Chapter 7.24 - Common Ion Effect and pH effect on Solubility

In the previous section, we saw solubility product constant. We saw some solved examples also. In this section we will see two more solved examples. After that, we will see common ion effect on solubility. We will also see the effect of pH on solubility

Solved example 7.92
Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_sp = 7.4 × 10^-8)
Solution:
1. Let V be the original volume of NaIO₃ (sodium iodate)
• In 1 liter, there will be 0.002 moles of NaIO₃
   ♦ So in V liters, there will be 0.002V moles of NaIO₃
2. Cu(ClO₃)₂ (cupric chlorate) will be having the same volume V
• In 1 liter, there will be 0.002 moles of Cu(ClO₃)₂
   ♦ So in V liters, there will be 0.002V moles of Cu(ClO₃)₂
3. The two solutions are mixed together. So the volume of the final mixture will be 2V
• Then molarity of NaIO₃ in the final mixture
= ^{No. of moles}⁄_Volume = ^0.002V⁄_2V = 0.001
• Similarly, molarity of Cu(ClO₃)₂ in the final mixture
= ^{No. of moles}⁄_Volume = ^0.002V⁄_2V = 0.001
4. Dissociation equations:
• NaIO₃ will dissociate as: NaIO₃ (s) ⇌ Na⁺ (aq) + IO₃^- (aq)
   ♦ NaIO₃ will completely dissociate to give the following concentrations:
         ✰ [Na⁺] = 0.001
         ✰ [IO₃^-] = 0.001
• Cu(ClO₃)₂ will dissociate as: Cu(ClO₃)₂ (s) ⇌ Cu²⁺ (aq) + 2ClO₃^- (aq)
   ♦ Cu(IO₃)₂ will completely dissociate to give the following concentrations:
         ✰ [Cu²⁺] = 0.001
         ✰ [ClO₃^-] = (2 × 0.001) = 0.002
5. Cu²⁺ and IO₃^- will combine together to give Cu(IO₃)₂ (cupric iodate):
Cu²⁺ + 2IO₃^- ⇌ Cu(IO₃)₂
6. The solid Cu(IO₃)₂ thus formed will be in equilibrium with it's two ions:
Cu(IO₃)₂ (s) ⇌ Cu²⁺ (aq)+ 2IO₃^- (aq)
6. Q of this reaction is given by: Q = [Cu²⁺][IO₃^-]² = (0.001)(0.001)² = 10^-9
◼  This Q is less than K_sp of Cu(IO₃)₂
7. So the system will try to increase Q
• That means, more and more Cu(IO₃)₂ will dissolve
• The precipitation of Cu(IO₃)₂ will not occur

Solved example 7.93
What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, K_sp = 6.3× 10^-18)
Solution:
1. Let V be the original volume of FeSO₄ (ferrous sulphate)
• Let x be the molarity of the solution
⇒ In 1 liter, there will be x moles of FeSO₄
⇒ In V liters, there will ve Vx moles of FeSO₄ 
2. Let the same V be the original volume of Na₂S (sodium sulfide)
• Let the molarity be the same x
⇒ In 1 liter, there will be x moles of Na₂S
⇒ In V liters, there will ve Vx moles of Na₂S
3. The two solutions are mixed together. So the volume of the final mixture will be 2V
• Then molarity of FeSO₄ in the final mixture
= ^{No. of moles}⁄_Volume = ^Vx⁄_2V = 0.5x
• Similarly, molarity of Na₂S in the final mixture
= ^{No. of moles}⁄_Volume = ^Vx⁄_2V = 0.5x
4. Dissociation equations:
• FeSO₄ will dissociate as: FeSO₄ (s) ⇌ Fe²⁺ (aq) + SO₄^2- (aq)
   ♦ FeSO₄ will completely dissociate to give the following concentrations:
         ✰ [Fe²⁺] = 0.5x
         ✰ [SO₄^2-] = 0.5x
• Na₂S will dissociate as: Na₂S (s) ⇌ 2Na⁺ + S^2-
   ♦ Cu(IO₃)₂ will completely dissociate to give the following concentrations:
         ✰ [Na⁺] = (2 × 0.5x) = x
         ✰ [S^2-] = 0.5x
5. Fe²⁺ and S^2- will combine together to give FeS (iron sulphide):
Fe²⁺ + S^2- ⇌ FeS
6. The solid FeS thus formed will be in equilibrium with it's two ions:
FeS (s) ⇌ Fe²⁺ (aq) + S^2- (aq)
6. Q of this reaction is given by: Q = [Fe²⁺][S^2-] = (0.5x)(0.5x) = 0.25x²
◼  This Q must be less than or equal to K_sp
If it is greater than K_sp, precipitation willtake place
7. To find the maximum aloowable value, we will equate the two:
Q = K_sp
⇒ 0.25x² = 6.3 × 10^-18
⇒ x = 5.02 × 10^-9

Solved example 7.94
The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen
sulphide is 1.0 × 10^-19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of
the following: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. In which of these solutions
precipitation will take place?
Solution:
1. Finding the concentration of S^2- (sulphide ions):
(i) The original concentration of S^2- is 1.0 × 10^-19 M
⇒ 1 liter will contain 1.0 × 10^-19 moles of S^2-
⇒ 1 mL will contain (1.0 × 10^-19 ÷ 1000) = 1.0 × 10^-22 moles of S^2-
⇒ 10 mL will contain (1.0 × 10^-22 × 10) = 1.0 × 10^-21 moles of S^2- ions
(ii) This 10 mL is added to 5 mL each of matallic solutions
⇒ The final volumes of each metallic solution is (10 + 5) = 15 mL
⇒ 1.0 × 10^-21 moles of S^2- is present in 15 mL of solution
⇒ No. of moles of S^2- in 1 mL = (1.0 × 10^-21 ÷ 15) = 6.67 × 10^-23
⇒ Concentration of S^2-
= No. of moles in 1 liter = (6.67 × 10^-23 × 1000) = 6.67 × 10^-20 moles
⇒ Concentration of S^2- in the 15 mL solution = [S^2-] = 6.67 × 10^-20 M
2. FeSO₄ will completely dissociate in aqueous solution
• So 0.04 M FeSO4 will give 0.04 M Fe²⁺ ions
• Let us find the concentration of Fe²⁺:
(i) The original concentration of Fe²⁺ is 0.04 M
⇒ 1 liter will contain 0.04 moles of Fe²⁺
⇒ 1 mL will contain (0.04 ÷ 1000) = 4.0 × 10^-5 moles of Fe²⁺
⇒ 5 mL will contain (4.0 × 10^-5 × 5) = 20.0 × 10^-5 moles of Fe²⁺ ions
(ii) This 5 mL is added to 10 mL of sulphide solutions
⇒ The final volumes of each metallic solution is (10 + 5) = 15 mL
⇒ 20.0 × 10^-5 moles of Fe²⁺ is present in 15 mL of solution
⇒ No. of moles of Fe²⁺ in 1 mL = (20.0 × 10^-5 ÷ 15) = 1.33 × 10^-5
⇒ Concentration of Fe²⁺ = No. of moles in 1 liter
= (1.33 × 10^-5 × 1000) = 1.33 × 10^-2 moles
⇒ Concentration of Fe²⁺ in the 15 mL solution = [Fe²⁺] = 1.33 × 10^-2 M
3. In a similar way, MnCl₂, ZnCl₂ and CdCl₂ will also completely dissociate into ions. So we get:
[Mn²⁺] = [Zn²⁺] = [Cd²⁺] = 1.33 × 10^-2 M
4. Consider the solution of FeSO₄
• The Fe²⁺ will combine with S^2- to form FeS
• The Q of this reaction = [Fe²⁺][S^2-] = (1.33 × 10^-2 × 6.67 × 10^-20) = 8.87 × 10^-22
◼ From the data book, we have: K_sp of FeS = 6.3 × 10^-18
    ♦ So Q has to increse. That means, the concentration of ions must increase
    ♦ So more and more FeS will dissolve. It will not precipitate
5. Consider the solution of MnCl₂
• The Mn²⁺ will combine with S^2- to form MnS
• The Q of this reaction = [Mn²⁺][S^2-] = (1.33 × 10^-2 × 6.67 × 10^-20) = 8.87 × 10^-22
◼ From the data book, we have: K_sp of MnS = 2.5 × 10^-13
    ♦ So Q has to increse. That means, the concentration of ions must increase
    ♦ So more and more MnS will dissolve. It will not precipitate
6. Consider the solution of ZnCl₂
• The Zn²⁺ will combine with S^2- to form ZnS
• The Q of this reaction = [Zn²⁺][S^2-] = (1.33 × 10^-2 × 6.67 × 10^-20) = 8.87 × 10^-22
◼ From the data book, we have: K_sp of ZnS = 1.6 × 10^-24
    ♦ So Q has to decrease. That means, the concentration of ions must decrease
    ♦ So more and more solid ZnS has to form. It will precipitate
7. Consider the solution of CdCl₂
• The Cd²⁺ will combine with S^2- to form CdS
• The Q of this reaction = [Cd²⁺][S^2-] = (1.33 × 10^-2 × 6.67 × 10^-20) = 8.87 × 10^-22
◼ From the data book, we have: K_sp of CdS = 8.0 × 10^-27
    ♦ So Q has to decrease. That means, the concentration of ions must decrease
    ♦ So more and more solid CdS has to form. It will precipitate

