Chapter 7.21 - The Buffer solution

In the previous section, we saw common ion effect and the hydrolysis of salts. In this section, we will see buffers

• A buffer solution (or simply buffer) is a specially prepared aqueous solution
   ♦ It is prepared in such a way as to have a certain pH value
         ✰ Even when some acid is added, it's pH will not decrease much
         ✰ Even when some base is added, it's pH will not increase much
• A buffer can be prepared by two methods:
Method 1:
By mixing a weak acid and it’s conjugate base
Method 2:
By mixing a weak base and it’s conjugate acid

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• We will first see how the buffer is prepared. After that we will see how buffer works
Preparation by method 1:
This can be written in 3 steps:
1. We need a weak acid
• Consider the weak acid CH₃COOH
   ♦ Take an aqueous solution of this CH₃COOH
   ♦ Let us call it solution 1
• We know that CH₃COOH will dissociate in aqueous solution as:
   ♦ CH₃COOH ⇌ CH₃COO^- + H⁺
   ♦ The solution 1 will be at equilibrium according to this equation
2. Next we want the conjugate base of CH₃COOH
   ♦ The conjugate base of CH₃COOH is CH₃COO^-
• We prepare a second solution
   ♦ Let us call it solution 2
• The solution 2 is an aqueous solution of CH₃COONa (sodium acetate)
• The CH₃COONa dissociates completely
   ♦ CH₃COONa  → CH₃COO^- + Na⁺
   ♦ So solution 2 will contain only CH₃COO^-  and Na⁺
         ✰ Thus solution 2 will contain the required conjugate base
3. Next we add solution 2 to solution 1
• That means, we are adding CH₃COO^- ions and Na⁺ ions to solution 1
   ♦ Na⁺ ions are stable. They do not take part in the reaction
• So the net effect is that, [CH₃COO^-] in the resulting solution increases
• Due to this increase in [CH₃COO^-], the equilibrium mentioned in (1) will shift towards the left
• Thus a new equilibrium will be reached. The buffer is ready

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Next we will see how this buffer works. It can be written in 5 steps:
1. Assume that, a strong acid is added to the buffer
• The strong acid will supply a large number of H⁺ ions
   ♦ So we would expect the pH of the buffer to fall
2. But the very purpose of the buffer is to keep the pH at the same value
• Indeed the buffer can do this because, it has a lot of excess CH₃COO^- ions
• Those ions will consume the incoming H⁺ ions. The equation is:
CH₃COO^- + H⁺ ⇌ CH₃COOH
3. Thus most of the incoming H⁺ ions are neutralized, keeping the pH from falling too much
4. At the beginning of this discussion, we said that:
A buffer (by method 1) is a mixture of a weak acid and it’s conjugate base
• For our present case. we chose CH₃COOH as the weak acid
• The pure solution of CH₃COOH is a mixture of CH₃COOH and it’s conjugate base CH₃COO^-
• Then why do we add CH₃COONa ?
• The answer is that:
The pure mixture will not have enough CH₃COO^- ions to neutralize the incoming H⁺ ions
5. To this prepared buffer, instead of adding a strong acid, we can add a strong base. Even then, the pH will not change much
• This is because, the incoming OH^- will react with CH₃COOH:
CH₃COOH + OH^- ⇌ CH₃COO^- + H₂O
• Thus the incoming OH^- ions are neutralized

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• Now we will see how the buffer is prepared by the second method. After that we will see how that buffer works
Preparation by method 2:
This can be written in 3 steps:
1. We need a weak base
• Consider the weak base NH₄OH
   ♦ Take an aqueous solution of this NH₄OH
   ♦ Let us call it solution 1
• We know that NH₄OH will dissociate in aqueous solution as:
   ♦ NH₄OH ⇌ NH₄⁺ + OH^-
   ♦ The solution 1 will be at equilibrium according to this equation
2. Next we want the conjugate acid of NH₄OH
   ♦ The conjugate acid of NH₄OH is NH₄⁺
• We prepare a second solution
   ♦ Let us call it solution 2
• The solution 2 is an aqueous solution of NH₄Cl (ammonium chloride)
• The NH₄Cl dissociates completely
   ♦ NH₄Cl  → NH₄⁺ + Cl^-
   ♦ So solution 2 will contain only NH₄⁺  and Cl^-
         ✰ Thus solution 2 will contain the required conjugate acid
3. Next we add solution 2 to solution 1
• That means, we are adding NH₄⁺ ions and Cl^- ions to solution 1
   ♦ Cl^- ions are stable. They do not take part in the reaction
• So the net effect is that, [NH₄⁺] in the resulting solution increases
• Due to this increase in [NH₄⁺], the equilibrium mentioned in (1) will shift towards the left
• Thus a new equilibrium will be reached. The buffer is ready

