Acids and Bases (Acid-Base Equilibria) - Strong Acids and Bases, Weak Acids and Bases, and pH Calculations

Acids and bases can be simply explained according to their ability to donate H⁺ or OH^- ions. There are few theories that describe what acids and bases are.

Arrhenius theory

According to Arrhenius theory acids are chemical substances that dissociate in water and donate H⁺ ions to the medium. And bases are chemical substances that dissociate in water and donate OH^- ions to the medium.

Bronsted Lowry theory

According to the Bronsted Lowry theory acid is any substance that has the ability to donate H⁺ ions. And bases are substances that have the ability to donate OH^- ions.

Lewis theory

According to the Lewis theory chemical substances that contain empty orbital and are capable to accept electron pair is known as Lewis acids. Lewis bases are substances that will donate electron pairs.

OH^- ions can donate electron pairs. Therefore, OH^- is a Lewis base. H⁺ accept electron pair. Therefore, H⁺ is a Lewis acid. According to the Lewis theory NH₃, H₂O, CN^-, etc are Lewis bases. AlCl₃, BCl₃, and BeCl3 are examples of Lewis acids.

Strong acids and weak acids

Acids are substances that donate H⁺ ions or accept electron pairs.

Strong acids

In an aqueous medium, strong acids will dissociate completely and donate H⁺ ions to the medium.

 E.g,

• Hydrochloric acid – HCl
• Sulfuric acid - H₂SO₄
• Nitric acid - HNO₃

Weak acids

But weak acids will not dissociate completely. They will dissociate partially, and the dissociation reaction is reversible. Therefore, there are undissociated acid molecules and dissociated ions in a weak acid solution.

e.g.,

• Acetic acid – CH₃COOH
• Oxalic acid
• Phosphoric acid

Let’s consider a weak acid, HA. It will dissociate in the aqueous medium as follows.

We can write the equilibrium constant for the concentration for the above reversible reaction as follows.

In the above equation, the concentration of the water remains constant. So, we can write a new constant as K_a as follows.

K_a is known as the dissociation constant of the acid. It only depends on the temperature. According to the K_a value of a particular acid, we can predict whether the acid is strong or weak. If the dissociation constant of an acid is very high, that means the acid is strong.

When the K_a value is decreasing, the acidic strength is also decreasing. Acids with K_a values larger than 1 are considered strong acids. Acids with K_a values less than 1 are considered weak acids.

Strong bases and weak bases

Bases are substances that donate OH^- ions to the medium or donate electron pairs.

Strong bases

Strong bases will completely dissociate in an aqueous medium and donate OH^- ions to the media.

e.g.

• Sodium hydroxide - NaOH
• Potassium hydroxide - KOH

Weak bases

Weak bases are bases that will dissociate partially in an aqueous media. The dissociation of a weak base is reversible. In a weak base solution, both undissociated base molecules and dissociated ions can be found at once.

e.g

• Ammonium hydroxide - NH₄OH

Let’s consider a weak base, B. It will dissociate in the aqueous medium as follows.

Let’s write the equilibrium constant of concentration for the above equilibrium.

In the above equation, the concentration of the water remains constant. Therefore, for both K_c and [H₂O] constants, we can write a single constant as K_b as follows.

K_b is the dissociation constant of the base. As same in acids, we can predict the basic properties of a base according to the K_b value.

If the k_b value is higher than 1 that means the base is strong. If the k_b value is less than 1, the base is weak. K_b only depends on the temperature.

Ionic product of water

Even pure water can conduct electricity. To conduct electricity there should be free electrons or free ions in the media. Therefore, even in pure water, there should be ions that are responsible for conducting electricity. Pure water can be ionized as follows,

or

Let’s write the equilibrium constant of concentration for the above equation.

As the concentration of the water remains constant, we can write a new constant as K_w as follows.

This constant is known as the ion product of water. At 25℃, the ion product of the pure water is 1.01×10⁻¹⁴ mol² dm^-6. K_w only depends on the temperature. When increasing the temperature, the dissociation is increased. Therefore, the ion concentration also increased. Hence the K_w is increased.

Let’s find the concentrations of H⁺ ions and OH^- ions of pure water at 25℃.

Since the concentrations of H⁺ ions and OH^- ions are equal. Let’s take it as “x”. so, the above equation can be written as follows.

In pure water the concentrations of H⁺ ions and OH^- ions are equal. When changing the temperature ion concentrations will be changed. But at any temperature, the concentration of H⁺ ions and OH^- ions will be equal in pure water.

If the concentration of H⁺ ions is higher than OH^- ions, the water is acidic. If the OH^- concentration is higher than the H⁺ ions, the water is basic. Pure water is neutral.

pH value

pH value is used to express the acidic, basic properties of water.

