Chapter 7.12 - Bronsted-Lowry Acids and Bases

In the previous section, we saw Arrhenius theory of acids and bases. In this section, we will see Brönsted-Lowry Acids and Bases

The Brönsted-Lowry Acids and Bases

• The Danish scientist Johannes Brönsted and the English scientist Thomas M. Lowry gave a more general definition of acids and bases
• According to their theory,
    ♦ Acids are substances which are capable of donating a H⁺ ion
    ♦ Bases are substances which are capable of accepting a H⁺ ion
• We know that, H⁺ ion is a proton. So we can write:
    ♦ Acids are proton donors
    ♦ Bases are proton acceptors
• This can be explained using an example
    ♦ We consider the solution of NH₃ in H₂O
    ♦ The equation is: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)
    ♦ It can be written in 14 steps:
1. H₂O making a donation:
(i) If H₂O donates a H⁺, we can write:
   ♦ One H atom is removed from H₂O
   ♦ Also, one proton is removed from H₂O
(ii) When one H atom is removed, what remains is OH
• When a proton is also removed, the remaining portion will have a net negative charge of -1
(iii) So when one H⁺ is removed from H₂O, what we get is: OH^-
• Since H₂O is able to donate one H⁺ in this way, it can be called a Lowry-Brönsted acid
2. NH₃ accepting a donation
(i) If NH₃ accepts a H⁺, we can write:
   ♦ One H atom is added to NH₃
   ♦ Also, one proton is added to NH₃
(ii) When one H atom is added, we get NH₄
• When a proton is also added, the new species will have a net positive charge of +1
(iii) So when one H⁺ is added to NH₃, what we get is: NH₄⁺
• Since NH₃ is able to accept one H⁺ in this way, it can be called a Lowry-Brönsted base
3. So when NH₃ is added to H₂O,
• H₂O will donate a proton
   ♦ So H₂O is the acid
• NH₃ will accept that proton
   ♦ So NH3 is the base
• The two processes can be written together as:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)
◼ This is a reversible reaction. So we must analyze the backward reaction also. The following steps from (4) to (6) will help us to analyze the backward reaction
4. NH₄⁺ making a donation:
• We see that, in the backward direction, NH₄⁺ is becoming NH₃
• This is possible only if NH₄⁺ donates one H⁺
• Let us see the steps:
(i) If NH₄⁺ donates a H⁺, we can write:
   ♦ One H atom is removed from NH₄⁺
   ♦ Also, one proton is removed from NH₄⁺
(ii) When one H atom is removed, what remains is NH₃⁺
• When a proton is also removed, the remaining portion will lose the +1 charge
(iii) So when one H⁺ is removed from NH₄⁺, what we get is: NH₃
• Since NH₄⁺ is able to donate one H⁺ in this way, it can be called a Lowry-Brönsted acid
5. OH^- accepting a donation
(i) We see that, in the backward direction, OH^- is becoming H₂O
• This is possible only if OH^- accepts one H⁺
• Let us see the steps:
(i) If OH^- accepts a H⁺, we can write:
   ♦ One H atom is added to OH^-
   ♦ Also, one proton is added to OH^-
(ii) When one H atom is added, we get HOH^-
• When a proton is also added, the charges will get cancelled
(iii) So when one H⁺ is added to OH^-, what we get is: H₂O
• Since OH^- is able to accept one H⁺ in this way, it can be called a Lowry-Brönsted base
6. So when NH₃ is added to water,
• NH₄⁺ will donate a proton
   ♦ Then according to the theory, NH₄⁺ will be called a Lowry-Brönsted acid
• OH^- will accept that proton
   ♦ Then according to the theory, OH^- will be called a Lowry-Brönsted base
7. Let us compare the results in (3) and (6)
• In (3) we have:
    ♦ H₂O is the acid
    ♦ NH₃ is the base
• In (6) we have:
    ♦ NH₄⁺ is the acid
    ♦ OH^- is the base
• So we have two pairs:
    ♦ In (3) we have a acid-base pair
    ♦ In (6) also, we have a acid-base pair
8. By making a simple rearrangement in (7), we can make two more pairs. The rearrangement is indicated by the cyan and yellow arrows in fig.7.8 below:

Fig.7.8

9. Thus the two new pairs are:
• Pair 1:
    ♦ H₂O is the acid
    ♦ OH^- is the base
• Pair 2:
    ♦ NH₄⁺ is the acid
    ♦ NH₃ is the base
10. Consider pair 1 in (9)
• H₂O can donate one H⁺
    ♦ So H₂O is an acid
• OH^- can accept that H⁺ and become H₂O
    ♦ So OH^- is a base
• The only difference between the above acid and base is the H⁺. That is:
    ♦ H₂O can become OH^- just by donating H⁺
        ✰ No other atoms or ions need to be donated
    ♦ OH^- can become H₂O just by accepting an H⁺
        ✰ No other atoms or ions need to be accepted
11. Consider pair 2 in (9)
• NH₄⁺ can donate one H⁺
    ♦ So NH₄⁺ is an acid
• NH₃ can accept that H⁺ and become NH₄⁺
    ♦ So NH₃ is a base
• The only difference between the above acid and base is the H⁺. That is:
    ♦ NH₄⁺ can become NH₃ just by donating H⁺
        ✰ No other atoms or ions need to be donated
    ♦ NH₃ can become NH₄⁺ just by accepting an H⁺
        ✰ No other atoms or ions need to be accepted
12. The difference of one proton
    ♦ In (10), we see that, the acid and base differ by only one H⁺
    ♦ In (11) also we see that, the acid and base differ by only one H⁺
◼  Such acid-base pair that differs only by one proton is called a conjugate acid-base pair
• So we can write:
    ♦ In a conjugate acid-base pair, there will be two items
        ✰ If one is an acid, the other will be it's conjugate base
        ✰ If one is a base, the other will be a conjugate acid
• An example:
    ♦ [H₂O, OH^-] is a conjugate acid-base pair
        ✰ H₂O is the acid
        ✰ OH^- is the conjugate base
• Another example:
    ♦ [NH₃, NH₄⁺] is a conjugate acid-base pair
        ✰ NH₃ is the base
        ✰ NH₄⁺ is the conjugate acid
13. The above information can be represented pictorially as shown below:

◼ We see an important point. It can be written in 2 steps:
(i) We wrote the acid-conjugate base pair together as: [H₂O, OH^-]
• But in the chemical equation, H₂O and OH^- are on opposite sides of the arrows
(ii) We wrote the base-conjugate acid pair together as: [NH₃, NH₄⁺]
• But in the chemical equation, NH₃ and NH₄⁺ are on opposite sides of the arrows
14. In such conjugate acid-base pairs,
    ♦ If the acid is strong, the base will be weak
    ♦ If the base is strong, acid will be weak
• In our present case:
    ♦ In the pair [H₂O, OH^-],
        ✰ H₂O is the acid. It is a weak acid
        ✰ OH^- is the conjugate base. It is a strong base
    ♦ In the pair [NH₃, NH₄⁺],
        ✰ NH₃ is the base. It is a strong base
        ✰ NH₄⁺ is the conjugate acid. It is a weak acid
• We will see the definitions for strong/weak acids and bases in later sections