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Next we will see how this buffer works. It can be written in 5 steps:
1. Assume that, a strong base is added to the buffer
• The strong base will supply a large number of OH^- ions
   ♦ So we would expect the pH of the buffer to rise
2. But the very purpose of the buffer is to keep the pH at the same value
• Indeed the buffer can do this because, it has a lot of excess NH₄⁺ ions
• Those ions will consume the incoming OH- ions. The equation is:
NH₄⁺ + OH^- ⇌ NH₄OH
3. Thus most of the incoming OH^- ions are neutralized, keeping the pH from rising too much
4. At the beginning of this discussion, we said that:
A buffer (by method 2) is a mixture of a weak base and it’s conjugate acid
• For our present case. we chose NH₄OH as the weak base
• The pure solution of NH₄OH is a mixture of NH₄OH and it’s conjugate acid NH₄⁺
• Then why do we add NH₄Cl ?
• The answer is that:
The pure solution will not have enough NH₄⁺ ions to neutralize the incoming OH^- ions
5. To this prepared buffer, instead of adding a strong base, we can add a strong acid. Even then, the pH will not change much
• This is because, the incoming H⁺ will react with NH₄OH:
NH₄OH + H⁺ ⇌ NH₄⁺ + H₂O
• Thus the incoming H⁺ ions are neutralized

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A summary of the discussion so far, can be given in a flowchart form. It is shown in the fig.7.21 below:

Fig.7.21

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• Next we will see two solved examples which will demonstrate the buffer action
Solved example 7.79
A buffer is prepared by adding 0.050 M CH₃COONa to 0.05 M CH₃COOH
(a) What is it’s pH ?
(b) What will be the new pH if 0.001 moles of HCl is added to 1 liter of the buffer. Assume that, volume remains at 1 liter even after adding the HCl
(c) What will be the pH of a solution obtained by adding 0.001 moles of HCl to 1 liter of pure water?
Solution:
Part (a):
1. Consider the original solution which is: 0.050 M CH₃COOH
• It’s dissociation can be written as:
CH₃COOH (aq) ⇌ CH₃COO^- + H⁺
2. Let at equilibrium, x moles of CH₃COO^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [CH₃COOH] = 0.050-x
         ✰ [CH₃COO^-] = x
         ✰ [H⁺] = x
3. To this equilibrium, we are adding 0.050 M CH₃COONa
• This CH₃COONa will undergo complete dissociation according to the equation:
CH₃COONa ⇌ CH₃COO^- + Na⁺
• Since there is complete dissociation, 0.050 moles of CH₃COO^- will be produced
4. So the various species in the resulting solution are:
CH₃COOH, CH₃COO^-, H⁺ and Na⁺
• Na⁺ is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
CH₃COOH (aq) ⇌ CH₃COO^- + H⁺
5. Based on (2) and (3), we can write:
   ♦ At the new equilibrium,
         ✰ [CH₃COOH] = 0.050-x
         ✰ [CH₃COO^-] = 0.050+x
         ✰ [H⁺] = x
6. Even when a new equilibrium is attained, the equilibrium constant will not change
• The equilibrium constant for this reaction can be obtained from the data book: K_a = 1.76  × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.76 \times 10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{(0.050+x)x}{(0.050-x)}}}$
7. x will be very small when compared to 0.050
• This is because, CH₃COOH is a weak acid. It will not give much CH₃COO^- ions
   ♦ So (0.050 + x) can be taken as 0.050
   ♦ For example, (0.050 + 0.0000001) can be taken as 0.050 for practical purposes
• Similarly, (0.050 – x) can be taken as 0.050
8. Thus the result in (6) becomes:
$\mathbf\small{\rm{1.76 \times 10^{-5}=\frac{(0.050)x}{(0.050)}=x}}$
(The reader can opt not to approximate (0.050+x) and (0.050-x) as 0.050. Then it will become a quadratic equation. The quadratic equation can be solved using a calculator or computer. But the result will be the same 1.76  × 10^-5)
9. Thus we get: [H⁺] = x = 1.76  × 10^-5
• So pH = -log₁₀([H⁺]) = -log₁₀(1.76  × 10^-5) = 4.75 