Where [H⁺] is the hydrogen ion concentration of the water. As same in the pH value, the pOH value is expressed using the concentration of OH^- ions in the water.

Most of the calculations included in the below sections are basic calculations that can use to find the pH value that is closer to the actual value. But to find a more accurate answer, we have to use some more advanced calculations. You can find out more advanced calculations that can be used to find more accurate answers for the pH in this article.

Problem 01

• Find the pH and pOH of pure water at 25℃.

At 25℃, pH and pOH are 7.

Relation between the pH and pOH

At 25℃,

Although the concentration of H⁺ ions and OH^- ions change, the sum of pH and pOH equals 14. (at 25℃)

Problem 02

• Find the pH of the following solutions at 25℃.
  • i. 0.1 mol dm^-3 HCl
  • ii. 0.01 mol dm^-3 HNO₃
  • iii. 0.025 mol dm^-3 H₂SO₄
  • iv. 1 × 10^-8 mol dm^-3 of HNO₃

• i. 0.1 mol dm^-3 HCl

Hydrochloric acid will completely dissociate and give H⁺ ions to the water. Water molecules in the medial also partially dissociated and give H⁺ ions to the medium.

H⁺ ions that are given by the dissociation of HCl = 0.1 mol dm^-3

H⁺ ions given by the partial dissociation of water = 1 × 10^-7 mol dm^-3

Compared to the H⁺ ions given by HCl, H⁺ ions from the dissociation of water is negligible. Therefore, we take the concentration of H⁺ ions given by HCl, as the total concentration of H⁺ ions.

1. ii. 0.01 mol dm^-3 HNO₃

In this situation also the H⁺ ions from the dissociation of water are negligible. Therefore, we take only the H⁺ ions which are given by the dissociation of HNO₃ acid.

Concentration of H⁺ ions = 0.01 mol dm^-3

1. iii. 0.025 mol dm^-3 H₂SO₄

H₂SO₄ will dissociate as follows. H⁺ ions given by the dissociation of water are neglected here.

Since one H₂SO₄ molecule gives two H⁺ ions to the media, the concentration of H⁺ ions will be = 0.05 mol dm^-3

1. iv. 1 × 10^-8 mol dm^-3 of HNO₃

In his situation, the H⁺ ions given by the dissociation of water are not negligible.

H⁺ ions given by the dissociation of water = 1.0 × 10^-7 mol dm^-3

H⁺ ions given by the dissociation of HNO₃ = 1.0 × 10^-8 mol dm^-3

Total H⁺ ion concentration = 1.1 × 10^-7 mol dm^-3

Problem 03

• Find the pH of the following basic solutions at 25℃.

1. i. 1 mol dm^-3 NaOH

OH^- ions given by the partial dissociation of water have been neglected here.

• ii. 0.01 mol dm^-3 KOH

OH^- ions given by the partial dissociation of water have been neglected here.

• iii. 0.009 mol dm^-3 Ba(OH)₂

Concentration of OH^- ions = 2 × 0.009 mol dm^-3 = 0.018 mol dm^-3

OH^- ions given by the partial dissociation of water have been neglected here.

1. iv. 1 × 10^-9 mol dm^-3 NaOH

Since the concentration of NaOH is very low, the OH^- ions given by the partial dissociation of water cannot be neglected here.

Finding the pH weak acids

Let’s consider a monoprotic weak acid of HA with an initial concentration of C mol dm^-3 and the dissociation of α mol dm^-3. The acid will be partially dissociated as follows.

The concentrations of each component can be expressed as follows.

α is the dissociation of the acid per 1 mol. If we have an acid solution with 1 mol dm^-3 concentration, the concentration of the acid after the dissociation would be 1- α.

Since we have a C mol dm^-3 solution, we should multiply 1- α by C, to get the concentration after dissociation. We can write the Ka expression for the above equilibrium as follows.

When compared to 1, α is very small. Therefore, 1- α is approximately equal to 1. So, the above expression can be recreated as follows,

Where,

• K_a = dissociation constant of the weak acid
• C = initial concentration of the acid
• α = dissociation of the weak acid.

If we neglect the H⁺ ions given from the partial dissociation of water, the concentration of the H⁺ ions would be Cα. So, the pH of the media can be found as follows,

Since the summation of pH and pOH is 14 at 25℃, we can find the pOH of the weak acid as well.

Ostwald’s dilution law

Ostwald’s dilution law states that the dissociation of a weak acid is inversely proportional to the concentration of the acid at a given temperature.

Problem 04

• Find the pH of the CH₃COOH solution with the initial concentration of 0.01 mol dm^-3 at 25℃. The dissociation of CH₃COOH at 25℃ is 1.8 × 10^-5 mol dm^-3. CH3COOH will partially dissociate in the water as follows,

Finding the pH and pOH of weak bases

Let’s consider a monoprotic weak base of BOH with an initial concentration of C mol dm^-3 and the dissociation of β mol dm^-3. The base will be partially dissociated as follows.

