How to calculate pKa from Ka? - ka to pka, Conversion, Formulas, Equations
pK_{a} is the negative logarithm of the acid dissociation constant (K_{a}). It is a more convenient way of measuring the strength of an acidic solution than K_{a}.
You will learn in this article how to calculate the pK_{a} value from K_{a} i.e. (K_{a} to pK_{a} conversion) by applying a very simple but quite valuable chemical formula.
So without any further delay, dive into the article, and let’s start reading!
What is pK_{a}?
The prefix p in pK_{a} stands for power. Just like pH determines the power of hydrogen ions present in an aqueous solution. pK_{a} measures the strength of an acidic solution as the power of the acid dissociation constant (K_{a}).
pK_{a} is calculated by taking the negative logarithm to the base 10 of the K_{a} value for acid, as shown in equation (i).
pK_{a} = -log_{10}K_{a}…………. Equation (i)
Weak organic acids have greater pK_{a} values than strong mineral acids.
pK_{a} is related to pK_{b} i.e., base dissociation constant for an aqueous solution as shown in equation (ii).
pK_{a} + pK_{b} = pK_{w} ……. Equation (ii)
Equation (ii) can be rewritten as equation (iii) by substituting the value of pK_{w} that stays fixed at room temperature and atmospheric pressure.
pK_{w} = water dissociation constant = 14 (at 25°C).
pK_{a} + pK_{b} = 14 ……. Equation (iii)
You may note that K_{a} measures the strength of the acid itself, while pK_{a }is an attribute associated with the strength of the aqueous solution that the acid forms upon ionization.
What is K_{a}?
K_{a} stands for acid dissociation constant.
The ionization equilibrium for the dissociation of a weak acid (HA) in an aqueous solution is represented as follows:
The acid dissociation constant (K_{a}) for the above reaction can be represented as equation (iv).
Ka = \frac{[H_{3}O^{+}][A^{-}]}{[HA][H_{2}O]}………. Equation (iv)
Where;
- [H_{3}O^{+}] = concentration of hydronium ions formed in the aqueous solution
- [A^{–}] = concentration of conjugate base of the acid
- [HA] = acid concentration at equilibrium
- [H_{2}O] = concentration of water
As water concentration stays constant throughout the reaction, while [H_{3}O^{+}] = [H^{+}], i.e., the concentration of H^{+} ions released in the aqueous solution.
So, equation (iv) can be rearranged as equation (v).
Ka = \frac{[H^{+}][A^{-}]}{[HA]}………. Equation (v)
The greater the strength of an acid, the higher the K_{a} value for its aqueous solution and vice versa.
What is the relationship between pK_{a} and K_{a}?
This equation (pK_{a} = -log_{10}K_{a}) tells us that pK_{a} and K_{a} are inversely proportional to each other. A higher K_{a} value results in a lower pK_{a} value and vice versa.
This implies that a strong acid that dissociates to a large extent in an aqueous solution possesses a higher K_{a} value; however, it has a smaller pK_{a}.
So pK_{a} is also inversely related to acidic strength.
How to find pK_{a }from K_{a}? – (K_{a} to pK_{a} conversion)
pK_{a} and K_{a} are interconvertible. The value of pK_{a} can be easily calculated if the value of K_{a} is known by using equation (i) which is (pK_{a} = -log_{10}K_{a}).
We can substitute the value of K_{a} into equation (i) and take the negative logarithm of this value to find pK_{a}.
pK_{a} = -log_{10}K_{a}…………. Equation (i)
It was highlighted at the beginning of the article that pK_{a} is a more convenient way of measuring the acidity of a solution as compared to K_{a}.
This is because K_{a} for acid is usually calculated in magnitudes of 10 raised to the power x. While pK_{a} is a small numerical value.
For instance, the K_{a} for acetic acid is 1.58 x 10^{-5}; contrarily, its pK_{a} is just 4.80.
K_{a} values are always positive. However, pK_{a} values can be both positive and negative.
We have provided you with the following solved examples through which you can learn how to calculate pK_{a} from K_{a} i.e. (K_{a} to pK_{a} conversion) by practically using the formulas given in this article.