Part (b):
1. We take 1 liter of the buffer prepared in part (a). To that 1 liter, we add 0.001 moles of HCl
• Given that, the volume does not change. So even after adding HCl, the volume is 1 liter
   ♦ Thus [HCl] = 0.001 M
2. HCl is a strong acid. It dissociates completely
• So we get: [H⁺] = [Cl^-] = 0.001
3. The 1 liter solution will contain the following species:
CH₃COOH, CH₃COO^-, H⁺, Na⁺ and Cl^-
• Na⁺ and Cl^- are stable. They will not take part in the reaction. We can ignore them
   ♦ The concentrations of the remaining species are:
         ✰ [CH₃COOH] = (0.050-x) = 0.050
         ✰ [CH₃COO^-] = (0.050+x) = 0.050
         ✰ [H⁺] = (x + 0.001) = 0.001
• We make the above approximations because x (calculated as 1.76  × 10^-5 in part a), is very small when compared to 0.050 and 0.001
4. The H⁺ will react with CH₃COO^- according to the equation:
CH₃COO^- + H⁺ ⇌ CH₃COOH (aq)
• Nearly all the 0.001 moles of H⁺ will be used up in this way
   ♦ So [CH₃COO^-] will decrease by 0.001
   ♦ Also [CH₃COOH] will increase by 0.001
   ♦ Let y be the final concentration of [H⁺]
5. Then we can write:
   ♦ At equilibrium,
         ✰ [CH₃COO^-] = (0.050 - 0.001) = 0.049
         ✰ [CH₃COOH] = (0.050 + 0.001) = 0.051
         ✰ [H⁺] = y
6. The reaction in (4) is the reverse of
CH₃COOH (aq) ⇌ CH₃COO^- + H⁺
• So for the reaction in (4), we have to take the reciprocal of K_a
• We get: $\mathbf\small{\rm{\frac{1}{K_a}= \frac{1}{1.76 \times 10^{-5}}=\frac{[CH_3COOH]}{[CH_3COO^-][H^+]}=\frac{0.051}{0.049y}}}$
⇒ y = 1.8318  × 10^-5
7. Thus we get: [H⁺] = y = 1.8318 × 10^-5
So pH = -log₁₀([H⁺]) = -log₁₀(1.8318  × 10^-5) = 4.74

Part (c):
1. When 0.001 moles of HCl is added to water, all those HCl molecules will dissociate into H⁺ and Cl^- ions
• So [H⁺] = 0.001
2. Then pH = -log₁₀([H⁺]) = -log₁₀(0.001) = 3.00
3. Let us compare the three pH values
• In part (a), we get:
pH of the buffer = 4.75
• In part (b) we get:
pH after adding 0.001 moles of HCl = 4.74
• In part (c) we get:
pH when 0.001 moles of HCl is added to pure water = 3.00
◼  That means, the buffer is effective in resisting pH change. If there was no buffer, the pH would have fallen from 4.75 to 3. But due to the buffer action, the pH falls from 4.75 to 4.74 only