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Common ion effect on solubility

Some basics can be written in 3 steps:
1. Consider the general form of an ionic equilibrium:
M_x^p+X_y^q- ⇌ M_x^p+ + X_y^q-
   ♦ On the left side of the equation, we have the salt
   ♦ On the right side of the equation, we have the ions
• The solid salt will be in equilibrium with the ions
2. If we increase the concentration of one of the ions, the equilibrium is disturbed
• Then according to Le Chatelier’s principle,the system will try to decrease the effect of those extra ions
• For that, some of the extra ions will combine with the opposite ions to give the salt
• This leads to precipitation of the salt
3. If we decrease the concentration of one of the ions, then also, equilibrium is disturbed
• Then according to Le Chatelier’s principle,the system will try to decrease the effect of those depleted ions
• For that, some of the solid will dissociate into the solution
• This leads to a decrease in the quantity of solids in the solution

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Let us see an example. It can be written in 7 steps:
1. A vessel contains Ag⁺ (silver ions) in a solution
• We want to know how much silver ions are present in that solution
2. For that, take about 10 mL of that solution
• To that 10 mL, add enough Cl^- ions
3. The equilibrium is disturbed due to the excess Cl^- ions
• In order to reduce the effect of excess Cl^- ions, they combine with Ag+ ions to form AgCl
• The reaction is: Ag⁺ + Cl^- ⇌ AgCl
    ♦ AgCl is insoluble in water. It will precipitate
4. If there is enough Cl^- ions, all the Ag⁺ in that 10 mL will precipitate as solid AgCl
• Solid AgCl is sparingly soluble in water. So it can be easily separated form the solution
5. The mass of that solid AgCl is measured
◼ From that mass, we get an important information:
The quantity of Ag⁺ ions in the 10 mL solution
6. If quantity in the 10 mL solution is known, we can easily calculate:
The quantity of Ag⁺ ions in the original solution
◼ This method of finding the quantity of a substance is called gravimetric estimation
7. Let us see two more examples:
• We can calculate the quantity of ferric ions
    ♦ For that, ferric ions are made to precipitate as ferric hydroxide
    ♦ Ferric hydroxide is sparingly soluble in water
• The quantity of barium ions
    ♦ For that, barium ions are made to precipitate as barium sulphate
    ♦ Barium sulphate is sparingly soluble in water

Solved example 7.95
Calculate the molar solubility of Ni(OH)₂ in 0.10 M NaOH. The ionic product of
Ni(OH)₂ is 2.0 × 10^-15
Solution:
1. The dissolution of Ni(OH)₂ will be according to the equation:
Ni(OH)₂ (s) ⇌ Ni^p+ (aq) + 2OH^q- (aq)
• The superscripts p+ and q- are given only to indicate that they are ions. Those charges do not have any effect on S
2. When one mole of Ni(OH)₂ dissolve in water,
   ♦ 1 mole of Ni^p+ is produced
   ♦ 2 moles of OH^q- are produced
3. So when S moles of Ni(OH)₂ dissolve in water,
   ♦ S moles of Ni^p+ are produced
   ♦ 2S moles of OH^q- are produced
4. But in our present case, 0.10 moles of OH^q- is already present
    ♦ This extra OH^q- ions is due to the dissociation of 0.10 M NaOH
• Then we can write:
K_sp = [Ni^p+]¹[OH^q-]² = (S)¹(2S+0.10)²
• Here, 2S will be very small (when compared to 0.10) because, the solubility of Ni(OH)₂ is very small
    ♦ So we can approximate (2S+0.10) as 0.10
5. So we can write:
K_sp = [Ni^p+]¹[OH^q-]² = (S)¹(2S+0.10)² = (S)¹(0.10)²
⇒ 2.0 × 10^-15 = 0.01 × S¹
⇒ S = 2.0 × 10^-13

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Effect of pH on solubility

This can be explained with the help of an example. It can be written in 11 steps:
1. Consider the sparingly soluble salt CaCO₃
• It’s dissociation can be written as:
CaCO₃ (s) ⇌ Ca²⁺ (aq) + CO₃^2- (aq)
2. CO₃^2- attracts one proton from the water molecule to become HCO₃^-
    ♦ When the water molecule loses one proton, it becomes OH^-
3. So there are two reactions taking place:
    ♦ CaCO₃ (s) ⇌ Ca²⁺ + CO₃^2-
    ♦ CO₃^2- + H₂O ⇌ HCO₃^- + OH^-
4. The net reaction can be obtained by adding the two. We get:
CaCO₃ (s) + H₂O ⇌ Ca²⁺ + HCO₃^- + OH^-
5. The presence of OH^- ions, make the solution basic
    ♦ So it’s pH will be greater than 7
6. If we add more OH^- ions, the system will try to reduce the effect by favoring the backward reaction
• So more and more solid CaCO₃ will be produced
• In effect, the CaCO₃ will become more and more insoluble
7. Remember that:
When we add more OH^- ions, we are increasing the pH
◼ So we can write:
When we increase the pH, CaCO₃ which is originally sparingly soluble becomes even more insoluble
8. If in step (6), instead of OH^- ions, we add H₃O⁺ ions, those extra H₃O⁺ ions will react with OH^- ions to form water
• The reaction is: H₃O⁺ + OH^- ⇌ 2H₂O
9. Thus the OH^- ions are removed
• The system will try to reduce the effect of this OH^- depletion by favoring the forward reaction
• So more and more solid CaCO₃ dissolves
10. Remember that:
When we add more H₃O⁺ ions, we are decreasing the pH
◼ So we can write:
When we decrease the pH, CaCO₃ which is originally sparingly soluble, becomes more soluble
11. This is the reason for the damages suffered by limestone (CaCO₃) statues due to acid rains

