Chapter 7.14 - Ionization of Acids and Bases

In the previous section, we saw Lewis acids and bases. In this section, we will see ionization of acids and bases

• We saw three theories about acids and bases:
   ♦ The Arrhenius theory
   ♦ The Brönsted-Lowry theory
   ♦ The Lewis theory
• The first two theories are simple and can be applied in most practical situations
• This is because, in most practical situations, acids and bases will be reacting with water
• We have seen that, both Arrhenius and Brönsted-Lowry theory help us to write accurate chemical equations of such reactions with water

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• Now we will see how the two theories help us to define strong acids and bases
• Let us first consider the Arrhenius theory. It can be written in 2 steps:
1. Acids like HCl, HNO₃ etc., dissociate almost completely when added to water
    ♦ HCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + Cl^-(aq)
    ♦ HNO₃ + H₂O ⇌ H₃O⁺(aq) + NO₃^-(aq)
(Recall that, the H⁺ ions combine with H₂O molecules to form H₃O⁺ ions)
• Since they dissociate almost completely, we get a large concentration of H⁺ ions
◼  So according to Arrhenius theory, such acids are strong acids
2. Similar is the case with bases like LiOH and NaOH. They dissociate almost completely
• Since they dissociate almost completely, we get a large concentration of OH^- ions
◼  So according to Arrhenius theory, such bases are strong bases

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Next, we will see how the Brönsted-Lowry theory helps us to define strong acids and bases. It can be written in 7 steps:
1. Acids like HCl, HNO₃ etc., dissociate almost completely when added to water
• Since they dissociate almost completely, we get a large concentration of H+ ions
• This is same as:
   ♦ They donate a large number of protons
   ♦ In other words, such acids are good proton donors
◼  So according to Brönsted-Lowry theory, such acids are strong acids
2. Similar is the case with bases like LiOH and NaOH. They dissociate almost completely
• Since they dissociate almost completely, we get a large concentration of OH^- ions
• OH^- ions accept H⁺ to become water
   ♦ If more OH^- ions are available, more H⁺ can be accepted
   ♦ This is same as:
         ✰ These bases can accept a large number of protons
• In other words, such bases are good proton acceptors
◼  So according to Brönsted-Lowry theory, such bases are strong bases

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Let us see how the Brönsted-Lowry theory can be used to analyze ionic equilibrium. It can be written in 6 steps
1. We have seen that, ionization reaction is a reversible reaction
    ♦ Both forward and backward reactions are taking place
    ♦ With the passage of time, the reaction will be moving towards equilibrium
2. So a question arises:
• At equilibrium, which of the following two conditions will be obtained?
    ♦ A condition in which there is a greater concentration of products
    ♦ A condition in which there is a greater concentration of reactants
3. We can make this question clearer using an example. It can be written in steps:
(i) Consider the reaction:
HX(aq) + H₂O(l) ⇌ H₃O⁺(aq) + X^-(aq)
(ii) At equilibrium which of the following two conditions will be obtained?
    ♦ A condition in which there is a greater concentration of H₃O⁺ and X^-
    ♦ A condition in which there is a greater concentration of HX and H₂O
4. At equilibrium,
• If there is a greater concentration of H₃O⁺ and X^-, We can say:
    ♦ Forward reaction is favored
    ♦ HX undergoes almost complete dissociation
• If there is a greater concentration of HX and H₂O, We can say:
    ♦ Backward reaction is favored
    ♦ HX undergoes very low dissociation
5. Let us write a summary. It can be written in two steps:
(i) If the forward reaction is favored,
    ♦ At equilibrium, there will be a large concentration of H₃O⁺ and X^-
    ♦ There will be only very low concentration of HX
    ♦ Then HX can be considered as a strong acid
(ii) If the backward reaction is favored,
    ♦ At equilibrium, there will be a large concentration of HX and H₂O
    ♦ There will be only very low concentration of H₃O⁺ and X^-
    ♦ Then HX can be considered as a weak acid
6. Recall that, Brönsted-Lowry theory gives us information about conjugate acids and bases also. Let us apply it to our present case. It can be written in 4 steps:
(i) We know how to identify conjugate acids and bases in a reaction
• For our present case, they can be written along with the chemical equation as shown below:

    ♦ So the [acid, conjugate base] pair is: [HX,X^-]
    ♦ Also, the [base, conjugate acid] pair is: [H₂O,H₃O⁺]
(ii) So we have two pairs:
    ♦ [acid, conjugate base], which is: [HX,X^-]
    ♦ [base, conjugate acid], which is: [H₂O,H₃O⁺]
• We have to analyze each pair when the forward reaction is favored
• We have to analyze each pair when the backward reaction is favored
(iii) First we will analyze the pairs when the forward reaction is favored:
◼ Consider the first pair [acid, conjugate base], which is: [HX,X^-]
    ♦ HX will dissociate completely
    ♦ The backward reaction will be slow
        ✰ X^- will be reluctant to accept protons
        ✰ That means X- is a weak base
• Thus in the pair [HX,X^-]:
    ♦ The acid HX is strong
    ♦ The conjugate base X^- is weak
◼ Consider the second pair [base, conjugate acid], which is: [H₂O,H₃O⁺]
    ♦ H₂O will dissociate completely
    ♦ The backward reaction will be slow
        ✰ H₃O⁺ will be reluctant to donate protons to X^-
        ✰ That means H₃O⁺ is a weak acid
• Thus in the pair [H₂O,H₃O⁺]:
    ♦ The base H₂O is strong
    ♦ The conjugate acid H₃O⁺ is weak
The two cases are shown in the fig.7.18 below:

Fig.7.18

(iv) Next we will analyze the pairs when the backward reaction is favored:
◼ Consider the first pair [acid, conjugate base], which is: [HX,X^-]
    ♦ HX will be reluctant to dissociate
    ♦ The backward reaction will be fast
        ✰ X^- will readily accept protons from H₃O⁺
        ✰ That means X^- is a strong base
• Thus in the pair [HX,X^-]:
    ♦ The acid HX is weak
    ♦ The conjugate base X^- is strong
◼ Consider the second pair [base, conjugate acid], which is: [H₂O,H₃O⁺]
    ♦ H₂O will be reluctant to dissociate
    ♦ The backward reaction will be fast
        ✰ H₃O⁺ will readily donate protons to X^-
        ✰ That means H₃O⁺ is a strong acid
• Thus in the pair [H₂O,H₃O⁺]:
    ♦ The base H₂O is weak
    ♦ The conjugate acid H₃O⁺ is strong
The two cases are shown in the fig.7.19 below:

Fig.7.19

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In the next section, we will see ionization constant of water and the pH scale


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Chapter 7.11 - Ionic Equilibrium

In the previous section, we saw the factors which affect equilibrium. In this section, we will see ionic equilibrium

• We have seen a large number of chemical equations showing reversible reactions
For example: H₂(g) + I₂(g) ⇌ 2HI(g)
• In some cases, 'ions' will be present in the place of 'reactants' or 'products'
For example: Fe³⁺(aq) + SCN^-(aq) ⇌ [Fe(SCN)]²⁺(aq)
• This equation indicates that:
Fe³⁺, SCN^- and [Fe(SCN)]²⁺ are present in an aqueous solution
   ♦ Fe³⁺ and SCN^- continuously combine together to form [Fe(SCN)]²⁺ ion
         ✰ This is the forward reaction
   ♦ [Fe(SCN)]²⁺ continuously dissociates into Fe³⁺ and SCN^-
         ✰ This is the backward reaction
• At equilibrium, the rate of forward and backward reactions become equal
• Before learning about ionic equilibrium, we must first learn about electrolytes. It can be written in 8 steps:
1. Pure water does not conduct electricity. But we do not have pure water in nature
• So all water that we see in our day to day life, will conduct electricity
2. Two opposite cases:
(i) If we dissolve sugar into pure water, that sugar solution will not conduct electricity
(ii) If we dissolve common salt (NaCl) in pure water, that salt solution will conduct electricity
3. The English scientist Michael Faraday did extensive research and discovered the reason for such opposing behaviour
• The following steps from (4) to (8) help us to get a clear idea about his findings
(4) Michael Faraday classified all substances into two categories
(i) The first category of substances are able to conduct electricity when dissolved in water
   ♦ These substances were called electrolytes
(ii) The other category are not able to conduct electricity when dissolved in water
   ♦ These substances were called non-electrolytes
5. Electrolytes were further classified into strong electrolytes and weak electrolytes
This is shown in fig.7.7 below:

Fig.7.7

6. When strong electrolytes are dissolved in water, they are ionized completely
• Let us see an example:
• When some NaCl is dissolved in water, all the NaCl molecules separates into Na⁺ and Cl^- ions
   ♦ We will not find any NaCl molecule in the solution
7. When weak electrolytes are dissolved in water, they are not ionized completely
• Let us see an example:
• When some CH₃COOH (acetic acid) is dissolved in water, all the CH₃COOH molecules do not separate into CH₃COO^- and H⁺ ions
   ♦ The solution will contain three items:
         ✰ CH₃COOH molecules
         ✰ CH₃COO^- ions
         ✰ H⁺ ions
8. Extent of ionization:
   ♦ Almost 100 % of the added NaCl will separate into ions
   ♦ Only 5% of the added CH₃COOH will separate into ions

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Now we can define ionic equilibrium. It can be written in 3 steps:
1. In the case of weak electrolytes, molecules as well as ions will be present in the solution
2. An equilibrium will be established between those ions and molecules
For example: CH₃COOH(aq) ⇌ CH₃COO^-(aq) + H⁺(aq)
3. This type of equilibrium involving ions in aqueous solution is called ionic equilibrium

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• Note the double arrow in the above step 2
   ♦ It indicates that, both forward and backward reactions occur simultaneously
         ✰ The acetic acid molecules continuously separate into ions
         ✰ The ions continuously combine to give back the acetic acid molecules
   ♦ When the rates of both the reactions are equal, equilibrium is attained
• There is no need for a double arrow in the case of NaCl
   ♦ Because, the Na⁺and Cl^- ions do not combine to give back NaCl molecules
   ♦ We can write: NaCl(s) → Na⁺(aq) + Cl^-(aq)

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• Let us recall some simple facts about acids, bases and salts that we saw in our earlier classes:
Acids
   ♦ Acids turn blue litmus paper into red
   ♦ Acids liberate H₂ when they react with metals
   ♦ Acids have a sour taste
   ♦ Examples of acids: HCl (Hydrochloric acid), CH₃COOH (acetic acid)
Bases
   ♦ Bases turn red litmus paper into blue
   ♦ Bases feel soapy
   ♦ Bases have a bitter taste
   ♦ Bases that dissolve in water are called alkalies
   ♦ Examples of bases: NaOH (Sodium hydroxide), KOH (Potassium hydroxide)
Salts
   ♦ Salts are formed when acids and bases are mixed together in right proportions
   ♦ Example: HCl + NaOH → NaCl + H₂O

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• NaCl (sodium chloride) is common salt
• Some basics about NaCl solution can be written in 3 steps:
1. In the solid state, NaCl exists as Na⁺ and Cl^- ions
   ♦ But those ions are not separated from each other
   ♦ They are held closely together by the electrostatic forces of attraction
2. When NaCl is dissolved in water, NaCl separates into Na⁺ and Cl^- ions
• But why don't they attract and become NaCl again?
• The answer is that, water has a special property (called high dielectric constant), which can reduce the force of attraction between Na⁺ and Cl^-
• We will see details about dielectric constant in higher classes
3. So the Na⁺ and Cl^- are separated from each other in the solution
• Now recall what we learnt in enthalpy of solution (Fig.6.15 in section 6.10 of the previous chapter)
• We saw that:
   ♦ Each Na+ will be surrounded by H₂O molecules
         ✰ Thus we obtain Na⁺(aq)
   ♦ Each Cl- will also be surrounded by H₂O molecules
         ✰ Thus we obtain Cl^-(aq)