Solved example 7.80
A buffer is prepared by adding 0.0350 M NH₄Cl to 0.0500 M NH₄OH
(a) What is it’s pH ?
(b) What will be the new pH if 0.001 moles of NaOH is added to 1 liter of the buffer. Assume that, volume remains at 1 liter even after adding the NaOH
(c) What will be the pH of a solution obtained by adding 0.001 moles of NaOH to 1 liter of pure water?
Solution:
Part (a):
1. Consider the original solution which is: 0.050 M NH₄OH
• It’s dissociation can be written as:
NH₄OH ⇌ NH₄⁺ + OH^-
2. Let at equilibrium, x moles of NH₄⁺ be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [NH₄OH] = 0.050-x
         ✰ [NH₄⁺] = x
         ✰ [OH^-] = x
3. To this equilibrium, we are adding 0.0350 M NH₄Cl
• This NH₄Cl will undergo complete dissociation according to the equation:
NH₄Cl ⇌ NH₄⁺ + Cl^-
• Since there is complete dissociation, 0.0350 moles of NH₄⁺ will be produced
4. So the various species in the resulting solution are:
NH₄OH, NH₄⁺, OH^- and Cl^-
• Cl^- is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
NH₄OH ⇌ NH₄⁺ + OH^-
5. Based on (2) and (3), we can write:
   ♦ At the new equilibrium,
         ✰ [NH₄OH] = 0.0500-x
         ✰ [NH₄⁺] = 0.0350+x
         ✰ [OH^-] = x
6. Even when a new equilibrium is attained, the equilibrium constant will not change
• The equilibrium constant for this reaction can be obtained from the data book: K_b = 1.77  × 10^-5
• So we can write: $\mathbf\small{\rm{K_b=1.75 \times 10^{-5}=\frac{[NH_4^+][OH^-]}{[NH_4OH]}=\frac{(0.0350+x)x}{(0.0500-x)}}}$
7. x will be very small when compared to 0.0500 and 0.0350
• This is because, NH₄OH is a weak base. It will not give much NH₄⁺ ions
   ♦ So (0.0500 + x) can be taken as 0.0500
   ♦ For example, (0.0500 + 0.0000001) can be taken as 0.0500 for practical purposes
• Similarly, (0.0350 – x) can be taken as 0.0350
8. Thus the result in (6) becomes:
$\mathbf\small{\rm{1.77 \times 10^{-5}=\frac{(0.0350)x}{(0.0500)}=0.7x}}$
x = 2.529 × 10^-5
(The reader can opt not to approximate (0.0500+x) as 0.0500 and (0.0350+x) as 0.0350. Then it will become a quadratic equation. The quadratic equation can be solved using a calculator or computer. But the result will be the same 2.529  × 10^-5)
9. Thus we get: [OH^-] = x = 2.529  × 10^-5
• So pOH = -log₁₀([OH^-]) = -log₁₀(2.529  × 10^-5) = 4.597
• So pH = (14 - pOH) = (14 - 4.597) = 9.403

Part (b):
1. We take 1 liter of the buffer prepared in part (a). To that 1 liter, we add 0.001 moles of NaOH
• Given that, the volume does not change. So even after adding NaOH, the volume is 1 liter
   ♦ Thus [NaOH] = 0.001 M
2. NaOH is a strong base. It dissociates completely
• So we get: [OH^-] = [Na⁺] = 0.001
3. The 1 liter solution will contain the following species:
NH₄OH, NH₄⁺, OH^-, Na⁺ and Cl^-
• Na⁺ and Cl^- are stable. They will not take part in the reaction. We can ignore them
   ♦ The concentrations of the remaining species are:
         ✰ [NH₄OH] = (0.050-x) = 0.0500
         ✰ [NH₄⁺] = (0.0350+x) = 0.0350
         ✰ [OH^-] = (x + 0.001) = 0.001
• We make the above approximations because x (calculated as 2.529  × 10^-5 in part a), is very small when compared to 0.050, 0.0350 and 0.001
4. The OH^- will react with NH₄⁺ according to the equation:
NH₄⁺ + OH^- ⇌ NH₄OH
• Nearly all the 0.001 moles of OH^- will be used up in this way
   ♦ So [NH₄⁺] will decrease by 0.001
   ♦ Also [NH₄OH] will increase by 0.001
   ♦ Let y be the final concentration of [OH^-]
5. Then we can write:
   ♦ At equilibrium,
         ✰ [NH₄⁺] = (0.0350 - 0.001) = 0.0340
         ✰ [NH₄OH] = (0.0500 + 0.001) = 0.0510
         ✰ [OH^-] = y
6. The reaction in (4) is the reverse of
NH₄OH ⇌ NH₄⁺ + OH^-
• So for the reaction in (4), we have to take the reciprocal of K_b
• We get: $\mathbf\small{\rm{\frac{1}{K_b}= \frac{1}{1.77 \times 10^{-5}}=\frac{[NH_4OH]}{[NH_4^+][OH^-]}=\frac{0.0510}{0.034 \, y}}}$
⇒ y = 2.655  × 10^-5
7. Thus we get: [OH^-] = y = 2.655 × 10^-5
• So pOH = -log₁₀([OH^-]) = -log₁₀(2.655 × 10^-5) = 4.576
• So pH = (14 - pOH) = (14 - 4.576) = 9.424