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• In the next chapter, we will see redox reactions


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Chapter 7.22 - Solved Examples Related to Buffer Solutions and pH

In the previous section, we saw two solved examples which showed us:
   ♦ How we can calculate the pH of buffers
   ♦ How buffer helps to avoid large changes in pH
• In this section, we will see a few more solved examples related to this category

Solved example 7.81
The ionization constant of phenol is 1 × 10^-10. What is the concentration of phenolate ion in 0.05 M solution of phenol? What will be the degree of ionization if the solution is also 0.01 M in sodium phenolate?
Solution:
Part (a):
1. Consider the original solution which is: 0.05 M C₆H₆O (phenol)
• It’s dissociation can be written as:
C₆H₆O ⇌ C₆H₅O^- + H⁺
2. Let at equilibrium, x moles of C₆H₅O^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [C₆H₆O] = 0.050-x
         ✰ [C₆H₅O^-] = x
         ✰ [H⁺] = x
3. The equilibrium constant for this reaction is given as: K_a = 1 × 10^-10
• So we can write: $\mathbf\small{\rm{K_a=1 \times 10^{-10}=\frac{[C_6H_5O^-][H^+]}{[C_6H_6O]}=\frac{(x)(x)}{(0.05-x)}}}$
4. x will be very small when compared to 0.050
• This is because, C₆H₆O is a weak acid. It will not give much C₆H₅O^- ions
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ For example, (0.05 - 0.0000001) can be taken as 0.05 for practical purposes
5. Thus the result in (3) becomes:
$\mathbf\small{\rm{1 \times 10^{-10}=\frac{x^2}{(0.050)}}}$
⇒ x = 2.236  × 10^-6
(The reader can opt not to approximate (0.050-x) as 0.050. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 2.236  × 10^-6)
6. Let us calculate 𝛼 (the degree of ionization) also
• We have: [H⁺] = [C₆H₅O^-] = x = 2.236  × 10^-6
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 2.236 × 10^-6 = 0.05 × 𝛼
⇒ 𝛼 = 4.472 × 10^-5

Part (b):
1. To the equilibrium that we saw in part (a), we are adding 0.01 M C₆H₅ONa (sodium phenolate)
• This C₆H₅ONa will undergo complete dissociation according to the equation:
C₆H₅ONa ⇌ C₆H₅O^- + Na⁺
• Since there is complete dissociation, 0.01 moles of C₆H₅O^- will be produced
2. So the various species in the resulting solution are:
C₆H₆O, C₆H₅O^-, H⁺ and Na⁺
• Na⁺ is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
C₆H₆O ⇌ C₆H₅O^- + H⁺
3. Let at equilibrium, x moles of C₆H₅O^- be produced from C₆H₆O
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [C₆H₆O] = 0.05-x
         ✰ [C₆H₅O^-] = 0.01+x
         ✰ [H⁺] = x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_a = 1 × 10^-10
• So we can write: $\mathbf\small{\rm{K_a=1 \times 10^{-10}=\frac{[C_6H_5O^-][H^+]}{[C_6H_6O]}=\frac{(0.01+x)x}{(0.05-x)}}}$
5. As before, x will be very small when compared to 0.05
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ Also, (0.01 + x) can be taken as 0.01
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{1 \times 10^{-10}=\frac{(0.01)x}{(0.05)}}}$
⇒ x = 5 × 10^-10
(The reader can opt not to approximate (0.050-x) and (0.01+x) as 0.05 and 0.01 respectively. 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 5  × 10^-10)
7. Thus we get: x = 5  × 10^-10
• We know that, x = c 𝛼
• So we get: 5 × 10^-10 = 0.05 × 𝛼
⇒ 𝛼 = 10^-8
◼ Let us compare the two results:
    ♦ In part (a), we have : 𝛼 = 4.472 × 10^-5
    ♦ In part (b), we have : 𝛼 = 10^-8
• It is clear that, when sodium phenolate is added, dissociation of phenol is suppressed
• The mixture of phenol and sodium phenolate can act as a buffer
• Note that, this example demonstrates common ion effect also

Solved example 7.82
Calculate the degree of ionization of 0.05 M acetic acid if it's pK_a value is 4.74. How is the degree of dissociation affected when it's solution also contains (a) 0.01 M (b) 0.1 M in HCl
Solution:
Part (a):
1. Consider the original solution which is: 0.05 M CH₃COOH (acetic acid)
• It’s dissociation can be written as:
CH₃COOH ⇌ CH₃COO^- + H⁺
2. Let at equilibrium, x moles of CH₃COO^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [CH₃COOH] = 0.05-x
         ✰ [CH₃COO^-] = x
         ✰ [H⁺] = x
3. The pK_a of this reaction is given as 4.74
• We know that, pK_a = -log₁₀(K_a)
   ♦ So K_a = antilog (-pK_a) = antilog (-4.74) = 1.82 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.82 \times 10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{(x)(x)}{(0.05-x)}}}$
4. x will be very small when compared to 0.05
• This is because, CH₃COOH is a weak acid. It will not give much CH₃COO^- ions
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ For example, (0.05 - 0.0000001) can be taken as 0.05 for practical purposes
5. Thus the result in (3) becomes:
$\mathbf\small{\rm{1.82 \times 10^{-5}=\frac{x^2}{(0.05)}}}$
⇒ x = 9.53  × 10^-4
(The reader can opt not to approximate (0.05-x) as 0.05. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 9.53 × 10^-4)
6. Let us calculate 𝛼 (the degree of ionization) also
• We have: [H⁺] = [CH₃COO^-] = x = 9.53  × 10^-4
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 9.53 × 10^-4 = 0.05 × 𝛼
⇒ 𝛼 = 0.01907

Part (b):
1. To the equilibrium that we saw in part (a), we are adding 0.01 M HCl (Hydrochloric acid)
• This HCl will undergo complete dissociation according to the equation:
HCl ⇌ H⁺ + Cl^-
• Since there is complete dissociation, 0.01 moles of H⁺ will be produced
2. So the various species in the resulting solution are:
CH₃COOH, CH₃COO^-, H⁺ and Cl^-
• Cl^- is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
CH₃COOH ⇌ CH₃COO^- + H⁺
3. Let at equilibrium, x moles of CH₃COO^- be produced from CH₃COOH
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [CH₃COOH] = 0.05-x
         ✰ [CH₃COO^-] = x
         ✰ [H⁺] = 0.01+ x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_a = 1.82 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.82 \times 10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{(0.01+x)x}{(0.05-x)}}}$
5. As before, x will be very small when compared to 0.05
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ Also, (0.01 + x) can be taken as 0.01
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{1.82 \times 10^{-5}=\frac{(0.01)x}{(0.05)}}}$
⇒ x = 9.0985 × 10^-5
(The reader can opt not to approximate (0.050-x) and (0.01+x) as 0.05 and 0.01 respectively. 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 9.0985 × 10^-5)
7. Thus we get: x = 9.0985  × 10^-5
• We know that, x = c 𝛼
• So we get: 9.0985 × 10^-5 = 0.05 × 𝛼
⇒ 𝛼 = 0.0018
◼ Let us compare the two results:
    ♦ In part (a), we have : 𝛼 = 0.01907
    ♦ In part (b), we have : 𝛼 = 0.0018
• It is clear that, when 0.01 M HCl is added, dissociation of acetic acid is suppressed

