Ka stands for acid dissociation constant. It determines the extent of ionization of an acid in an aqueous solution. In contrast, pH refers to the power of hydrogen. It is a measure of the H+ ions released by an acid dissociation.
The greater the extent of ionization of an acid, the more H+ ions will be released. The aqueous solution of the given acid will thus have a higher Ka value but a smaller pH.
Strong mineral acids such as HCl ionize completely and therefore have a low pH. Contrarily, weak organic acids such as acetic acid break down to a small extent in an aqueous solution and therefore possess comparatively higher pH values.
In this article, you will learn how to find the Ka of an acidic solution if its pH is known.
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).
………. Equation (ii)
The greater the strength of an acid, the higher its Ka value, and vice versa.
What is pH?
As mentioned at the beginning of the article, pH refers to the power of hydrogen. It is used to determine the acidity or the basicity of a solution.
Acidic solutions have low pH values ranging from 0-6, while basic solutions have a high pH value ranging from 8-14.
A neutral solution in which [H+] = [OH–] has a pH =7.
Equation (iii) given below relates the pH of a solution to the concentration of H+ ions present in it.
pH = -log [H+] …. Equation (iii)
If the pH of a solution is known, we can determine the concentration of hydrogen ions present in it by taking antilog and rearranging equation (iii) as equation (iv).
[H+] = 10-pH ……… Equation (iv)
What is the relationship between Ka and pH?
The acid dissociation constant (Ka) is inversely proportional to the pH of an acidic solution.
As per equation (iv), the [H+] increases as the value of pH decreases. An increase in [H+] denotes the acid ionizes to a large extent; thus, it possessed a higher Ka value.
How to calculate Ka from pH? (pH to Ka)
You may notice that both equation (iv) and equation (ii) have a common entity, i.e., [H+]. So if the pH of a solution is known, we can find its [H+].
This [H+], along with the concentration of acid (HA) and its conjugate base (A–) at equilibrium, can be used to determine the Ka value for the given acid.
You may understand this concept more clearly through the examples given below.
Solved examples for finding Ka from pH?
Example # 1:The pH of 1 M solution of acetic acid (CH3COOH) is 2.4. Find the Ka value for CH3COOH.
Acetic acid (CH3COOH) dissociates to produce acetate (CH3COO–) ions and hydrogen (H+) ions in an aqueous solution.
The initial concentration of CH3COOH = [HA]initial = 1 M.
Step I:
As the pH of the solution is given so, we can easily find [H+] by using equation (iv).
[H+] = 10-pH ……… Equation (iv)
[H+] = 10-2.4 = 3.98 x 10-3 M.
The reaction equation above shows that 1 mole acetic acid breaks down to produce 1 mole H+ and 1 mole CH3COO– ions.
Thus at equilibrium, [H+] = [CH3COO–] = 3.98 x 10-3 M.
Step II:
Now the concentration of acetic acid left undissociated at equilibrium can be determined by subtracting [A–]equilibrium from [HA]Initial
As calculated in the previous step, [A–]equilibrium = [CH3COO–] equilibrium = 3.98 x 10-3 M.
So, [CH3COOH] equilibrium = 1- (3.98 x 10-3) = 0.996 M.
Step III:
Result: The Ka value for 1 M acetic acid under the above conditions is 1.59 x 10-5.
Example # 2:Calculate the Ka value of 0.2 M hydrofluoric acid (HF) with a pH of 4.88.
In the question statement, the initial concentration of HF is given, i.e., 0.2 M.
Step I:
As the pH of the solution is also given so, we can easily find [H+] using equation (iv).
[H+] = 10-pH ……… Equation (iv)
[H+] = 10-4.88 = 1.32 x 10-5 M.
The reaction equation above shows that 1 mole HF breaks down to produce 1 mole H+ and 1 mole F– ions. Thus at equilibrium, [H+] = [F–] = 1.32 x 10-5 M.