Part (c):
1. When 0.001 moles of NaOH is added to water, all those NaOH molecules will dissociate into Na⁺ and OH^- ions
• So [OH^-] = 0.001
2. Then pOH = -log₁₀([OH^-]) = -log₁₀(0.001) = 3.00
So PH = (14 - pH) = (14 - 3) = 11
3. Let us compare the three pH values
• In part (a), we get:
pH of the buffer = 9.403
• In part (b) we get:
pH after adding 0.001 moles of NaOH = 9.424
• In part (c) we get:
pH when 0.001 moles of NaOH is added to pure water = 11
◼  That means, the buffer is effective in resisting pH change. If there was no buffer, the pH would have increased from 9.403 to 11. But due to the buffer action, the pH increased from 9.403 to 9.424 only

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• In the above two solved examples:
   ♦ Part (a) demonstrates how we can calculate the pH of buffers
   ♦ Parts (b) and (c) demonstrate how buffer helps to avoid large changes in pH
• In the next section, we will see a few more solved examples related to this category


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Chapter 7.18 - Ionization Constants of Weak Bases

In the previous section, we saw ionization constant (K_a) of weak acids. In this section, we will see ionization constant (K_b) of weak bases. Later in this section, we will also see the relation between K_a and K_b

• In an earlier section, we saw that:
    ♦ Strong bases like NaOH undergo complete ionization when added to water
    ♦ Weak bases do not undergo complete ionization when added to water
• We can write the main features of the ionization constant of weak bases in 7 steps:
1. Let XOH be a weak base. It's ionization process can be represented by the following equation:
XOH(aq) ⇌ X⁺(aq) + OH^-(aq)
2. Let c be the initial concentration of XOH
    ♦ That is., at t = 0, [XOH] = c moles L^-1
3. Let 𝛼 be the extent of ionization
    ♦ That is., 𝛼 is the 'fraction of c' which undergoes dissociation
    ♦ 𝛼 is a fraction like ¹⁄₄, ²⁄₃ etc.,
        ✰ (Remember that, fractions can be expressed as decimals also)  
• Then number of moles of XOH undergoing ionization = c𝛼
4. When c𝛼 moles of XOH undergoes ionization, the number of moles of XOH remaining will be equal to (c - c𝛼)
• Also, from the stoitiometric coefficients, we can write:
    ♦ When c𝛼 moles of XOH undergoes dissociation,
        ✰ c𝛼 moles of X⁺ will be formed
        ✰ c𝛼 moles of OH^- will be formed
5. So we obtained the concentrations at equilibrium
• Using those concentrations, we can write the expression for the equilibrium constant
    ♦ It is called the ionization constant of the base XOH
    ♦ It is denoted as K_b
• So we can write $\mathbf\small{\rm{K_b=\frac{(c\alpha)^2}{c-c\alpha}}}$
• Thus we get Eq.7.10: $\mathbf\small{\rm{K_b=\frac{c^2\alpha^2}{c(1-\alpha)}}}$
(Note that, we do not consider the concentration of H₂O because, it is a pure liquid)
6. To write K_a, we can use the earlier method also
• We get Eq.7.11: $\mathbf\small{\rm{K_b=\frac{[X^+][OH^-]}{[XOH]}}}$
7. It is clear that, if [X⁺] and [OH^-] are larger, K_b will be larger
• [X⁺] and [OH^-] will be larger for strong acids
◼ So we can write:
Strong acids will have a large value of K_b

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• In previous sections, we have seen that:
If the equilibrium constant K is known, the concentrations at equilibrium can be calculated (see solved example 7.6 in section 7.5). We saw similar solved examples involving K_a in the previous section also
• In our present case, we have K_b in place of K_a
    ♦ If this K_b is known, we can calculate the concentrations at equilibrium
    ♦ Once the concentrations are known, we can calculate pH and 𝛼
         ✰ (Recall that, basic solutions have pH greater than 7)
    ♦ The following solved examples demonstrate the procedure