Part (c):
1. To the equilibrium in part (a), we are adding 0.1 M HCl (Hydrochloric acid)
• This HCl will undergo complete dissociation according to the equation:
HCl ⇌ H⁺ + Cl^-
• Since there is complete dissociation, 0.1 moles of H⁺ will be produced
2. So the various species in the resulting solution are:
CH₃COOH, CH₃COO^-, H⁺ and Cl^-
• Cl^- is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
CH₃COOH ⇌ CH₃COO^- + H⁺
3. Let at equilibrium, x moles of CH₃COO^- be produced from CH₃COOH
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [CH₃COOH] = 0.05-x
         ✰ [CH₃COO^-] = x
         ✰ [H⁺] = 0.1+ x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_a = 1.82 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.82 \times 10^{-5}=\frac{[CH_3COO^-][H^+]}{[CH_3COOH]}=\frac{(0.01+x)x}{(0.05-x)}}}$
5. As before, x will be very small when compared to 0.05
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ Also, (0.1 + x) can be taken as 0.1
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{1.82 \times 10^{-5}=\frac{(0.1)x}{(0.05)}}}$
⇒ x = 9.0985 × 10^-6
(The reader can opt not to approximate (0.05-x) and (0.1+x) as 0.05 and 0.1 respectively. 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 9.0985 × 10^-6)
7. Thus we get: x = 9.0985  × 10^-6
• We know that, x = c 𝛼
• So we get: 9.0985 × 10^-6 = 0.05 × 𝛼
⇒ 𝛼 = 0.00018
◼ Let us compare the three results:
    ♦ In part (a), we have : 𝛼 = 0.01907
    ♦ In part (b), we have : 𝛼 = 0.0018
    ♦ In part (c), we have : 𝛼 = 0.00018
• It is clear that:
   ♦ When 0.01 M HCl is added, dissociation of acetic acid is suppressed
   ♦ When 0.1 M HCl is added, dissociation of acetic acid is even more suppressed
• Note that, this example demonstrates common ion effect also

Solved example 7.83
The ionization constant of dimethylamine is 5.4 × 10^-4. Calculate it's degree of ionization in it's 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1 M NaOH
Solution:
Part (a):
1. Consider the original solution which is: 0.02 M NH₄OH (dimethylamine)
• It’s dissociation can be written as:
(CH₃)₂NH + H₂O ⇌ (CH₃)₂NH₂⁺ + OH^-
2. Let at equilibrium, x moles of (CH₃)₂NH₂⁺ be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [(CH₃)₂NH] = 0.02-x
         ✰ [(CH₃)₂NH₂⁺] = x
         ✰ [OH^-] = x
3. The equilibrium constant of this reaction is given as: K_b = 5.4 × 10^-4
• So we can write: $\mathbf\small{\rm{K_b=1 \times 5.4^{-4}=\frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}=\frac{(x)(x)}{(0.02-x)}}}$
4. x will be very small when compared to 0.02
• This is because, (CH₃)₂NH is a weak base. It will not give much (CH₃)₂NH₂⁺ ions
   ♦ So (0.02 - x) can be taken as 0.02
   ♦ For example, (0.02 - 0.0000001) can be taken as 0.02 for practical purposes
5. Thus the result in (3) becomes:
$\mathbf\small{\rm{5.4 \times 10^{-4}=\frac{x^2}{(0.02)}}}$
⇒ x = 3.286 × 10^-3
(The reader can opt not to approximate (0.02-x) as 0.02. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 3.286  × 10^-3)
6. Let us calculate 𝛼 (the degree of ionization) also
• We have: [OH^-] = [(CH₃)₂NH₂⁺] = x = 3.286 × 10^-3
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 3.286 × 10^-3 = 0.02 × 𝛼
⇒ 𝛼 = 0.1643

Part (b):
1. To the equilibrium that we saw in part (a), we are adding 0.1 M NaOH (sodium hydroxide)
• This NaOH will undergo complete dissociation according to the equation:
NaOH ⇌ Na⁺ + OH^-
• Since there is complete dissociation, 0.1 moles of OH^- will be produced
2. So the various species in the resulting solution are:
(CH₃)₂NH, (CH₃)₂NH₂⁺, OH^- and Na⁺
• Na⁺ is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
(CH₃)₂NH + H₂O ⇌ (CH₃)₂NH₂⁺ + OH^-
3. Let at equilibrium, x moles of (CH₃)₂NH₂⁺  be produced from (CH3)(CH₃)₂NH
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [(CH₃)₂NH] = 0.02-x
         ✰ [(CH₃)₂NH₂⁺] = x
         ✰ [OH^-] = 0.1+x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_b = 5.4 × 10^-4
• So we can write: $\mathbf\small{\rm{K_b=5.4 \times 10^{-4}=\frac{[(CH_3)_2NH_2^+][OH^-]}{[(CH_3)_2NH]}=\frac{(0.1+x)x}{(0.02-x)}}}$
5. As before, x will be very small when compared to 0.02
   ♦ So (0.02 - x) can be taken as 0.02
   ♦ Also, (0.1 + x) can be taken as 0.1
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{5.4 \times 10^{-4}=\frac{(0.1)x}{(0.02)}}}$
⇒ x = 1.08 × 10^-4
(The reader can opt not to approximate (0.02-x) and (0.1+x) as 0.02 and 0.1 respectively. 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 1.08 × 10^-4)
7. Thus we get: x = 1.08 × 10^-4
• We know that, x = c 𝛼
• So we get: 1.08 × 10^-4 = 0.02 × 𝛼
⇒ 𝛼 = 0.0054
◼ Let us compare the two results:
    ♦ In part (a), we have : 𝛼 = 0.1643
    ♦ In part (b), we have : 𝛼 = 0.0054
• It is clear that, when sodium hydroxide is added, dissociation of dimethylamine is suppressed
• Note that, this example demonstrates common ion effect also

Solved example 7.84
The ionization constant of propanoic acid is 1.32 × 10^-5. Calculate the degree of
ionization of the acid in its 0.05M solution and also its pH. What will be its
degree of ionization if the solution is 0.01M in HCl also?
Solution:
Part (a):
1. Consider the original solution which is: 0.05 M CH₃CH₂COOH (acetic acid)
• It’s dissociation can be written as:
CH₃CH₂COOH ⇌ CH₃CH₂COO^- + H⁺
2. Let at equilibrium, x moles of CH₃CH₂COO^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [CH₃CH₂COOH] = 0.05-x
         ✰ [CH₃CH₂COO^-] = x
         ✰ [H⁺] = x
3. The K_a of this reaction is given as 1.32 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.32 \times 10^{-5}=\frac{[CH_3CH_2COO^-][H^+]}{[CH_3CH_2COOH]}=\frac{(x)(x)}{(0.05-x)}}}$
4. x will be very small when compared to 0.05
• This is because, CH₃CH₂COOH is a weak acid. It will not give much CH₃CH₂COO^- ions
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ For example, (0.05 - 0.0000001) can be taken as 0.05 for practical purposes
5. Thus the result in (3) becomes:
$\mathbf\small{\rm{1.32 \times 10^{-5}=\frac{x^2}{(0.05)}}}$
⇒ x = 0.000812 = 8.12  × 10^-4
(The reader can opt not to approximate (0.05-x) as 0.05. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 8.12 × 10^-4)
6. Let us calculate 𝛼 (the degree of ionization) also
• We have: [H⁺] = [CH₃CH₂COO^-] = x = 8.12  × 10^-4
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 8.12 × 10^-4 = 0.05 × 𝛼
⇒ 𝛼 = 0.0162
7. pH = -log₁₀([H⁺]) = -log₁₀([8.12 × 10^-4]) = 3.09