The concentrations of each component can be expressed as follows.

β is the dissociation of the base per 1 mol. If we have a base solution with 1 mol dm^-3 concentration, the concentration of the base after the dissociation would be 1 - α.

Since we have a C mol dm^-3 solution, we should multiply 1 - α by C, to get the concentration after dissociation. We can write the K_b expression for the above equilibrium as follows.

When compared to 1, β is very small. Therefore, 1 - α is approximately equal to 1. So, the above expression can be recreated as follows,

Where,

• K_b = dissociation constant of the weak base
• C = initial concentration of the base
• β = dissociation of the weak base.

If we neglect the OH^- ions given from the partial dissociation of water, the concentration of the OH^- ions would be Cβ. So, the pOH of the media can be found as follows,

Since the summation of pH and pOH is 14 at 25℃, we can find the pH of the weak acid as well.

Problem 05

• Find the pOH of a CH₃NH₂ solution with the initial concentration of 0.05 mol dm^-3 at 25℃. The dissociation of CH₃NH₂ at 25℃ is 4.4 × 10^-4 mol dm^-3.

CH₃NH₂ will partially dissociate in the water as follows,

At 25℃,

Conjugate acids and bases

According to the Bronsted-Lowry definition of acids and bases, acids are proton donors and bases are proton acceptors. When an acid donated a proton, it will create an ion that is capable of accepting a proton. Since the proton acceptors are bases, that ion can be known as the conjugated base of the particular acid.

In the above equilibrium, H⁺ ions are donated in a forward reaction. Therefore, CH₃COOH is an acid. In a backward reaction CH₃COO^- are accepting H⁺ ions and acts as a base. Therefore, CH₃COO^- is the conjugated base of the CH₃COOH acid.

Bases accept protons. After accepting a proton, they will create ions that can donate a proton. That means an acid. That ion is the conjugated acid of the above base.

In the above equilibrium, H⁺ ions are accepted in a forward reaction. Therefore, NH₃ is a base. In a backward reaction, NH₄⁺ ions are donating H⁺ ions and act as an acid. Therefore, the NH₄⁺ ion is the conjugated acid of the NH₃ base.

Relation between the dissociation constants of an acid-base pair

In an aqueous media, acetic acid (CH₃COOH) will dissociate as follows.

The forward reaction is a proton donor reaction. So, we can write the K_a expression for the forward reaction.

When water I added CH₃COO^- ion, the following hydrolysis reaction will occur. This reaction is similar to the back reaction of the above equilibrium.

By multiplying the above K_b and K_a equations,

The above relation shows that the multiple of K_a and K_b of an acid and its conjugated base at a given temperature equals the ion product of water (K_w).

Problem 06

• Find the pH at 25℃, of an HCOONa solution with the initial concentration of 0.02 mol dm^-3. The dissociation constant (K_a) of HCOOH acid at 25℃ is 1.6 × 10^-4 mol dm^-3.

HCOONa salt will completely dissociate in water and result in HCOO^- ions and Na⁺ ions in the media. HCOO^- ions show an equilibrium with the water. let’s take α as the amount of dissociation at the equilibrium.

At 25℃,

We can write the K_b expression for the above equilibrium.

When compared to 0.02, α is too small. Therefore, 0.02 - α is approximately equal to 0.02. according to that, the equation will be recreated as follows.

α is the OH^- ion concentration of the media. Therefore, we can find the pOH of the media.

At 25℃,

The pH of the above salt solution is greater than 7. That means the solution is basic. Salts which are resulted from reacting strong bases and weak acids are basic.

Problem 07

• Find the pH of an NH₄Cl solution at 25℃ with the initial concentration of 0.02 mol dm^-3. The dissociation constant (K_b) of NH₃ at 25℃ is 1.8 × 10^-5 mol dm^-3.

NH₄Cl salt will completely dissociate in water and result in NH₄⁺ ions and Cl^- ions in the media. NH₄⁺ ions will partially dissociate with NH₃ and H⁺ ions in water. If β is the amount of dissociation at equilibrium, concentrations of each component would be as follows,

At 25℃,

We can write the K_a expression for the above equilibrium.

When compared to 0.02, β is too small. Therefore, 0.02 - β is approximately equal to 0.02. according to that, the equation will be recreated as follows.

β is the H⁺ ion concentration of the media. Therefore, we can find the pH of the media.

The pH of the above salt solution is less than 7. That means the solution is acidic. Salts which are resulted from reacting strong acids and weak bases are acidic. An aqueous solution of a salt is formed by reacting strong acid and a strong base is neutral. At 25℃ the pH of such solution is 7.