Solved example 7.68
The pH of 0.004 M hydrazine solution is 9.7. Calculate it's ionization constant K_b and pK_b
Solution:
1. The balanced equation for the dissociation of hydrazine is:
NH₂NH₂(aq) + H₂O(l) ⇌ NH₂NH₃⁺(aq) + OH^-(aq)
2. Given that, pH is 9.7
• We have: pH = -log₁₀([H⁺])
• Substituting the given pH, we get: 9.7 = -log₁₀([H⁺])
2. This is same as: log₁₀([H⁺]) = -9.7
• So [H⁺] will be equal to the antilog of -9.7
• Thus we get: [H⁺] = antilog (-9.7) = 1.9952 × 10^-10 M
3. We have: [H⁺][OH^-] = 10^-14
• So [OH^-] = 5.012 × 10^-5 M
4. From the stoitiometric coefficients, it is clear that:
[NH₂NH₃⁺] will be same as [OH^-]
5. So we get: [NH₂NH₃⁺] = 5.012 × 10^-5
• So [NH₂NH₂] = (0.004 - 5.012 × 10^-5) = 0.00394 ≃ 0.004
6. Next we calculate K_b. We have:
$\mathbf\small{\rm{K_b=\frac{[NH_2NH_3^+][OH^-]}{[NH_2NH_2]}=\frac{(5.012 \times 10^{-5})^2}{0.004}}}$ = 6.36 × 10^-7
7. So pK_b = -log₁₀(K_b) = -log₁₀(6.36 × 10^-7) = 6.2

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Relation between K_a and K_b

Relation between K_a and K_b can be explained with the help of an example. It can be written in 9 steps:
1. Consider the reaction between NH₃ and H₂O. The balanced equation is:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)
• Here NH₃ acts as the base because, it accepts a proton and also produces OH^- ions in the aqueous solution
• So we can write the base dissociation constant as: $\mathbf\small{\rm{K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}}}$
2. In this reaction, we can identify the conjugate acid and conjugate base
(See step 13 in section 7.12)
• We can write them as:
    ♦ NH₃ is the base, H₂O is the acid
    ♦ NH₄⁺ is the conjugate acid, OH^- is the conjugate base
3. We see that, NH₄⁺ is the conjugate acid
• Let us see it's reaction with water. The balanced equation is:
NH₄⁺(aq) + H₂O(l) ⇌ H₃O⁺(aq) + NH₃(aq)
• Here NH₄⁺ indeed acts as the acid because, it donates a proton and also produces H₃O⁺ ions in the aqueous solution
• So we can write the acid dissociation constant as: $\mathbf\small{\rm{K_a=\frac{[H_3O^+][NH_3]}{[NH_4^+]}}}$
4. Let us add the reactions in (1) and (3)
• It can be written in 4 steps:
(i) On the left side, we will be having four items:
NH₃(aq) + H₂O(l) + NH₄⁺(aq) + H₂O(l)
(ii) On the right side, we will be having four items:
NH₄⁺(aq) + OH^-(aq) + H₃O⁺(aq) + NH₃(aq)
(iii) So after addition, we will get:
NH₃(aq) + H₂O(l) + NH₄⁺(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq) + H₃O⁺(aq) + NH₃(aq)
(iv) NH₃(aq) and NH₄⁺(aq) are common on both sides. So the net reaction is:
H₂O(l) + H₂O(l) ⇌ OH^-(aq) + H₃O⁺(aq)
5. The above net reaction is familiar to us
• We know the ionization constant of that reaction. It is: K_w = 10^-14
6. Now let us find the product of two items:
    ♦ K_b that we wrote in (1)
    ♦ K_a that we wrote in (3)
• We get: K_aK_b = $\mathbf\small{\rm{\frac{[H_3O^+][NH_3]}{[NH_4^+]}\times \frac{[NH_4^+][OH^-]}{[NH_3]}}}$
⇒ K_aK_b = [H₃O⁺][OH^-]
• We have seen this product before. It is equal to K_w
    ♦ Also, K_w is the ionisation constant of the net reaction obtained in (4)
• So we can write: K_aK_b = K_w
7. Let us write a summary of the above steps:
• We added the two reactions in (1) and (3) and obtained the net reaction
• We obtained the product of the ionization constants of the reactions in (1) and (3)
• We found that:
    ♦ The product
    ♦ is equal to
    ♦ The ionization constant of the net reaction
• The reactions in (1) and (3) are related to conjugate acid-base pair
◼ So we can write:
For a conjugate acid-base pair, K_aK_b = K_w
8. We can write this information as a general rule. It can be written in 3 steps:
(i) We have a few reactions
    ♦ We write the equilibrium constants of those reactions : K₁, K₂, K₃ . . .
    ♦ We write the product of those constants: K₁ × K₂ × K₃ . . .
(ii) We add the reactions and write the net reaction
• We denote the equilibrium constant of this net reaction as K_NET
(iii) Then K_NET will be equal to the product in (i)
• Thus we get Eq.7.10: K_NET = K₁ × K₂ × K₃ . . .
9. Consider the result that we wrote in 6 : K_aK_b = K_w
• We know that, the K values involve negative powers. In order to avoid those negative powers and thus make it more presentable, we can use logarithms
• We get:
log₁₀(K_a) + log₁₀(K_b) = log₁₀(K_w) = log₁₀(10^-14)
• Multiplying throughout by -1, we get:
-log₁₀(K_a) + -log₁₀(K_b) = -log₁₀(K_w) = -log₁₀(10^-14)
• But we have:
    ♦ -log₁₀(K_a)= pK_a
    ♦ -log₁₀(K_b)= pK_b
    ♦ -log₁₀(K_w)= pK_w
    ♦ -log₁₀(10^-14)= -(-14) = 14
• Thus we get Eq.7.11: pK_a + pK_b = pK_w = 14