Part (b):
1. To the equilibrium that we saw in part (a), we are adding 0.01 M HCl (Hydrochloric acid)
• This HCl will undergo complete dissociation according to the equation:
HCl ⇌ H⁺ + Cl^-
• Since there is complete dissociation, 0.01 moles of H⁺ will be produced
2. So the various species in the resulting solution are:
CH₃CH₂COOH, CH₃CH₂COO^-, H⁺ and Cl^-
• Cl^- is stable. It will not take part in reaction
• So the new equilibrium will be according to the equation:
CH₃CH₂COOH ⇌ CH₃CH₂COO^- + H⁺
3. Let at equilibrium, x moles of CH₃CH₂COO^- be produced from CH₃CH₂COOH
• Then we can write:
   ♦ At the new equilibrium,
         ✰ [CH₃CH₂COOH] = 0.05-x
         ✰ [CH₃CH₂COO^-] = x
         ✰ [H⁺] = 0.01+ x
4. Even when a new equilibrium is attained, the equilibrium constant will not change. It will be the same K_a = 1.32 × 10^-5
• So we can write: $\mathbf\small{\rm{K_a=1.32 \times 10^{-5}=\frac{[CH_3CH_2COO^-][H^+]}{[CH_3CH_2OOH]}=\frac{(0.01+x)x}{(0.05-x)}}}$
5. As before, x will be very small when compared to 0.05
   ♦ So (0.05 - x) can be taken as 0.05
   ♦ Also, (0.01 + x) can be taken as 0.01
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{1.32 \times 10^{-5}=\frac{(0.01)x}{(0.05)}}}$
⇒ x = 6.6 × 10^-5
(The reader can opt not to approximate (0.05-x) and (0.01+x) as 0.05 and 0.01 respectively. 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 6.6 × 10^-5)
7. Thus we get: x = 6.6  × 10^-5
• We know that, x = c 𝛼
• So we get: 6.6 × 10^-5 = 0.05 × 𝛼
⇒ 𝛼 = 0.00132
◼ Let us compare the two results:
    ♦ In part (a), we have : 𝛼 = 0.0162
    ♦ In part (b), we have : 𝛼 = 0.00132
• It is clear that, when 0.01 M HCl is added, dissociation of propanoic acid is suppressed
• Note that, this example demonstrates common ion effect also

Solved example 7.85
The pH of 0.1M solution of cyanic acid (HCNO) is 2.34. Calculate the ionization
constant of the acid and its degree of ionization in the solution
Solution:
1. Consider the solution which is: 0.01 M HCNO (cyanic acid)
• It’s dissociation can be written as:
HCNO ⇌ OCN^- + H⁺
2. Let at equilibrium, x moles of CH₃CH₂COO^- be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [HCNO] = 0.1-x
         ✰ [OCN^-] = x
         ✰ [H⁺] = x
3. To find x, we can make use of the given pH value
We have: x = [H⁺] = antilog(-pH) = antilog(-2.34) = 0.00457
4. So we can write: $\mathbf\small{\rm{K_a=\frac{[OCN^-][H^+]}{[HCNO]}=\frac{(x)(x)}{(0.01-x)}}}$
5. x will be very small when compared to 0.01
• This is because, HCNO is a weak acid. It will not give much OCN^- ions
   ♦ So (0.1 - x) can be taken as 0.1
   ♦ For example, (0.1 - 0.0000001) can be taken as 0.1 for practical purposes
6. Thus the result in (4) becomes:
$\mathbf\small{\rm{K_a=\frac{(0.00457)^2}{(0.1)}=0.000209}}$ = 2.09 × 10^-4
(The reader can opt not to approximate (0.1-x) as 0.1. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 2.09 × 10^-4)
7. Let us calculate 𝛼 (the degree of ionization) also
• We have: [H⁺] = [OCN^-] = x = 0.00457
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 0.00457 = 0.1 × 𝛼
⇒ 𝛼 = 0.0457

---

In the next section, we will see solubility equilibria


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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

---

• 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

---

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

---

---

• 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

---

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

---

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

---

• 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 7.20 - Common Ion Effect

In the previous section, we saw the factors affecting acid strength. In this section, we will see common ion effect. Later in this section, we will see hydrolysis of salts also

• Common ion effect can be explained using an example. It can be written in 7 steps:
1. Consider the dissociation of acetic acid:
CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO^-(aq)
• CH₃COOH (acetic acid) dissociates into H⁺ ions and CH₃COO^- (acetate ions)
• The equilibrium constant of this reaction is given by: $\mathbf\small{\rm{K_a=\frac{[H^+][CH_3COO^-]}{[CH_3COOH]}}}$
2. Let us take 0.05 M acetic acid solution
• From the data book, we can obtain the K_a value of CH₃COOH
• Using that K_a value and the initial concentration (0.05 M), we can calculate the pH of the solution
• We have already seen the calculation in solved example 7.65 in section 7.17
3. Now assume that, a 0.05 M CH₃COONa (sodium acetate) solution is added to the original 0.05 M solution mentioned in (2) above
◼  But what is 0.05 M CH₃COONa solution?
• Answer can be written in 3 steps:
(i) CH₃COONa  is a salt
(ii) Sufficient quantity of that salt is dissolved in water so that, 0.05 moles of CH₃COONa is present in 1 liter of the resulting solution
(iii) When dissolved in water, CH₃COONa dissociates completely into CH₃COO^- and Na+ (We will see the reason for this complete dissociation later)
4. Since there is complete dissociation, the 0.05 M solution of CH₃COOH will contain 0.05 moles of CH₃COO^- ions
• So when we add the 0.05 M CH₃COONa solution, we are adding an extra 0.05 moles of acetate ions
5. Consider the equilibrium in (1)
• Due to the addition of extra acetate ions, the equilibrium will shift towards the left
(This is in accordance with the Le Chatelier's principle)
   ♦ That means, rate of forward reaction becomes low
   ♦ That means, dissociation of CH₃COOH becomes low
6. Here we supplied CH₃COO^-, which was already present in the solution
   ♦ So CH₃COO^- is a common ion
   ♦ The supply of this common ion, caused a shift in equilibrium
• In the same way, if we supply extra H⁺ ions, then also the equilibrium will shift towards the left
◼  This shift in equilibrium is called common ion effect
7. We can define it in 5 steps:
(i) We add a suitable substance to a solution
(ii) This substance provides an ionic species which is already present in the solution
(iii) This ionic species which is already present is called the common ion
(iv) Due to the addition of extra common ions, the equilibrium shifts
(v) This shift in equilibrium is called common ion effect