Solved examples of determining pK_{a }when K_{a} given
Example #1: The acid dissociation constant (K_{a}) value for propanoic acid (CH_{3}CH_{2}COOH) is 1.34 x 10^{-5}. What is its pK_{a} value? |
CH_{3}CH_{2}COOH is a weak acid that partially ionizes to yield H^{+} and CH_{3}CH_{2}COO^{–} ions in water. As the K_{a} value for CH_{3}CH_{2}COOH is given in the question statement so we can find its pK_{a} by applying the equation as shown below. ∴ pK_{a} = -log_{10}K_{a} ∴ pK_{a} = -log_{10}(1.34 x 10^{-5}) = 4.87 Result: The pK_{a} value for propanoic acid (CH_{3}CH_{2}COOH) is 4.87. |
Example #2: The acid dissociation constant (K_{a}) value for nitric acid (HNO_{3}) is 2.4 x 10^{1}. What is its pK_{a} value? |
HNO_{3} is a strong acid that completely ionizes to give H^{+} and NO_{3}^{–} ions in an aqueous solution. As the K_{a} value for HNO_{3} is given in the question statement so we can find its pK_{a} by applying the equation as shown below. ∴ pK_{a} = -log_{10}K_{a} ∴ pK_{a} = -log_{10}(2.4 x 10^{1}) = -1.38. Result: The pK_{a} value for nitric acid (HNO_{3}) is -1.38. A negative pK_{a} signifies that HNO_{3} is an extremely strong acid. The proton is only weakly held to the acid, which it readily liberates in an aqueous solution. |
Example # 3: A chemist prepared 0.01 M butyric acid solution in his lab. How can you help the chemist find the pK_{a} of the acidic solution if its pH at the equilibrium point is reported in the literature as pH= 4.63? |
Butyric acid, also known as butanoic acid (CH_{3}CH_{2}CH_{2}COOH), is a weak acid that partially dissociates to release H^{+} and butanoate (CH_{3}CH_{2}CH_{2}COO^{–}) ions in an aqueous solution. As the pH of the solution is given in the question statement, so we can find its [H^{+}] at equilibrium using equation (vi). ∴ [H^{+}]_{equilibrium} = 10^{-pH} ………. Equation (vi) ∴ [H^{+}]_{equilibrium} = 10^{-4.63} = 2.34 x 10^{-5} M. As per the balanced chemical equation shown above; ⇒ [H^{+}]_{equilibrium }= [CH_{3}CH_{2}CH_{2}COO^{–}]_{equilibrium } So [CH_{3}CH_{2}CH_{2}COO^{–}]_{equilibrium} = [H^{+}]_{equilibrium }= 2.34 x 10^{-5} M. The initial butyric acid concentration is also given in the question, so [CH_{3}CH_{2}CH_{2}COOH]_{equilibrium }= [CH_{3}CH_{2}CH_{2}COOH]_{initial }– [CH_{3}CH_{2}CH_{2}COO^{–}]_{equilibrium} = 0.01 – (2.34 x 10^{-5}) = 9.98 x 10^{-3} M. Now that we know the acid dissociation constant (K_{a}) value, we can easily find pK_{a} by applying equation (i). ∴ pK_{a} = -log_{10}K_{a}…………. Equation (i) ∴ pK_{a} = -log_{10}(5.49 x 10^{-8}) = 7.26 Result: The pK_{a} value for the butyric acid solution is 7.26. A comparatively high pK_{a} denotes a low acidic strength. |
Example # 4: The K_{a} value for a weak acid is given to be 2.33 x 10^{-11}. Which of the following options provides the correct pK_{a} value for its acidic solution? A) 11.23 B) 10.93 C) 10.63 D) 11.83 E) 9.93 |
Answer: Option C (pK_{a} = 10.63) is the correct answer. Explanation: pK_{a} can be determined from the already given K_{a} value by applying equation (i). pK_{a} = -log_{10}K_{a} …………. Equation (i) pK_{a} = -log_{10}(2.33 x 10^{-11}) = 10.63 |
Example # 5: A 0.025 M solution of a weak acid dissociates to produce 0.005 mol/L hydrogen ions at the equilibrium stage. Find its pK_{a}. |
A weak acid (HA) partially dissociates to liberate H^{+} and A^{–} ions in water. The concentration of hydrogen ions released at equilibrium is given in the question statement i.e., [H^{+}]_{ equilibrium }= 0.005 M. ∴ [H^{+}]_{ equilibrium }= [A^{–}]_{equilibrium }= 0.005 M The original concentration of HA is also given in the question statement, so [HA]_{equilibrium }= [HA]_{initial }– [H^{+}]_{ equilibrium }= 0.025 – 0.005 = 0.02 M. As a final step, we just need to substitute the above value into equation (i) to find the required pK_{a} value. ∴ pK_{a} = -log_{10}K_{a} …………. Equation (i) ∴ pK_{a} = -log_{10}(1.25 x 10^{-3}) = 2.90 Result: The pK_{a} value for the given acidic solution is 2.90. |
Also check:
FAQ
What is K_{a}? |
