pH tells us whether a solution is acidic or basic in nature, while pKa determines the extent of acidity present in an aqueous solution. Acidity is a common attribute in both of these concepts. So if we know the pH of a solution, we can find its pKa. But how?
You will find out how to find pH from pKa while reading the article further on and learn to do that yourself, too, through the solved examples we have provided for you.
An acid dissociates to release hydrogen ions (H+) in an aqueous solution. The amount of H+ ions released in the aqueous solution determines the strength of the acid.
The concentration of H+ ions released can be measured against a parameter called pH (power of hydrogen).
The pH of a solution is related to [H+] by the formula given in equation (i).
pH = -log10 [H+] ….Equation (i)
The hydrogen ion concentration of a solution usually varies from 1 to 10-14 g eq./L. When converted into pH, it is represented in numbers from 0 to 14. The greater the acidity of a solution, the higher [H+], so as per equation (i), so it has a lower pH value.
On the pH scale, acidic solutions have a pH ranging from 0 to 6. pH 7 represents a purely neutral solution such as water, while a pH above 7 denotes the basicity of an aqueous solution.
What is pKa?
Ka stands for acid dissociation constant. pKa is the negative logarithm of Ka to the base 10.
A weak acid (HA) partially dissociates to produce H+ and A– ions in water. H+ ions combine with H2O molecules to form hydronium (H3O+) ions. A– is known as the conjugate base of the acid. HA and A– together are known as a conjugate acid-base pair.
The ionization equilibrium for the dissociation of HA in an aqueous solution can be represented as follows:
The equilibrium constant (Ka) for the above reaction can be represented as equation (ii)
Ka = \frac{[H_{3}O^{+}][A^{-}]}{[HA][H_{2}O]}………. Equation (ii)
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 (ii) can be rearranged as equation (iii).
Ka = \frac{[H^{+}][A^{-}]}{[HA]}……….Equation (iii)
pKa is calculated from Ka by taking the negative logarithm of the latter, as shown in equation (iv).
pKa = -log10 Ka ……….Equation (iv)
The greater the strength of an acid, it undergoes dissociation to a large extent. Thus a large number of H+ ions are released in the aqueous solution. The acid thus possesses a high Ka value; however, it has a low pKa value as per equation (iv).
In this way, pKa is inversely related to the strength of an acid, just like pH.
What is the relationship between pH and pKa?
As discussed above, both pH and pKa are inversely related to acidic strength. This implies that pH and pKa are directly related to each other.
Strong mineral acids such as HCl and HNO3 possess low pH and pKa values. Contrarily, weak organic acids such as acetic acid that undergo partial ionization in water possess high pH and pKa values.
How to find pH from pKa? – (pKa to pH)
A diverse range of chemical reactions are performed in the chemistry laboratory. Most of them require constant pH maintenance. Therefore, certain chemical solutions called buffers are used to keep their pH constant. Buffer solutions resist small changes in pH.
Our human body also maintains its pH constant using blood buffers.
A buffer solution is usually composed of a weak acid and a highly ionizable salt solution.
HA only partially dissociates into H+ and A– ions, while the highly ionizable salt MA completely dissociates into M+ and A– ions.
A high concentration of A– ions in the mixture shifts the reversible equilibrium (I) backward by the common ion effect.
This results in a high concentration of the weak acid (HA) and its conjugate base (A–) in the buffer solution.
The pH of an acidic buffer solution is related to its pKa via the Henderson Hasselbalch equation i.e., equation (v) given below.
…………….Equation (v)
Thus, if the pKa and the required concentrations of a weakly acidic solution are known, we can find its pH by applying equation (v) shown above.
You must remember that [A–] and [HA] represent equilibrium concentrations of the conjugate base and the acid, respectively.
Now let’s see through the examples given below; how to use equation (v) and all the other equations given in this article to find pH from pKa in different situations.
Solved examples for finding pH from pKa
Example # 1: Find the pH of 0.200 M benzoic acid (C6H5COOH) that dissociates to produce 0.05 M C6H5COO– ions at equilibrium. Ka for benzoic acid is 6.3 x 10-5.
Benzoic acid (C6H5COOH) is a weak acid that partially ionizes to produce H+ and benzoate (C6H5COO–) ions in water.
As per the balanced chemical equation, 1 mole of C6H5COOH breaks down to yield 1 mole of each of the C6H5COO– and H+ ions at equilibrium.
As per the question statement; [C6H5COOH] initial = 0.200 M, [C6H5COO–] equilibrium = 0.05 M.
Hence; [C6H5COOH] equilibrium = 0.200 – 0.05 = 0.15 M.
Result: The pH of the given benzoic acid solution is 3.72.