Solved example 7.73
In the above discussion, the original solution was 0.05 M CH₃COOH. To this solution, 0.05 M CH3COONa was added. Calculate the pH of the resulting solution
Solution:
1. The balanced equation for the dissociation of acetic acid is:
CH₃COOH(aq) ⇌ H⁺(aq) + CH₃COO^-(aq)
• Let at equilibrium, x moles each of H⁺ 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. To this equilibrium, we are adding 0.05 moles of CH₃COO^- ions
So the total concentration of CH₃COO^- ions will be (0.05 + x)
3. For this reaction, K_a can be obtained as: $\mathbf\small{\rm{K_a=\frac{[H^+][CH3COO^-]}{[CH3COOH]}}}$
• Substituting the values, we get:
1.8 × 10^-5 = $\mathbf\small{\rm{\frac{x(0.05+x)}{(0.05-x)}}}$
⇒ x² + (0.050018)x - 0.09 × 10^-5 = 0
4. Solving this quadratic equation, we get: x = 1.798 × 10^-5 or -0.05
• negative value is not acceptable. So we take x = 1.798 × 10^-5 = 1.8 × 10^-5
5. So we can write:
• The concentration of H⁺ at equilibrium = [H⁺] = x = 1.8 × 10^-5 M
6. Once we know [H⁺], we can calculate the pH
We have: pH = -log₁₀([H⁺]) = -log₁₀(1.8 × 10^-5) = 4.74
7. Earlier, we calculated the pH of original solution as 3.03 (solved example 7.65 of section 7.17)
• When acetate ions are added, the pH increases to 4.74
• Increase in pH means, a decrease in [H⁺]
• The decrease in [H⁺] is expected. This is because, the backward reaction is favored when extra acetate ions are added

◼ We will see more solved examples demonstrating 'common ion effect' in later sections

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Hydrolysis of salts and pH of their solutions

This can be explained in 19 steps:
1. We know that salts are formed when an acid reacts with a base
    ♦ For example: HCl + NaOH → NaCl + H₂O
• Here, HCl is a strong acid and NaOH is a strong base
2. Salts can be formed from weak acids and weak bases also. In fact, there are four combinations possible:
    ♦ Strong acid – Strong base
    ♦ Weak acid – Strong base
    ♦ Strong acid - Weak base
    ♦ Weak acid – Weak base
• We will first consider the salts from Strong acid – Strong base combination. The following steps from (3) to (5) explain the hydrolysis of such salts
3. Consider the salt formed from HCl (strong acid) and NaOH (strong base):
HCl + NaOH → NaCl + H₂O
• So NaCl is a salt formed from a strong acid and a strong base
4. Take some NaCl and add it to water
• The salt will dissociate completely. The equation is:
NaCl → Na⁺ (aq) + Cl^- (aq)
• This is a complete dissociation. That is., all molecules of NaCl will dissociate into separate ions
5. So in the solution, we will be having three items:
Na⁺, Cl^- and H₂O
• Both Na⁺ and Cl^- have octet. They will not react with H₂O
• Since there is no reaction with H₂O, the ions H⁺ and OH^- will not be produced
• So the solution will be neutral. It's pH will be 7

---

• Next we will consider the salts from Weak acid – Strong base combination. The following steps from (6) to (9) explain the hydrolysis of such salts
6. Consider the salt formed from CH₃COOH (weak acid) and NaOH (strong base):
CH₃COOH + NaOH ⇌ CH₃COONa + H₂O
• So CH₃COONa is a salt formed from a weak acid and a strong base
7. Take some CH₃COONa and add it to water
• The salt will dissociate completely. The equation is:
CH₃COONa (aq) → CH₃COO^- (aq) + Na⁺ (aq)
• This is a complete dissociation. That is., all molecules of CH₃COONa will dissociate into separate ions
8. So in the solution, we will be having three items:
CH₃COO^-, Na⁺ and H₂O
• Na⁺ will not react with H₂O because it has octet
• CH₃COO^- will react with H₂O:
CH₃COO^- (aq) ⇌ CH₃COOH (aq) + OH^- (aq)
9. We know that, CH₃COOH is a weak acid. So it will not dissociate so much as to give appreciable quantities of H⁺ ions
• That means, according to the equation in (8), we will be having an excess quantity of OH^- ions
• Those excess OH^- ions will make the solution basic
• Due to the basic character, pH of that solution will be greater than 7

---

• Next we consider the salts from strong acid – weak base combination. The following steps from (10) to (13) explain the hydrolysis of such salts
10. Consider the salt formed from HCl (strong acid) and NH₄OH (weak base):
HCl + NH₄OH ⇌ NH₄Cl + HCl
• So NH₄Cl is a salt formed from a strong acid and a weak base
11. Take some NH₄Cl and add it to water
• The salt will dissociate completely. The equation is:
NH₄Cl (aq) → NH₄⁺ (aq) + Cl^- (aq)
• This is a complete dissociation. That is., all molecules of NH₄Cl will dissociate into separate ions
12. So in the solution, we will be having three items:
NH₄⁺, Cl^- and H₂O
• Cl^- will not react with H₂O because it has octet
• NH₄⁺ will react with H₂O:
NH₄⁺ (aq) ⇌ NH₄OH (aq) + H⁺ (aq)
13. We know that, NH₄OH is a weak base. So it will not dissociate so much as to give appreciable quantities of OH^- ions
• That means, according to the equation in (12), we will be having an excess quantity of H⁺ ions
• Those excess H⁺ ions will make the solution acidic
• Due to the acidic character, pH of that solution will be less than 7

---

• Finally we consider the salts from Weak acid – Weak base combination. The following steps from (14) to (18) explain the hydrolysis of such salts
14. Consider the salt formed from CH₃COOH (weak acid) and NH₄OH (weak base):
CH₃COOH + NH4OH ⇌ CH₃COONH₄ + H₂O
• So CH₃COONH₄ (ammonium acetate) is a salt formed from a weak acid and a weak base
15. Take some CH₃COONH₄ and add it to water
• The salt will not dissociate completely. The equation is:
CH₃COONH₄ (aq) ⇌ CH₃COO^- (aq) + NH₄⁺ (aq)
• This is not a complete dissociation. That is., all molecules of CH₃COONH₄ will not dissociate into separate ions
16. So in the solution, we will be having four items:
CH₃COONH₄, CH₃COO^-, NH₄⁺ and H₂O
• CH₃COO^- will react with H₂O:
CH₃COO^- (aq) + H₂O (l) ⇌ CH₃COOH, (aq) + OH^- (aq)
• NH₄⁺ will also react with H₂O:
NH₄⁺ (aq) ⇌ NH₄OH (aq) + H⁺ (aq)
17. We can write:
    ♦ The CH₃COO^- produces some excess OH^- ions
    ♦ The NH₄⁺ produces some excess H⁺ ions
18. So there will be a competition between OH^- and H⁺
    ♦ If [OH^-] is greater, the solution will be basic
    ♦ If [H⁺] is greater, the solution will be acidic
19. In such a situation, the pH is calculated using the following equation:
$\mathbf\small{\rm{pH=7+\frac{pK_a - pK_b}{2}}}$
• Where,
    ♦ K_a is the ionization constant of the weak acid CH₃COOH
    ♦ K_b is the ionization constant of the weak base NH₄OH
• We see that:
    ♦ If the difference (pK_a - pKb) is positive, the pH will be greater than 7
    ♦ If the difference (pK_a - pKb) is negative, the pH will be less than 7
• Also note that:
The equation contains constants only. So the concentrations can vary. But the pH will be constant because, pK_a and pK_b are constants

