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 7.17 - Ionization Constants of Weak Acids

In the previous section, we completed a discussion on pH scale. In this section, we will see ionization constants of weak acids

• In an earlier section, we saw that:
    ♦ Strong acids like HCl undergo complete ionization when added to water
    ♦ Weak acids do not undergo complete ionization when added to water
• We can write the main features of the ionization constant of weak acids in 7 steps:
1. Let HX be a weak acid. It's ionization process can be represented by the following equation:
HX(aq) + H₂O(l) ⇌ H₃O⁺(aq) + X^-(aq)
2. Let c be the initial concentration of HX
    ♦ That is., at t = 0, [HX] = 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 HX undergoing ionization = c𝛼
4. When c𝛼 moles of HX undergoes ionization, the number of moles of HX remaining will be equal to (c - c𝛼)
• Also, from the stoitiometric coefficients, we can write:
    ♦ When c𝛼 moles of HX undergoes dissociation,
        ✰ c𝛼 moles of H₃O⁺ will be formed
        ✰ c𝛼 moles of X^- 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 acid HX
    ♦ It is denoted as K_a
• So we can write $\mathbf\small{\rm{K_a=\frac{(c\alpha)^2}{c-c\alpha}}}$
• Thus we get Eq.7.8: $\mathbf\small{\rm{K_a=\frac{c^2\alpha^2}{c(1-\alpha)}}}$
(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.7.9: $\mathbf\small{\rm{K_a=\frac{[H^+][X^-]}{[HX]}}}$
7. It is clear that, if [H⁺] and [X^-] are larger, K_a will be larger
• [H⁺] and [X^-] will be larger for strong acids
◼ So we can write:
Strong acids will have a large value of K_a

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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)
• In our present case, we have K_a in place of K
    ♦ If this K_a is known, we can calculate the concentrations at equilibrium
    ♦ Once the concentrations are known, we can calculate pH and 𝛼
    ♦ The following solved examples demonstrate the procedure

Solved example 7.62
The ionization constant of HF is 3.2 × 10^-4 . Calculate the degree of dissociation of HF in its 0.02 M solution. Calculate the concentration of all species present (H₃O⁺, F^- and HF) in the solution and its pH.
Solution:
1. The balanced equation for the dissociation of HF is:
HF(aq) + H₂O(l) ⇌ H₃O⁺(aq) + F^-(aq)
• Let at equilibrium, x moles each of H₃O⁺ and F^- be present
• From the balanced equation, it is clear that, if x mol each of H₃O⁺ and F^- are formed, the same x mol of HF would be consumed
• So the concentration of HF at equilibrium would be (0.02 - x) mol
2. For this reaction, K_a can be obtained as: $\mathbf\small{\rm{K_a=\frac{[H_3O^+][F^-]}{[HF]}}}$
• Substituting the values from (1), we get:
3.2 × 10^-4 = $\mathbf\small{\rm{\frac{x^2}{0.02-x}}}$
3. Solving this quadratic equation, we get: x = 0.002375 or -0.00269
• negative value is not acceptable. So we take x = 0.002375
4. So we can write:
• The concentrations at equilibrium are:
[H₃O⁺] = [F^-] = x = 0.0024 M
[HF] = (0.02 - x) = 0.01763
5. Once we know [H₃O⁺], we can calculate the pH
We have: pH = -log₁₀([H₃O⁺]) = -log₁₀(0.0024) = 2.62
6. We can find 𝛼 as follows:
• We have: c𝛼 = x = 0.0024
• c = 0.002 M
• So 𝛼 = ^0.0024⁄_0.02 = 0.12