Step II:
Now the concentration of HF left undissociated at equilibrium can be determined by subtracting [A–]equilibrium from [HA]Initial
As determined in the previous step, [A–]equilibrium = [F–] equilibrium = 1.32 x 10-5 M.
So, [HF] equilibrium = 0.2- (1.32 x 10-5) = 0.199 M.
Step III:
Result: The Ka value for 0.2 M HF as per the above data is 8.8 x 10-10.
Example # 3:John recorded that the pH of a 0.1 M solution of carbonic acid equals 4.68. How can he find the acid dissociation constant (Ka) for this acid using its pH value?
So a unique aspect of this example is that H2CO3 is a diprotic acid.
1 mole H2CO3 releases 2 mole H+ ions.
The initial concentration of H2CO3 is given, i.e., 0.1 M.
Step I:
As the pH of the solution is also given so, we can easily find [H+] using equation (iv).
[H+] = 10-pH ……… Equation (iv)
[H+] = 10-4.68 = 2.09 x 10-5 M.
The reaction equation above shows that 1 mole of H2CO3 breaks down to produce 1 mole of CO32- ions and 2 moles of H+ ions. Thus at equilibrium, 2 [H+] = [CO32-]
[CO32-] = ½ [H+] = ½ (2.09 x 10-5) = 1.04 x 10-5 M.
Step II:
Now the concentration of H2CO3 left undissociated at equilibrium can be determined by subtracting [A–]equilibrium from [HA]Initial
As calculated in the previous step, [A–]equilibrium = [CO32-] equilibrium = 1.04 x 10-5 M.
So, [H2CO3] equilibrium = 0.1 – (1.04 x 10-5) = 0.099 M.
Step III:
Result: The Ka value for 0.1 M carbonic acid, as per the above data, is 4.59 x 10-14.
You may note that the extremely small Ka value calculated above shows that carbonic acid is a very weak acid. So instead of losing both of its protons together, under normal conditions, the carbonic acid usually loses a single proton at a time to produce HCO3– and H+.
This is known as the first ionization of carbonic acid, whose Ka value is 4.2 x 10-7.
Ka stands for acid dissociation constant. It represents the extent of ionization of usually a weak acid in an aqueous solution.
What is an acid?
An acid is defined as a chemical substance that releases H+ ions in an aqueous solution.
It is also defined as a proton donor as per the Bronsted-Lowry theory. However, the Lewis theory defines acids as electron pair acceptors.
How Ka relates to the strength of an acid?
The acid dissociation constant (Ka) is directly related to the strength of an acid.
The greater the acidic strength, the higher the Ka value, as a strong acid undergoes dissociation to a large extent in an aqueous solution.
What is pH?
pH refers to the power of hydrogen. It is a numerical representation of the acidity or basicity of a solution. The greater the acidity of a solution, the lower its pH.
How do you find the concentration of H+ ions in a solution if its pH is given?
[H+] is related to pH by the formula given below.
pH = -log [H+] or [H+] = 10-pH
If the pH of a solution is known, we can easily determine the concentration of H+ ions present in it by taking the antilog of the numerical value, as shown in the above formula.
What is the relation between Ka and pH?
Ka is inversely related to pH.
The greater the strength of an acid, it undergoes dissociation to a large extent; therefore, it has a high Ka value. However, it has a low pH value.
How to determine Ka from pH?
The following simple steps can be followed to calculate the Ka of acid from the pH of its aqueous solution.
i) Determine [H+] by applying the formula: [H+] = 10-pH
ii) Determine the equilibrium concentration of the acid (HA) and its conjugate base (A–) by using the mole ratio from the balanced chemical equation and [H+] calculated in the first step.
iii) Find Ka by plugging in all the required values into the equation given below.
Summary
Ka stands for acid dissociation constant. It represents the extent of ionization of an acid in an aqueous solution.
pH denotes the power of hydrogen. It measures the concentration of H+ ions present in an aqueous solution.
pH = -log [H+].
If the pH of an acidic solution is known, we can easily determine its Ka by plugging in the required concentration values into the equation given below.
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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………. Equation (ii)