Solved example 7.74
The pK_a of acetic acid and pK_b of ammonium hydroxide are 4.76 and 4.75 respectively. Calculate the pH of ammonium acetate solution
Solution:
• We have: $\mathbf\small{\rm{pH=7+\frac{pK_a - pK_b}{2}}}$
• Substituting the values, we get:
$\mathbf\small{\rm{pH=7+\frac{4.76 - 4.75}{2}}}$ = 7.005

Solved example 7.75
The ionization constant of nitrous acid is 4.5 × 10^-4. Calculate the pH of 0.04 M
sodium nitrite solution and also its degree of hydrolysis
Solution:
1. Consider the dissociation of HNO₂ (nitrous acid)
HNO₂ ⇌ NO₂^- + H⁺
• The ionization constant for this process is given as 4.5 × 10^-4
2. NaNO₂ (sodium nitrite) is a salt prepared from a weak acid HNO₂ and strong base NaOH:
HNO₂ + NaOH  → NaNO₂
• So when this salt is added to water, it completely dissociates into Na⁺ and NO₂^-:
NaNO₂  → Na⁺ + NO₂^-
3. The Na⁺ is stable so it will not react with water
• But NO₂^- will react with water:
NO₂^- + H₂O ⇌ HNO₂ + OH^-
   ♦ This reaction is called hydrolysis of nitrite ion
• The HNO₂ thus produced is a weak acid. It will not give much H⁺
   ♦ So the solution will have an excess of OH^-. This will make the solution basic
4. We are asked to find the degree of this hydrolysis of nitrite ion
• That is, we are asked to find the fraction of NO₂^- , that will become HNO₂
• For that, we want the ionization constant for the reaction in (3)
• That is., we want the ionization constant of the reaction:
NO₂^- + H₂O ⇌ HNO₂ + OH^-
5. This reaction is the net of two reactions:
(i) NO₂^- + H⁺ ⇌ HNO₂
(ii) H₂O ⇌ H⁺ + OH^-
6. Reaction (i) is the reverse of the reaction that we wrote in (1)
   ♦ So it's K will be the reciprocal: $\mathbf\small{\rm{\frac{1}{4.5 \times 10^{-4}}}}$
• For the reaction (ii), we know the value of K: 10^-14
7. So, for the reaction in (3), the value of K will be:
$\mathbf\small{\rm{\frac{1}{4.5 \times 10^{-4}}\times 10^{-14}}}$ = 2.22 × 10^-11
8. Consider the reaction in (3):
NO₂^- + H₂O ⇌ HNO₂ + OH^-
9. Let at equilibrium, x moles of HNO₂         be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [NO₂^-] = 0.04-x
         ✰ [HNO₂] = x
         ✰ [OH^-] = x
10. We obtained the equilibrium constant for this reaction as: K = 2.22 × 10^-11
• So we can write: $\mathbf\small{\rm{K=2.22 \times 10^{-11}=\frac{[HNO_2][OH^-]}{[NO_2^-]}=\frac{(x)(x)}{(0.04-x)}}}$
11. x will be very small when compared to 0.04
   ♦ So (0.04 - x) can be taken as 0.04
   ♦ For example, (0.04 - 0.0000001) can be taken as 0.04 for practical purposes
12. Thus the result in (10) becomes:
$\mathbf\small{\rm{2.22 \times 10^{-11}=\frac{x^2}{(0.04)}}}$
⇒ x = 9.428  × 10^-7
(The reader can opt not to approximate (0.04-x) as 0.04. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 9.428  × 10^-7)
13. Let us calculate 𝛼 (the degree of ionization) also
• We have: [OH^-] = [HNO₂] = x = 9.428  × 10^-7
• We know that, x = c 𝛼
    ♦ Where  c is the initial concentration of the solute
• So we get: 9.428  × 10^-7 = 0.04 × 𝛼
⇒ 𝛼 = 2.357 × 10^-5
14. pH = (14 - pOH) = (14 + log₁₀([OH^-]) = (14 + log₁₀([9.428  × 10^-7]) = 7.97

Solved example 7.76
A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the
ionization constant of pyridine
Solution:
1. C₆H₆NCl (pyridinium hydrochloride) is a salt prepared from a weak base C₆H₅N and strong acid HCl:
C₆H₅N + HCl  → C₆H₆NCl
• So when this salt is added to water, it completely dissociates into Cl^- and C₆H₆N⁺:
C₆H₆NCl  → C₆H₆N⁺ + Cl^-
2. The Cl^- is stable. So it will not react with water
• But C₆H₆N⁺ will react with water:
C₆H₆N⁺ + H₂O ⇌ C₆H₆NOH + H⁺
   ♦ This reaction is called hydrolysis of C₆H₆N⁺ ion
• The C₆H₆NOH (pyridine) thus produced is a weak base. It will not give much OH^-
   ♦ So the solution will have an excess of H⁺ This will make the solution acidic
3. Consider the above hydrolysis reaction
• We are given that, if the [C₆H₆N⁺] is 0.02, the solution will have a pH of 3.44
• This pH of 3.44 is produced by the H⁺ ions
• From the pH value, we will get [H⁺]
    ♦ We have: [H⁺] = antilog (-pH) = antilog (-3.44) = 3.63 × 10^-4
4. In the hydrolysis reaction in (2), let x moles each of C₆H₆NOH and H⁺ be produced at equilibrium
    ♦ That is., at equilibrium, [C₆H₆NOH] = [H⁺] = x
• Then [C₆H₆N⁺] at equilibrium will be (0.02-x)  
5. So the ionization constant K_a will be given by: $\mathbf\small{\rm{K_a=\frac{[H^+][C_6H_6NOH]}{[C_6H_6N^+]}=\frac{x^2}{(0.02-x)}}}$
6. x will be very small when compared to 0.02
   ♦ So (0.02 - x) can be taken as 0.02
   ♦ For example, (0.02 - 0.0000001) can be taken as 0.02 for practical purposes
7. Thus the result in (5) becomes:
$\mathbf\small{\rm{K_a=\frac{x^2}{0.02}}}$
• But from (3), we have: x = [H⁺] = 3.63 × 10^-4
• Substituting this value of x, we get: K_a = 6.59 × 10^-6
(The reader can opt not to approximate (0.02-x) as 0.02. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 6.59  × 10^-6)
8. The hydrolysis reaction in (2) is the net of two reactions:
(i) C₆H₆N⁺ + OH^- ⇌ C₆H₆NOH
(ii) H₂O ⇌ H⁺ + OH^-
9. The reaction (i) is the reverse of: C₆H₆NOH ⇌ C₆H₆N⁺ + OH^-
    ♦ This reaction is the ionization of pyridine
• Let the ionization constant of this reaction be K_b
    ♦ Then the ionization constant of reaction (i) will be $\mathbf\small{\rm{\frac{1}{K_b}}}$
10. Consider the product of the following two items:
    ♦ Ionization constant of reaction 8(i), which is: $\mathbf\small{\rm{\frac{1}{K_b}}}$
    ♦ Ionization constant of reaction 8(ii), which is: 10^-14
• This product will be the ionization constant of the reaction in (2)
    ♦ We have already calculated this constant in (7)
11. So we can write: $\mathbf\small{\rm{\frac{1}{K_b}\times 10^{-14}=6.59 \times 10^{-6}}}$
⇒ K_b = 1.517 × 10^-9

Solved example 7.77
Predict if the solutions of the following salts are neutral, acidic or basic:
NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF
Solution:
NaCl:
1. NaCl is formed from NaOH (strong base) and HCl (strong acid)
• When added to water, NaCl will dissociate completely as:
NaCl → Na⁺ + Cl^-
2. Both Na⁺ and Cl^- are stable ions. They will not react with water
• So the solution will be neutral