Solved example 7.63
The pH of 0.1 M monobasic acid is 4.50. Calculate the concentration of species H⁺, A^- and HA at equilibrium. Also, determine the value of K_a and pK_a of the monobasic acid
Solution:
1. Monobasic acid will have only one H atom
• We have: pH = -log₁₀([H⁺])
• Substituting the given pH, we get: 4.50 = -log₁₀([H⁺])
2. This is same as: log₁₀([H⁺]) = -4.50
• So [H⁺] will be equal to the antilog of -4.50
3. Thus we get: [H⁺] = antilog (-4.50) = 0.00003162 = 3.16 × 10^-5 M
4. The balanced equation is: HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A^-(aq)
• From the stoitiometric coefficients, it is clear that:
[A^-] will be equal to [H⁺]
• So we get: [A^-] = 3.16 × 10^-5 M
5. If 3.16 × 10^-5 M of [H⁺] and [A^-] are produced, the [HA] will be equal to:
(0.1 - 3.16 × 10^-5) = 0.0999684 ≃ 0.1 M
6. Once we know the concentrations, we can calculate K_a:
$\mathbf\small{\rm{K_a=\frac{[H_3O^+][A^-]}{[HA]}=\frac{(3.16 \times 10^{-5})^2}{0.1}}}$ = 1.0 × 10^-8
7. We have: pH = -log₁₀([H⁺])
• In a similar ways, we can write: pK_a = -log₁₀([H⁺])
• Thus we get: pK_a = -log₁₀(10⁸) = 8
8. Significance of pK_a can be written in 4 steps:
(i) We see that, if the 'negative power' of K_a is large, pK_a will be large
(ii) If the K_a has a large negative power, it indicates that, K_a is small
(iii) If K_a is small, it indicates that, the acid is weak
(iv) So we can write:
If pK_a is large, it will be a weak acid
9. Alternatively, we can use 𝛼 (expressed as a percentage) also to express the strength of weak acid. It can be written in 3 steps:
(i) We have: $\mathbf\small{\rm{\alpha=\frac{c \alpha}{c}=\frac{[HA]_{dissociated}}{[HA]_{initial}}}}$
(ii) So we get: $\mathbf\small{\rm{\alpha \, (percentage)=\frac{[HA]_{dissociated}}{[HA]_{initial}}\times 100}}$
(iii) When 𝛼 is expressed as a percentage, it is called percent dissociation

Solved example 7.64
Calculate the pH of 0.08 M solution of hypochlorous acid HOCl. The ionization constant of the acid is 2.5 × 10^-5. Determine percentage dissociation of HOCl
Solution:
1. The balanced equation for the dissociation of HOCl is:
HOCl(aq) + H₂O(l) ⇌ H₃O⁺(aq) + ClO^-(aq)
• Let at equilibrium, x moles each of H₃O⁺ and ClO^- be present
• From the balanced equation, it is clear that, if x mol each of H₃O⁺ and ClO^- are formed, the same x mol of HOCl would be consumed
• So the concentration of HOCl at equilibrium would be (0.08 - x) mol
2. For this reaction, K_a can be obtained as: $\mathbf\small{\rm{K_a=\frac{[H_3O^+][ClO^-]}{[HOCl]}}}$
• Substituting the values from (1), we get:
2.5× 10^-5 = $\mathbf\small{\rm{\frac{x^2}{0.08-x}}}$
⇒ x² + (2.5 × 10^-5)x - 0.2 × 10^-5 = 0
3. Solving this quadratic equation, we get: x = 0.001402 or -0.001427
• negative value is not acceptable. So we take x = 0.001402
4. So we can write:
• The concentrations at equilibrium are:
   ♦ [H₃O⁺] = [ClO-] = x = 0.001402 M
   ♦ [HOCl] = (0.08 - x) = 0.0786
5. Once we know [H₃O⁺], we can calculate the pH
• We have: pH = -log₁₀([H₃O⁺]) = -log₁₀(0.001402) = 2.85
6. We can find 𝛼 as follows:
• We have: c𝛼 = x = 0.001402
• c = 0.08 M
• So 𝛼 = ^0.001402⁄_0.08 = 0.0175
7. Thus we get:
Percent dissociation = 𝛼 expressed as percentage = (0.0175 × 100) = 1.75%

