weak acid (HA) in an aqueous solution is represented as follows:
The acid dissociation constant (Ka) for the above reaction can be represented as equation (i).
Ka = \frac{[H_{3}O^{+}][A^{-}]}{[HA][H_{2}O]}………. Equation (i)
Where;
[H3O+] = concentration of hydronium ions formed in the aqueous solution
[A–] = concentration of conjugate base of the acid
[HA] = acid concentration at equilibrium
[H2O] = concentration of water
As water concentration stays constant throughout the reaction, while [H3O+] = [H+], i.e., the concentration of H+ ions released in the aqueous solution. So, equation (i) can be rearranged as equation (ii).
Ka = \frac{[H^{+}][A^{-}]}{[HA]}……….Equation (ii)
The greater the strength of an acid, the higher the Ka value for its aqueous solution and vice versa.
What is Kb?
Kb represents the base dissociation constant.
The ionization equilibrium for the dissociation of a weak base (B) in an aqueous solution is represented as:
The base dissociation constant (Kb) for the above reaction can be represented as equation (iii).
The greater the strength of a base, the higher the Kb value for its aqueous solution.
What is the relationship between Ka and Kb?
As discussed at the beginning of the article, Ka and Kb are interconvertible. For any acid-base chemical reaction, if the value of Kb for the base is known, we can easily determine Ka for its conjugate acid and vice versa.
The chemical entity that relates Ka to Kb is known as the water dissociation constant (Kw), as shown in equation (v) given below.
Ka. Kb = Kw………. Equation (v)
⇒ Kw= [H+] [OH–]
The value of Kw is fixed at 25°C, i.e., room temperature. Kw = 1.0 x 10-14.
As per equation (v), Ka and Kb are inversely proportional to each other. This means the stronger an acid, the weaker its conjugate base, and vice versa.
Taking the negative log on both sides of equation (v) gives us:
-log (Ka.Kb) = -log Kw
-[log Ka + log Kb] = -log Kw
-log Ka – log Kb = – log Kw
As –log Ka = pKa, -log Kb = pKb and –log Kw = pKw = 14.
So another relationship between Ka, Kb, and Kw is
pKa + pKb = 14….Equation (vi)
How to find Ka from Kb and Kw?
Equation (v) can be rearranged to make Ka the subject of the formula. As the value of Kw is fixed so, if the value of Kb is known, we can easily calculate the acid dissociation constant, i.e., Ka. Let’s see how through the examples given below.
Solved examples for calculating Ka from Kb?
Example # 1:Sodium propanoate (CH3CH2COO–Na+) is the conjugate base of propanoic acid (CH3CH2COOH). The Kb value for CH3CH2COO–Na+ = 7.46 x 10-10 at 25°C. How can we find the Ka for propanoic acid using this information?
CH3CH2COOH and CH3CH2COO–Na+ are a conjugate acid-base pair.
The Kb value for the conjugate base is given in the question statement. We know that Kw = 1.0 x 10-14 at 25°C, so we can use the equation given below to determine Ka.
⇒ Ka = Kw/Kb
Plugging in the given values.
∴ Ka =(1.0 × 10-14)/(7.46 x 10-10) = 1.34 x 10-5.
Result: The acid dissociation constant (Ka) value for propanoic acid is 1.34 x 10-5.
Example # 2: The Kb value for acetate (CH3COO–) ions produced by the dissociation of 1 M acetic acid is 5.56 x 10-10. What is the Ka for CH3COOH at room temperature (25°C)?
The Kb value for acetate ions is given. CH3COO– is the conjugate base of acetic acid (CH3COOH).
So we can easily apply the following formula to determine Ka.
⇒ Ka = Kw/Kb
Plugging in the given values.
∴ Ka = (1.0 × 10-14)/(5.56 x 10-10) = 1.8 x 10-5.
Result: The acid dissociation constant (Ka) value for 1M acetic acid is 1.8 x 10-5, as per the data given above.
Example # 3: F– is the conjugate base of HF. The pKb value for F– at 25°C is 10.83. Use this information to find Ka and pKa for HF.
The pKa value for HF can be calculated by applying equation (vi) given below.
pKa + pKb = 14…………. Equation (vi)
The pKb is given in the question statement, i.e., 10.83, so pKa = 14 –pKb = 14-10.83 = 3.17
Now that the pKa for HF is known, we can either directly find its Ka by applying the following formula:
Method # 1:
pKa = -log Ka or Ka = 10-pKa
Ka = 10-3.17 = 6.76 x 10-4.