K_{a} stands for acid dissociation constant. A weak acid (HA) partially dissociates to produce H^{+} and A^{–} ions in water. H^{+} ions combine with H_{2}O molecules to form hydronium (H_{3}O^{+}) ions. A^{–} is known as the conjugate base of the acid. HA and A^{–} together are known as a conjugate acid-base pair. The equilibrium constant for a reversible reaction is the ratio of the product of the concentration of products to the product of reactant concentrations. The ionization equilibrium for the dissociation of HA in an aqueous solution can be represented as follows: ⇒ HA + H_{2}O ⇌ H_{3}O^{+} + A^{–} The equilibrium constant (K_{a}) for the above reaction can be represented as equation (i) ⇒ Ka = \frac{[H_{3}O^{+}][A^{-}]}{[HA][H_{2}O]}………. Equation (i) Where;
As water concentration stays constant throughout the reaction, while [H_{3}O^{+}] = [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, it undergoes dissociation to a larger extent in the aqueous solution; thus, it possesses a higher K_{a} value. Weak organic acids such as acetic acid and citric acid have K_{a} values below 1. However, strong mineral acids such as HCl and HNO_{3} that completely dissociates to release a large number of H^{+} ions in an aqueous solution have K_{a} values above 1. |
What is pK_{a}? |
pK_{a} measures the strength of an acidic solution. It is the negative logarithm of the K_{a} value for an acid. |
How is pK_{a} related to the strength of a Bronsted acid? |
As per the Bronsted-Lowry acid-base theory, an acid is defined as a proton donor. The proton is loosely held to a strong acid which can readily liberate it in an aqueous solution. However, weak acids strongly hold their protons, not ready to release in water. Greater the pK_{a} value, the more strongly held the proton is by the Bronsted acid indicating a weak acidic strength and vice versa. So pK_{a} is inversely related to the strength of a Bronsted acid. |
How is K_{a} related to the strength of an acid? |
Higher the K_{a} value, the greater the strength of an acid. So the acid dissociation constant is directly related to the strength of an acid. Strong acids completely ionize in water to release H^{+} ions. Thus, they have high K_{a} values. Weak acids only partially ionize in water to release a limited number of H^{+} ions, so they have low K_{a} values. Also read: How to tell if an acid or base is strong or weak? |
What is the relationship between pK_{a} and K_{a}? |
pK_{a} is inversely related to K_{a}. The higher the K_{a} value of an acid, the lower its pK_{a} and vice versa. |
What is the formula to calculate pK_{a} from K_{a}? |
pK_{a} can be calculated by taking the negative logarithm of K_{a} as follows; ∴ pK_{a} = -log_{10}K_{a}. |
How do you calculate K_{a} from pK_{a}? |
K_{a} and pK_{a} are interconvertible chemical parameters. K_{a} can be calculated from pK_{a} by taking the antilog of pK_{a} as follows; ∴ K_{a} = 10^{-pKa} |
Summary
- pK_{a} stands for the power of K_{a}. It measures the acidic strength of an aqueous solution.
- It is calculated as a negative logarithm of K_{a} to the base 10.
- pK_{a }= -log_{10}K_{a}.
- The greater the strength of an acid, the lower its pK_{a} value, and vice versa.
- K_{a} denotes acid dissociation constant. It measures the extent of ionization of an acid in an aqueous solution.
- Greater the K_{a} value, the higher the strength of the acid.
- If the value of K_{a} for acid is known, we can easily find its pK_{a} value by taking the negative logarithm of K_{a}. The formula to determine pK_{a} from K_{a} is [pK_{a }= -log_{10}K_{a}].
References
- Master Organic Chemistry. (2012, May 9). Acid-Base Reactions: Ka and pKa. Retrieved from https://www.masterorganicchemistry.com/2012/05/09/acid-base-reactions-ka-and-pka/
- wikiHow. (n.d.). How to Find Ka from pKa. Retrieved from https://www.wikihow.com/Find-Ka-from-pKa
- Study.com. (n.d.). Converting Between Ka and pKa: Questions. Retrieved from https://study.com/skill/practice/converting-between-ka-and-pka-questions.html
- Chegg. (n.d.). The formula for calculating pKa from Ka value. Retrieved from https://www.chegg.com/homework-help/questions-and-answers/3-formula-calculating-pka-ka-value-pka-logka
About the author
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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