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• Let us see another example:
    ♦ We consider the solution of HCl in H₂O
    ♦ The equation is: HCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + Cl^-(aq)
    ♦ It can be written in 13 steps:
1. HCl making a donation
(i) If HCl donates a H⁺, we can write:
   ♦ One H atom is removed from HCl
   ♦ Also, one proton is removed from HCl
(ii) When one H atom is removed, what remains is Cl
• When a proton is also removed, the remaining portion will have a net negative charge of -1
(iii) So when one H⁺ is removed from HCl, what we get is: Cl^-
• Since HCl is able to donate one H⁺ in this way, it can be called a Lowry-Brönsted acid
2. H₂O accepting a donation
(i) If H₂O accepts a H⁺, we can write:
   ♦ One H atom is added to H₂O
   ♦ Also, one proton is added to H₂O
(ii) When one H atom is added, we get H₃O
• When a proton is also added, the new species will have a net positive charge of +1
(iii) So when one H⁺ is added to H₂O, what we get is: H₃O⁺
• Since H₂O is able to accept one H⁺ in this way, it can be called a Lowry-Brönsted base
3. So when HCl is added to H₂O,
• HCl will donate a proton
   ♦ So HCl is the acid
• H₂O will accept that proton
   ♦ So H₂O is the base
• The two processes can be written together as:
HCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + Cl^-(aq)
◼ This is a reversible reaction. So we must analyze the backward reaction also. The following steps from (4) to (6) will help us to analyze the backward reaction
4. H₃O⁺ making a donation:
(i) We see that, in the backward direction, H₃O⁺ is becoming H₂O
• This is possible only if H₃O⁺ donates one H⁺
• Let us see the steps:
(i) If H₃O⁺ donates a H⁺, we can write:
   ♦ One H atom is removed from H₃O⁺
   ♦ Also, one proton is removed from H₃O⁺
(ii) When one H atom is removed, what remains is H₂O⁺
• When a proton is also removed, the remaining portion will lose the +1 charge
(iii) So when one H⁺ is removed from H₃O⁺, what we get is: H₂O
• Since H₃O⁺ is able to donate one H⁺ in this way, it can be called a Lowry-Brönsted acid
5. Cl^- accepting a donation
(i) We see that, in the backward direction, Cl^- is becoming HCl
• This is possible only if Cl^- accepts one H⁺
• Let us see the steps:
(i) If Cl^- accepts a H⁺, we can write:
   ♦ One H atom is added to Cl^-
   ♦ Also, one proton is added to Cl^-
(ii) When one H atom is added, we get HCl^-
• When a proton is also added, the charges will get cancelled
(iii) So when one H⁺ is added to Cl^-, what we get is: HCl
• Since Cl^- is able to accept one H⁺ in this way, it can be called a Lowry-Brönsted base
6. So when HCl is added to water,
• H₃O⁺ will donate a proton
   ♦ Then according to the theory, H₃O⁺ will be called a Lowry-Brönsted acid
• Cl^- will accept that proton
   ♦ Then according to the theory, Cl^- will be called a Lowry-Brönsted base
7. Let us compare the results in (3) and (6)
• In (3) we have:
    ♦ HCl is the acid
    ♦ H₂O is the base
• In (6) we have:
    ♦ H₃O⁺ is the acid
    ♦ Cl^- is the base
• So we have two pairs:
    ♦ In (3) we have a acid-base pair
    ♦ In (6) also, we have a acid-base pair
8. By making a simple rearrangement in (7), we can make two more pairs. The rearrangement is indicated by the cyan and yellow arrows in fig.7.9 below:

Fig.7.9

9. Thus the two new pairs are:
• Pair 1:
    ♦ HCl is the acid
    ♦ Cl^- is the base
• Pair 2:
    ♦ H₃O⁺ is the acid
    ♦ H₂O is the base
10. Consider pair 1 in (9)
• HCl can donate one H⁺
    ♦ So HCl is an acid
• Cl^- can accept that H⁺ and become HCl
    ♦ So Cl^- is a base
• The only difference between the above acid and base is the H⁺. That is:
    ♦ HCl can become Cl^- just by donating H⁺
        ✰ No other atoms or ions need to be donated
    ♦ Cl^- can become HCl just by accepting an H⁺
        ✰ No other atoms or ions need to be accepted
11. Consider pair 2 in (9)
• H₃O⁺ can donate one H⁺
    ♦ So H₃O⁺ is an acid
• H₂O can accept that H⁺ and become H₃O⁺
    ♦ So H₂O is a base
• The only difference between the above acid and base is the H⁺. That is:
    ♦ H₃O⁺ can become H₂O just by donating H⁺
        ✰ No other atoms or ions need to be donated
    ♦ H₂O can become H₃O⁺ just by accepting an H⁺
        ✰ No other atoms or ions need to be accepted
12. The difference of one proton
    ♦ In (10), we see that, the acid and base differ by only one H⁺
    ♦ In (11) also we see that, the acid and base differ by only one H⁺
◼  Such acid-base pair that differs only by one proton is called a conjugate acid-base pair
• So we can write:
    ♦ In a conjugate acid-base pair, there will be two items
        ✰ If one is an acid, the other will be it's conjugate base
        ✰ If one is a base, the other will be a conjugate acid
• An example:
    ♦ [HCl, Cl^-] is a conjugate acid-base pair
        ✰ HCl is the acid
        ✰ Cl^- is the conjugate base
• Another example:
    ♦ [H₂O, H₃O⁺] is a conjugate acid-base pair
        ✰ H₂O is the base
        ✰ H₃O⁺ is the conjugate acid
13. The above information can be represented pictorially as shown below:

◼ We see an important point. It can be written in 2 steps:
(i) We wrote the acid-conjugate base pair together as: [HCl, Cl^-]
• But in the chemical equation, HCl and Cl^- are on opposite sides of the arrows
(ii) We wrote the base-conjugate acid pair together as: [H₂O, H₃O⁺]
• But in the chemical equation, H₂O and H₃O⁺ are on opposite sides of the arrows
14. In such conjugate acid-base pairs,
    ♦ If the acid is strong, the base will be weak
    ♦ If the base is strong, acid will be weak
• In our present case:
    ♦ In the pair [HCl, Cl^-],
        ✰ HCl is the acid. It is a strong acid
        ✰ Cl^- is the conjugate base. It is a weak base
    ♦ In the pair [H₂O, H₃O⁺],
        ✰ H₂O is the base. It is a weak base
        ✰ H₃O⁺ is the conjugate acid. It is a strong acid
• We will see the definitions for strong/weak acids and bases in later sections

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◼ Here we see an interesting point. It can be written in two steps:
(i) When NH₃ is dissolved in H₂O, the H₂O acts as an acid
(ii) When HCl is dissolved in H₂O, the H₂O acts as a base

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We will now see some solved examples
Solved example 7.46
What will be the conjugate bases for the following Brönsted acids: HF, H₂SO₄ and HCO₃^- ?
Solution:
Part (i): HF
1. Given that, HF is a Brönsted acid
2. The conjugate base of this acid will accept a proton and give back this acid
• So we want to find the species which will accept a H⁺ and give back HF
3. Let the species be X
• Then we can write: X + H⁺ → HF
4. So to get X, we have to remove H⁺ from both sides of the equation
   ♦ Removing one H⁺ means, removing a proton
   ♦ So after removal, there will be an extra electron
   ♦ So after removal, there will be an extra negative charge
   ♦ Also, one H atom will be removed
• Thus when H⁺ is removed from HF, it becomes F^-
5. So the equation in (3) becomes: X → F^-
• That is., X is same as F^-
6. So the species F^- will accept a proton and give back HF
• Thus F^- is the conjugate base

Part (ii): H₂SO₄
1. Given that, H₂SO₄ is a Brönsted acid
2. The conjugate base of this acid will accept a proton and give back this acid
• So we want to find the species which will accept a H⁺ and give back H₂SO₄
3. Let the species be X
• Then we can write: X + H⁺ → H₂SO₄
4. So to get X, we have to remove H⁺ from both sides of the equation
   ♦ Removing one H⁺ means, removing a proton
   ♦ So after removal, there will be an extra electron
   ♦ So after removal, there will be an extra negative charge
   ♦ Also, one H atom will be removed
• Thus when H⁺ is removed from H₂SO₄, it becomes HSO₄^-
5. So the equation in (3) becomes: X → HSO₄^-
• That is., X is same as HSO₄^-
6. So the species HSO₄^- will accept a proton and give back H₂SO₄
• Thus HSO₄^- is the conjugate base