KBr:
1. KBr is formed from KOH (strong base) and HBr (strong acid)
• When added to water, KBr will dissociate completely as:
KBr → K⁺ + Br-/
2. Both K⁺ and Br-/ are stable ions. They will not react with water
• So the solution will be neutral

NaCN:
1. NaCN is formed from NaOH (strong base) and HCN (weak acid)
• When added to water, NaCN will not dissociate completely:
NaCN ⇌ Na⁺ + CN^-
2. Na⁺ is a stable ion. It will not react with water
• But CN^- will react with water:
CN^- + H₂O ⇌ HCN + OH^-
3. The HCN thus formed, being weak, will not give much H⁺ ions
    ♦ So there will be an excess of OH^- ions
• Thus the solution will be basic

NH₄NO₃:
1. NH₄NO₃ is formed from NH₄OH (weak base) and HNO2 (strong acid)
• When added to water, NH₄NO₃ will not dissociate completely:
NH₄NO₃ ⇌ NH₄⁺ + NO₃^-
2. NO₃^- is a stable ion. It will not react with water
• But NH₄⁺ will react with water:
NH₄⁺ + H₂O ⇌ NH₄OH + H⁺
3. The NH₄OH thus formed, being weak, will not give much OH^- ions
    ♦ So there will be an excess of H⁺ ions
• Thus the solution will be acidic

NaNO₂:
1. NaNO₂ is formed from NaOH (strong base) and HNO₂ (weak acid)
• When added to water, NaNO₂ will not dissociate completely:
NaNO₂ ⇌ Na⁺ + NO₂^-
2. Na⁺ is a stable ion. It will not react with water
• But NO₂^- will react with water:
NO₂^- + H₂O ⇌ HNO₂ + OH^-
3. The HNO₂ thus formed, being weak, will not give much H⁺ ions
    ♦ So there will be an excess of OH^- ions
• Thus the solution will be basic

KF:
1. KF is formed from KOH (strong base) and HF (weak acid)
• When added to water, KF will not dissociate completely:
KF ⇌ K⁺ + F^-
2. K⁺ is a stable ion. It will not react with water
• But F^- will react with water:
F^- + H₂O ⇌ HF + OH^-
3. The HF thus formed, being weak, will not give much H⁺ ions
    ♦ So there will be an excess of OH^- ions
• Thus the solution will be basic

Solved example 7.78
The ionization constant of chloroacetic acid is 1.35 × 10^-3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?
Solution:
1. Consider the dissociation of ClCH₂COOH (chloroacetic acid)
ClCH₂COOH ⇌ ClCH₂COO^- + H⁺
• The ionization constant for this process is given as 1.35 × 10^-3
2. ClCH₂COONa (sodium salt of the acid ClCH₂COOH) is a salt prepared from this weak acid ClCH₂COOH and strong base NaOH:
ClCH₂COOH + NaOH  → ClCH₂COONa + H₂O
• So when this salt is added to water, it completely dissociates into Na⁺ and ClCH₂COO^-:
ClCH₂COONa  → Na⁺ + ClCH₂COO^-
3. The Na⁺ is stable so it will not react with water
• But ClCH₂COO^- will react with water:
ClCH₂COO^- + H₂O ⇌ ClCH₂COOH + OH^-
   ♦ This reaction is called hydrolysis of chloroacetate ion
• The ClCH₂COOH thus produced is a weak acid. It will not give much H⁺
   ♦ So the solution will have an excess of OH^-. This will make the solution basic
4. We are asked to find the pH of the solution in (3)
• For that, we want the ionization constant for the reaction in (3)
• That is., we want the ionization constant of the reaction:
ClCH₂COO^- + H₂O ⇌ ClCH₂COOH + OH^-
5. This reaction is the net of two reactions:
(i) ClCH₂COO^- + H⁺ ⇌ ClCH₂COOH
(ii) H₂O ⇌ H⁺ + OH^-
6. Reaction (i) is the reverse of the reaction that we wrote in (1)
   ♦ So it's K will be the reciprocal: $\mathbf\small{\rm{\frac{1}{1.35 \times 10^{-3}}}}$
• For the reaction (ii), we know the value of K: 10^-14
7. So, for the reaction in (3), the value of K will be:
$\mathbf\small{\rm{\frac{1}{1.35 \times 10^{-3}}\times 10^{-14}}}$ = 7.407 × 10^-12
8. Consider the reaction in (3):
ClCH₂COO^- + H₂O ⇌ ClCH₂COOH + OH^-
• Let at equilibrium, x moles of ClCH₂COOH be produced
• Then we can write:
   ♦ At equilibrium,
         ✰ [ClCH₂COO^-] = 0.1-x
         ✰ [ClCH₂COOH] = x
         ✰ [OH^-] = x
9. We obtained the equilibrium constant for this reaction as: K = 7.407 × 10^-12
• So we can write: $\mathbf\small{\rm{K=7.407 \times 10^{-12}=\frac{[ClCH_2COOH][OH^-]}{[ClCH_2COO^-]}=\frac{(x)(x)}{(0.1-x)}}}$
10. x will be very small when compared to 0.1
   ♦ So (0.1 - x) can be taken as 0.1
   ♦ For example, (0.1 - 0.0000001) can be taken as 0.1 for practical purposes
11. Thus the result in (10) becomes:
$\mathbf\small{\rm{7.407 \times 10^{-12}=\frac{x^2}{(0.1)}}}$
⇒ x = 8.607 × 10^-7
(The reader can opt not to approximate (0.1-x) as 0.1. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 8.607 × 10^-7)
12. pH = (14 - pOH) = (14 + log₁₀([OH^-]) = (14 + log₁₀([8.607 × 10^-7]) = 7.93
• This is the answer for part (b)
13. In part (a), we are asked to find the pH of 0.1 M solution of the original acid, which does not contain the salt solution
• For that, we consider the dissociation:
ClCH₂COOH ⇌ ClCH₂COO^- + H⁺
• Let at equilibrium, x moles each of ClCH₂COO^- and H+/ be produced
Then we get:
   ♦ At equilibrium,
         ✰ [ClCH₂COO^-] = 0.1-x
         ✰ [ClCH₂COOH] = x
         ✰ [H⁺] = x
14. We are given the equilibrium constant for this reaction as: K = 1.35 × 10^-3
• So we can write: $\mathbf\small{\rm{K=1.35 \times 10^{-3}=\frac{[ClCH_2COOH][OH^-]}{[ClCH_2COO^-]}=\frac{(x)(x)}{(0.1-x)}}}$
15. x will be very small when compared to 0.1
   ♦ So (0.1 - x) can be taken as 0.1
   ♦ For example, (0.1 - 0.0000001) can be taken as 0.1 for practical purposes
16. Thus the result in (10) becomes:
$\mathbf\small{\rm{1.35 \times 10^{-3}=\frac{x^2}{(0.1)}}}$
⇒ x = 0.0116
(The reader can opt not to approximate (0.1-x) as 0.1. Then it will become a more complex quadratic equation. That quadratic equation can be solved using a calculator or computer. But the result will be the same 0.0116)
17. So pH = log₁₀([H⁺]) = log₁₀(x) = log₁₀(0.0116) = 1.93 

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In the next section, we will see buffer solutions


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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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