Solved example 7.65
The ionization constant of acetic acid is 1.74 × 10^-5. Calculate the degree of
dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of
acetate ion in the solution and its pH.
Solution:
1. The balanced equation for the dissociation of acetic acid is:
CH₃COOH(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃COO^-(aq)
• Let at equilibrium, x moles each of H₃O⁺ and CH₃COO^- be present
• From the balanced equation, it is clear that, if x mol each of H₃O⁺ and CH₃COO^- are formed, the same x mol of CH₃COOH would be consumed
• So the concentration of CH₃COOH at equilibrium would be (0.05 - x) mol
2. For this reaction, K_a can be obtained as: $\mathbf\small{\rm{K_a=\frac{[H_3O^+][CH3COO^-]}{[CH3COOH]}}}$
• Substituting the values from (1), we get:
1.74 × 10^-5 = $\mathbf\small{\rm{\frac{x^2}{0.05-x}}}$
⇒ x² + (1.74 × 10^-5)x - 0.087 × 10^-5 = 0
3. Solving this quadratic equation, we get: x = 0.000924 or -0.000941
• negative value is not acceptable. So we take x = 0.000924
4. So we can write:
• The concentrations at equilibrium are:
   ♦ [H₃O⁺] = [CH3COO-] = x = 0.000924 M
   ♦ [CH3COOH] = (0.05 - x) = 0.0491 ≃ 0.05 M
5. Once we know [H₃O⁺], we can calculate the pH
We have: pH = -log₁₀([H₃O⁺]) = -log₁₀(0.000924) = 3.03

Solved example 7.66
It has been found that the pH of a 0.01M solution of an organic acid is 4.15. Calculate the concentration of the anion, the ionization constant of the acid and its pK_a .
Solution:
1. We have: pH = -log₁₀([H⁺])
• Substituting the given pH, we get: 4.15 = -log₁₀([H⁺])
2. This is same as: log₁₀([H⁺]) = -4.15
• So [H⁺] will be equal to the antilog of -4.15
3. Thus we get: [H⁺] = antilog (-4.15) = 7.079 × 10^-5 M
4. The balanced equation is: HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A^-(aq)
• From the stoitiometric coefficients, it is clear that:
[A^-] will be equal to [H⁺]
• So we get: [A^-] = 7.079 × 10^-5 M
5. If 7.079 × 10^-5 M of [H⁺] and [A^-] are produced, the [HA] will be equal to:
(0.01 - 7.079 × 10^-5) = 0.0099 ≃ 0.01 M
6. Once we know the concentrations, we can calculate K_a:
$\mathbf\small{\rm{K_a=\frac{[H_3O^+][A^-]}{[HA]}=\frac{(7.079 \times 10^{-5})^2}{0.01}}}$ = 5.01 × 10^-7
7. We have: pH = -log₁₀([H⁺])
• In a similar ways, we can write: pK_a = -log₁₀([K_a])
• Thus we get: pK_a = -log₁₀(5.01 × 10^-7) = 6.3

Solved example 7.67
The degree of ionization of a 0.1M bromoacetic acid solution is 0.132. Calculate
the pH of the solution and the pK_a of bromoacetic acid.
Solution:
1. Given that c = 0.1 and 𝛼 = 0.132
So [H⁺] = c𝛼 = (0.1 0.132) = 0.0132
2. pH = -log₁₀([H⁺]) = -log₁₀(0.0132) = 1.88
3. We have: $\mathbf\small{\rm{K_a=\frac{c^2\alpha^2}{c-c \alpha)}}}$
• Substituting the values, we get: $\mathbf\small{\rm{K_a=\frac{(0.0132)^2}{0.1-0.0132}}}$ = 0.002
4. pK_a = -log₁₀([H⁺]) = -log₁₀(0.002) = 2.7

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In the next section, we will see ionization constants of weak bases


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