Part (iii): HCO₃^-
1. Given that, HCO₃^- is a Brönsted acid
2. The conjugate base of this acid will accept a proton and give back this acid
• So we want to find the species which will accept a H⁺ and give back HCO₃^-
3. Let the species be X
• Then we can write: X + H⁺ → HCO₃^-
4. So to get X, we have to remove H⁺ from both sides of the equation
   ♦ Removing one H⁺ means, removing a proton
   ♦ So after removal, there will be an extra electron
   ♦ So after removal, there will be an extra negative charge
   ♦ Also, one H atom will be removed
• Thus when H⁺ is removed from HCO₃^-, it becomes CO₃^2-
5. So the equation in (3) becomes: X → CO₃^2-
• That is., X is same as CO₃^2-
6. So the species CO₃^2- will accept a proton and give back HCO₃^-
• Thus CO₃^2- is the conjugate base

Solved example 7.47
Write the conjugate acids for the following Brönsted bases: NH₂^-, NH₃ and HCOO-
Solution:
Part (i): NH₂^-
1. Given that, NH₂^- is a Brönsted base
2. The conjugate acid of this base will donate a proton and give back this base
• So we want to find the species which will donate a H⁺ and give back NH₂^-
3. Let the species be X
• Then we can write: X - H⁺ → NH₂^-
4. So to get X, we have to add H⁺ to both sides of the equation
   ♦ Adding one H⁺ means, adding a proton
   ♦ So after addition, there will be an extra positive charge
   ♦ Also, one H atom will be added
• Thus when H⁺ is added to NH₂^-, it becomes NH₃
5. So the equation in (3) becomes: X → NH₃
• That is., X is same as NH₃
6. So the species NH₃ will donate a proton and give back the base NH₂^-
• Thus NH₃ is the conjugate acid

Part (ii): NH₃
1. Given that, NH₃ is a Brönsted base
2. The conjugate acid of this base will donate a proton and give back this base
• So we want to find the species which will donate a H⁺ and give back NH₃
3. Let the species be X
• Then we can write: X - H⁺ → NH₃
4. So to get X, we have to add H⁺ to both sides of the equation
   ♦ Adding one H⁺ means, adding a proton
   ♦ So after addition, there will be an extra positive charge
   ♦ Also, one H atom will be added
• Thus when H⁺ is added to NH₃, it becomes NH₄⁺
5. So the equation in (3) becomes: X → NH₄⁺
• That is., X is same as NH₄⁺
6. So the species NH₄⁺ will donate a proton and give back the base NH₃
• Thus NH₄⁺ is the conjugate acid

Part (iii): HCOO^-
1. Given that, HCOO^- is a Brönsted base
2. The conjugate acid of this base will donate a proton and give back this base
• So we want to find the species which will donate a H⁺ and give back HCOO^-
3. Let the species be X
• Then we can write: X - H⁺ → HCOO^-
4. So to get X, we have to add H⁺ to both sides of the equation
   ♦ Adding one H⁺ means, adding a proton
   ♦ So after addition, there will be an extra positive charge
   ♦ Also, one H atom will be added
• Thus when H⁺ is added to HCOO^-, it becomes HCOOH
5. So the equation in (3) becomes: X → HCOOH
• That is., X is same as HCOOH
6. So the species HCOOH will donate a proton and give back the base HCOO^-
• Thus HCOOH is the conjugate acid

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From the above two solved examples, we get an easy method to find conjugate acid and conjugate base. It can be written in 2 steps:
1. Given an acid. It's conjugate base can be written by removing H⁺ from that acid
2. Given a base. It's conjugate acid can be written by adding H⁺ to that base

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In the next section, we will see a few more solved examples


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