In contrast, we can also find Ka by applying the formula Ka = Kw/Kb. But for that, we first need to find Kb from pKb given in the question statement.
Method # 2:
Kb = 10-pKb
Kb = 10-10.83 = 1.48 x 10-11
Now,Ka = (1.0 × 10-14)/(1.48 x 10-11) = 6.76 x 10-4.
The same answer in both cases ensures that both the methods used above are correct.
Result: The Ka for hydrofluoric acid (HF) is 6.76 x 10-4, as per the above information.
Example # 4: A chemist recorded the pOH of a 0.50 M aqueous solution to be 11.5. He says the concentration of the conjugate acid formed in this solution is 0.25 M. Can you help him determine the Ka of the conjugate acid?
The information provided in the question above may seem unrelated at first. But don’t worry; we can definitely help the chemist. However, for that, we first need to know some more formulae.
The pOH of an aqueous solution is related to pKb by equation vii given below.
pOH = pKb + log[BH+]/[B] …………..Equation vii
You may note that the base (B) concentration and the concentration of conjugate acid (BH+) formed are already given in the question statement.
[B] = 0.50 M and [BH+] = 0.25 M
The value of pOH is also given. pOH = 11.5
So we can plug in all the given information in equation vii to determine pKb.
Result: The Ka for the conjugate acid formed by the acid-base reaction in the above example is 6.3 x 10-3.
You may note that a high pOH value such as 11.5 represents a weak base; thus, the conjugate acid formed from this base is strong with a comparatively high Ka value than all the other example questions.
Example # 5:The Kb value for the hypobromite (BrO–) ion is 4.0 x 10-6. Which of the following options gives the correct Ka value for hypobromous acid?
A) 1 x 10-14 C) 8.5 x 10-8
B) 5.6 x 10-9 D) 2.5 x 10-9 E) 2.5 x 10-6
Answer:Option D is the correct answer. The acid dissociation constant (Ka) for hypobromous acid is 2.5 x 10-9.
Explanation:
Hypobromite (BrO–) ion is the conjugate base of hypobromous (HBrO) acid.
The Kb value for BrO– is given in the question statement. So we can apply the following formula to determine the Ka value for HBrO as shown below.
Ka is defined as the acid dissociation constant. It determines the extent of ionization of usually a weak acid in an aqueous solution.
A higher Ka value denotes that the respective acid breaks down to a large extent in water, i.e., it has a higher strength.
What is Kb?
Kb stands for the base dissociation constant that determines the extent of ionization of a base in an aqueous solution.
What is Kw?
Kw represents the water dissociation constant. The autoionization of water produces H+ and OH– ions. The ionization equilibrium for the dissociation of water into its constituent ions can be represented as follows:
The value of Kw is fixed at room temperature (25 °C) i.e., Kw = 1.0 x 10-14.
How can we calculate Ka from Kb and Kw?
Ka is related to Kb and Kw by the formula: Kw = Ka.Kb.
The above equation can be rearranged to make Ka the subject of the formula.
∴ Ka = Kw/Kb
As the value of Kw is fixed, so if we know the value of Kb we can easily determine Ka by the above formula.
How are Ka and Kb related to the strength of an acid or a base?
The higher the Ka or Kb value, the greater the strength of the respective acid or base.
Summary
The extent of ionization of a weak acid or base in an aqueous solution can be determined using their Ka and Kb values, respectively.
Ka represents the acid dissociation constant.
Kb stands for base dissociation constant.
Greater the strength of an acid or base, the higher the respective dissociation constant values because the latter ionizes to a larger extent in an aqueous solution.
Ka is related to Kb by the equation Kw = Ka.Kb.
If the value of Kb for a base is known, the Ka value for its conjugate acid can be determined by rearranging the expression:Kb = Kw/Ka.
Ammara Waheed is a highly qualified and experienced chemist, whose passion for Chemistry is evident in her writing. With a Bachelor of Science (Hons.) and Master of Philosophy (M. Phil) in Physical and Analytical Chemistry from Government College University (GCU) Lahore, Pakistan, with a hands-on laboratory experience in the Pakistan Council of Scientific and Industrial Research (PCSIR), Ammara has a solid educational foundation in her field. She comes from a distinguished research background and she documents her research endeavors for reputable journals such as Wiley and Elsevier. Her deep knowledge and expertise in the field of Chemistry make her a trusted and reliable authority in her profession. Let's connect - https://www.researchgate.net/profile/Ammara-Waheed
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