Example # 2: A 0.2 M solution of a weak acid undergoes 10 % ionization till it reaches the equilibrium point. Find the pH of the acidic solution.
Acid dissociation constant Ka = 1.0 x 10-5.
10 % ionization implies the ratio[A–]/[HA] = 1/10
The value of Ka is given in the question statement so we can find pKa using equation (iv), as shown below.
pKa = -log10 Ka ……….Equation (iv)
pKa = -log10 (1 x 10-5) = 5
Now, let’s substitute all the above data into equation (v) to find the required pH.
……….Equation (v)
pH = 5 + log10(1/10)
pH = 5 + 1 = 6
Result: The pH of the given acidic solution is 6.
Such a high pH value denotes an extremely weak acidic solution.
Example # 3: A buffer solution consists of 0.65 M acetic acid and 0.25 M sodium acetate at equilibrium. Find the pH of the buffer given that Ka (CH3COOH) = 1.58 x 10-5.
The Ka value is given in the question statement so we can find pKa using equation (iv), as shown below.
pKa = -log10 Ka ……….Equation (iv)
pKa = -log10 (1.58 x 10-5) = 4.80
Result: The pH of the buffer solution is 4.39.
Example # 4: The nitrous acid (HNO2) buffer solution consists of 1.0 M HNO2 and 0.225 M NaNO2 at equilibrium. Find the pH of the buffer solution.
Ka (HNO2) = 5.6 x 10-4.
The Ka value is given in the question statement so we can find pKa using equation (iv), as shown below.
pH stands for the power of hydrogen. It measures the concentration of hydrogen ions (H+) released in an aqueous solution by the formula:
∴ pH = -log10 [H+].
How is pH related to the strength of an acid?
pH is inversely related to the strength of an acid. The stronger the acid, the more H+ ions are released in the aqueous solution; therefore, the negative logarithm of [H+] decreases i.e., the pH decreases.
What is pKa?
pKa is the negative logarithm of the acid dissociation constant i.e., Ka.
pKa = -log10 Ka.
How is pKa related to the strength of an acid?
pKa is inversely related to the strength of an acid. The greater the acidic strength, the higher the Ka value; thus, the negative logarithm of Ka i.e., pKa, decreases.
Are pH and pKa the same?
No, pH and pKa are closely related but have two different chemical parameters.
What is the difference between pH and pKa?
pH determines whether an aqueous solution is acidic or basic in nature. However, pKa tells us whether an acidic solution is strongly acidic or weakly acidic.
Does pKa change with pH?
Yes. A change in pH signifies the concentration of hydrogen ions [H+] present in the aqueous solution changes. An increase or decrease in [H+] directly affects Ka and thus pKa.
What is a buffer solution and how it works?
A buffer solution is a weakly acidic or basic solution that resists changes in pH.
For example, the acetate buffer consists of a large amount of CH3COOH molecules and CH3COO– ions.
CH3COO–Na+ completely ionizes to release CH3COO– ions by an irreversible chemical reaction.
CH3COOH partially ionizes to release CH3COO– and H+ ions, as shown by the reversible reaction below.
When a small amount of an acid is added to the buffer solution, the pH of the solution decreases.
As per Le Chatelier’s principles, the above equilibrium shifts backward to decrease [H+] and thus increasing the pH to the original value. In this way, buffers resist small changes in pH.
How is the pH of a solution related to pKa?
pH is directly related to pKa as per Henderson Hasselbalch’s equation given below.
where [HA]= concentration of the acid and [A–]= concentration of conjugate base of the acid at equilibrium.
Summary
pH stands for the power of hydrogen. It measures the concentration of hydrogen ions (H+) present in an aqueous solution.
pH is related to [H+] by the formula pH = -log10 [H+].
pKa is defined as the negative logarithm to the base 10 of the acid dissociation constant (Ka). pKa = – log10 Ka.
pH tells us whether an aqueous solution is acidic or basic in nature, while pKa determines the strength of an acidic solution.
pH is directly related to pKa.
Both pH and pKa are inversely related to the strength of an acid.
Greater the acidic strength, lower the pH and pKa values and vice versa.
A buffer solution usually consists of a weak acid or a weak base and its highly ionizable salt.
Buffers resist changes in pH by shifting the direction of the dynamic equilibrium, such as to oppose the change.
pH is related to pKa by the Henderson Hasselbalch equation: pH = pKa + log10[A–]/[HA] where [HA]= concentration of the acid and [A–]= concentration of conjugate base of the acid at equilibrium.
If the pKa and required equilibrium concentrations of an aqueous solution are known, we can find pH from pKa by substituting the given values into the Henderson Hasselbalch equation.
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 (v)