Solved example 7.69
Determine the degree of ionization and pH of a 0.05 M ammonia solution. The ionization constant of ammonia is 1.77 × 10^-5. Also calculate the ionization constant of the conjugate acid of ammonia
Solution:
1. The balanced equation is:
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH^-(aq)
• Let at equilibrium, x moles each of NH₄⁺ and OH^- be present
• From the balanced equation, it is clear that, if x mol each of NH₄⁺ and OH^- are formed, the same x mol of NH₃ would be consumed
• So the concentration of NH₃ at equilibrium would be (0.05 - x) mol
2. For this reaction, K_b can be obtained as: $\mathbf\small{\rm{K_b=\frac{[NH_4^+][OH^-]}{[NH_3]}}}$
• Substituting the values from (1), we get:
1.77 × 10^-5 = $\mathbf\small{\rm{\frac{x^2}{0.05-x}}}$
⇒ x² + (1.77 × 10^-5)x - 0.0885 × 10^-5 = 0
3. Solving this quadratic equation, we get: x = 0.000932 or -0.000949
• negative value is not acceptable. So we take x = 0.000932
4. So we can write:
• The concentrations at equilibrium are:
   ♦ [NH₄⁺] = [OH^-] = x = 0.000932 M
   ♦ [NH₃] = (0.05 - x) = 0.049
5. Once we know [OH^-], we can calculate the pH
• We have: pH = (14 - pOH)= [14 - -log₁₀([OH^-])]
= [14 - -log₁₀(0.00932)] = [14 - 3.031] = 10.97
6. We can find 𝛼 as follows:
• We have: c𝛼 = x = 0.000932
• c = 0.05 M
• So 𝛼 = ^0.00932⁄_0.05 = 0.018
7. We can find K_a of the conjugate acid of ammonia as follows:
• We have seen that, for a conjugate acid-base pair, K_aK_b = K_w = 10^-14
• So we get: $\mathbf\small{\rm{K_b=\frac{10^{-14}}{K_a}= \frac{10^{-14}}{1.77 \times 10^{-5}}}}$ = 5.64 × 10^-10

Solved example 7.70
The ionization constant of HF, HCOOH and HCN at 298 K are 6.8 × 10^-4, 1.8 × 10^-4 and 4.8 × 10^-9 respectively. Calculate the ionization constant of the corresponding conjugate base
Solution:
• For a conjugate acid-base pair, K_aK_b = K_w = 10^-14
• In this problem, K_a values are given
• So we get: $\mathbf\small{\rm{K_b=\frac{10^{-14}}{K_a}}}$
(i) K_b of the conjugate base of HF = $\mathbf\small{\rm{\frac{10^{-14}}{6.8 \times 10^{-4}}}}$ = 1.47 × 10^-11
(i) K_b of the conjugate base of HCOOH = $\mathbf\small{\rm{\frac{10^{-14}}{1.8 \times 10^{-4}}}}$ = 5.55 × 10^-11
(i) K_b of the conjugate base of HCN = $\mathbf\small{\rm{\frac{10^{-14}}{4.8 \times 10^{-9}}}}$ = 2.08 × 10^-6