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• Now we can compare NaCl and CH₃COOH. The comparison can be written 6 in steps:
1. The original samples:
• In the original sample of NaCl, there are separate Na⁺ and Cl^- ions (held together by strong electrosattic forces)
• In the original sample of CH₃COOH, there are no separate ions. Only CH₃COOH molecules
2. Separation into ions
• When NaCl is added to water, all NaCl molecules get completely separated into Na⁺ and Cl^- ions
• When CH₃COOH is added to water, only a small number of molecules get separated into CH₃COO^- and H⁺ ions
3. Recombination
• In the case of NaCl, the Na⁺ and Cl^- ions cannot combine again to give back NaCl because, water has a high dielectric constant
• In the case of CH₃COOH, inspite of the high dielectric constant of water, some CH₃COO^- and H⁺ ions can combine again to give back CH3COOH. We will see the reason in later sections
4. Hydration by water
• In the case of NaCl, the Na⁺ and Cl^- ions get hydrated by water. Thus they become Na⁺(aq) and Cl^-(aq)
• In the case of CH₃COOH also, the CH₃COO^- and H⁺ ions get hydrated by water. Thus they become CH₃COO^-(aq) and H⁺(aq)
5. Dissociation and Ionization
• If the original sample exists as separate ions (as in the case of NaCl), the process of 'separation into ions' is called dissociation
• If the original sample exists as separate molecules (as in the case of CH₃COOH), the process of 'separation into ions' is called ionization
◼  In this present chapter, both the words 'dissociation' and 'ionization' will mean the same thing: The separation of original molecules into ions
6. So now a question arises:
◼  NaCl undergoes complete ionization when dissolved in water. But CH₃COOH undergoes only partial ionization. Why is there such a difference?
• The answer can be given based on bond strength. We will see the details in later chapters
• At present, all we need to know is this:
When partial ionization takes place, an ionic equilibrium will be established between the ions and the original molecules

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Arrhenius concept of Acids and Bases

• First we will see how the Arrhenius concept is applied to acids. It can be written in 5 steps:
1. According to Arrhenius theory, acids can be represented as HX
• This is because, H (hydrogen atom) will be present in all acids
2. So the ionization of an acid can be represented as:
HX(aq) → H⁺(aq) + X^-(aq)
3. We know that, the H atom has only one electron and one proton
• So H⁺ ion will be having only one proton
• That means, an H⁺ ion consists of only the nucleus with the one proton
• We know that, nucleus is very small when compared to the total size of the atom
    ♦ So the H⁺ ion will be very small in size
• The +ve charge will be concentrated in a very small space
    ♦ So the charge density is high
4. This makes the H⁺ ion very reactive
• It attaches itself to the lone pairs of electrons (see fig.4.104 of section 4.17) in the H₂O molecule
◼ The H₂O molecule thus becomes H₃O⁺ ion
• This can be represented as: H₂O(l) + H⁺ → H₃O⁺(aq)
5. Adding equations in (2) and (4), we get:
HX(aq) + H₂O(l) + H⁺(aq) → H⁺(aq) + X^-(aq) + H₃O⁺(aq)
• Cancelling H⁺ on either sides, we get:
HX(aq) + H₂O(l) → H₃O⁺(aq) + X^-(aq)
◼ This simply means that:
When the acid HX is dissolved in water, H₃O⁺(aq) and X^-(aq) ions are produced
• H₃O⁺ is known as the hydrated proton. It is also known as hydronium ion

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• Next we will see how the concept is applied to bases. It can be written in 2 steps:
1. According to Arrhenius theory, bases can be represented as XOH
• This is because, OH will be present in all bases
2. So the ionization of a base can be represented as:
XOH(aq) → X⁺(aq) + OH^-(aq)
◼ This simply means that:
When the base XOH is dissolved in water, X⁺ and OH^- ions are produced
• OH^- ion is called hydroxyl ion

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◼ In general, we can write:
• According to Arrhenius theory:
    ♦ Acids are substances that dissociates in water to give H₃O⁺(aq) ions
    ♦ Bases are substances that produce OH^-(aq) ions

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The Arrhenius theory suffers some limitations. They can be written in 2 steps:
1. The theory is applicable only to aqueous solutions
2. Substances like ammonia do not possess the hydroxyl group. But they exhibit basic nature. Arrhenius theory cannot explain the basic nature of such substances

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In the next section, we will see Brönsted-Lowry Acids and Bases


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