Solved example 7.71
The pH of 0.005M codeine (C₁₈H₂₁NO₃) solution is 9.95. Calculate its ionization constant and pK_b
Solution:
1. Given that pH = 9.95
• This is greater than 14. So it is a basic solution
• pOH will be (14 - 9.95) = 4.05
2. From pOH, we can calculate [OH^-]
[OH^-] = antilog of -4.05 = 8.913 × 10^-5
3. We are asked to find the ionization constant K_b
• We have: $\mathbf\small{\rm{K_b=\frac{[positive \; ion][OH^-]}{[C_{18}H_{21}NO_3]}}}$
• [positive ion] will be same as [OH^-]
• Thus we get [positive ion] = 8.913 × 10^-5 M
4. Next we want [C₁₈H₂₁NO₃] at equilibrium
• Assume that, only a very small portion of the solute undergoes dissociation
    ♦ That is., 𝛼 is very small
• Then [C₁₈H₂₁NO₃] will be same as the initial concentration, which is 0.005
5. Thus we get: $\mathbf\small{\rm{K_b=\frac{(8.913 \times 10^{-5})^2}{0.005}}}$ = 1.588 × 10^-6
6. pK_b = -log(K_b) = -log(1.588 × 10^-6) = 5.79

Solved example 7.72
What is the pH of 0.001 M aniline solution? The ionization constant of aniline is 4.27 × 10^-10. Calculate the degree of ionization of aniline in the solution. Also calculate the ionization constant of the conjugate acid of aniline
Solution:
1. The balanced equation is:
C₆H₅NH₂(aq) + H₂O(l) ⇌ C₆H₅NH₄⁺(aq) + OH^-(aq) 
• Let at equilibrium, x moles each of C₆H₅NH₄⁺ and OH^- be present
• From the balanced equation, it is clear that, if x mol each of C₆H₅NH₄⁺ and OH^- are formed, the same x mol of C₆H₅NH₂ would be consumed
• So the concentration of C₆H₅NH₂ at equilibrium would be (0.001 - x) mol
2. For this reaction, K_b can be obtained as: $\mathbf\small{\rm{K_b=\frac{[C_6H_5NH_4^+][OH^-]}{[C_6H_5NH_2]}}}$
• Substituting the values from (1), we get:
4.27 × 10^-10 = $\mathbf\small{\rm{\frac{x^2}{0.001-x}}}$
⇒ x² + (4.27 × 10^-10)x - 4.27 × 10^-13 = 0
3. Solving this quadratic equation, we get: x = 6.532 × 10^-7 or -6.536 × 10^-7
• negative value is not acceptable. So we take x = 6.532 × 10^-7
4. So we can write:
• The concentrations at equilibrium are:
   ♦ [C₆H₅NH₄⁺] = [OH^-] = x = 6.532 × 10^-7 M
   ♦ [NH₃] = (0.001 - x) = 0.00099
5. Once we know [OH^-], we can calculate the pH
• We have: pH = (14 - pOH)= [14 - -log₁₀([OH^-])]
= [14 - -log₁₀(6.532 × 10^-7)] = [14 - 6.18] = 7.82
6. We can find 𝛼 as follows:
• We have: c𝛼 = x = 6.532 × 10^-7
• c = 0.001 M
• So 𝛼 = 6.532 × 10^-7/0.001 = 0.000653
7. We can find K_a of the conjugate acid of ammonia as follows:
• We have seen that, for a conjugate acid-base pair, K_aK_b = K_w = 10^-14
• So we get: $\mathbf\small{\rm{K_b=\frac{10^{-14}}{K_a}= \frac{10^{-14}}{4.27 \times 10^{-10}}}}$ = 2.34 × 10^-5

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In the next section, we will see polybasic acids and polyacidic bases


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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 7.12 - 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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