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 7.20 - 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

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• 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

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• 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 7.10 - Effect of Pressure and Temperature on Equilibrium

In the previous section, we saw how change in concentration affects equilibrium. In this section, we will see three more factors which  affect equilibrium

Factor 2: Pressure
• This can be written in 8 steps:
1. Changes in pressure affects equilibrium of reactions in which at least one reactant or product is a gas
2. Suppose that, the system is inside a cylinder fitted with a piston
• Let P₁ and V₁ be the pressure and volume at equilibrium
• If we push down the piston to make the volume $\mathbf\small{\rm{\frac{V_1}{2}}}$, what will be the change in pressure?
3. We know that PV = a constant
• So P₁V₁ = P₂V₂
⇒ $\mathbf\small{\rm{p_1V_1=p_2 \times \frac{V_1}{2}}}$
⇒ P₂ = 2P₁
• That means, pressure becomes double the initial value
4. We know that:
• P₁ and P₂ are the total pressures
    ♦ Total pressure of the system is the sum of the partial pressures
    ♦ Partial pressure of a gas depends on it’s number of moles
• Now, total pressure has increased to a new value P₂
    ♦ So partial pressures will also increase
5. According to Le Chatelier’s principle, the system will try to reduce the effect of this increased pressure
• That means, the pressure must decrease from P₂
• For that, the reaction will proceed in that direction which will result in a lesser number of moles
6. An example:
• In the reaction CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), there are four gaseous moles on the reactant side, but only two gaseous moles in the product side
• So, when the pressure is increased, the reaction moves in the forward direction
7. Another example:
• In the reaction C(s) + CO₂(g) ⇌ 2CO(g), there is only one gaseous mole on the reactant side, but two gaseous moles in the product side
• So, when the pressure is increased, the reaction moves in the backward direction
8. Note that, with regard to 'change in pressure':
• We need not consider solids and liquids because, a change in pressure will not affect their volumes
• If volumes are not affected, concentrations are also not affected

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Effect of adding an inert gas

• This can be written in 2 steps:
1. Consider a system in equilibrium
• If an inert gas is added into the system, it will not take part in the reaction
• So the concentrations of reactants and products will not change
2. Thus we can write:
In a system at equilibrium, if we add an inert gas with out changing the pressure, the equilibrium will not be affected

---

Factor 3: Temperature
• Recall the two factors that we have already seen:
(i) When concentrations are changed, the K_c does not change. The concentrations re-adjust to bring back K_c to the original value
(ii) When pressure is changed, the K_c(or K_p) does not change. The concentrations re-adjust to bring back K_c(or K_p) to the original value

• But the third factor is different. When temperature changes, K_c(or K_p) attains a new value
• This can be explained using an example. It can be written in 5 steps:
1. Consider the reaction: H₂(g) + I₂(g) ⇌ 2HI(g); ΔH = -10.4 kJ mol^-1
• It is clear that, the forward reaction is exothermic. 10.4 kJ mol^-1 will be released
2. At 500 K, the Kp of this reaction is: 160
• If the temperature is increased to 700 K, the K_p becomes 54
◼ We see that, when temperature is increased, the equilibrium constant decreases
the following steps from (3) to (5) will give the reason:
3. According to Le Chatelier’s principle, the reaction will move in that direction which reduces the effect
• In our present case, temperature is increased. So the system tries to reduce the temperature
4. For that, the reaction will move in that direction which absorbs heat
    ♦ The forward reaction in (1) is exothermic which releases heat
    ♦ The backward reaction is endothermic which absorbs heat
• So the backward reaction is preferred
• So more reactants will be produced and more products will be consumed
5. That means, numerator (in the expression $\mathbf\small{\rm{K_p=\frac{(p_C)^c (p_D)^d}{(p_A)^a (p_B)^b}}}$) decreases
• Thus the equilibrium constant decreases

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Factor 4: Catalyst
This can be written in 4 steps:
1. A catalyst is added to speed up the reaction
2. But it speeds up both forward and backward reactions
• Thus the equilibrium is attained in a smaller time
3. Since the rates of both forward and backward reactions are increased, the concentrations of reactants and products remain the same
4. So the value of equilibrium constant does not change

---

Now we will see some solved examples:
Solved example 7.36
Does the number of moles of reaction products increase, decrease or remain
same when each of the following equilibria is subjected to a decrease in pressure
by increasing the volume?
(a) PCl₅ (g) ⇌ PCl₃(g) + Cl₂(g)
(b) CaO(s) + CO₂(g) ⇌ CaCO₃(s)
(c) 3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)
Solution:
Part (a):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure decreases, the system will try to increase the number of gaseous moles
    ♦ So the reaction will proceed in the forward direction
    ♦ Thus the number of moles of reaction products will increase
Part (b):
• Total number of gaseous moles in the reactant side = 2
• Total number of gaseous moles in the product side = 1
◼ When pressure decreases, the system will try to increase the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction
    ♦ Thus the number of moles of reaction products will decrease
Part (c):
• Total number of gaseous moles in the reactant side = 4
• Total number of gaseous moles in the product side = 4
◼ When pressure decreases, the system will try to increase the number of gaseous moles. But in this case, the number of gaseous moles is same for both reactants and products
    ♦ So the decrease in pressure has no effect in the direction
    ♦ Thus the number of moles of reaction products will remain the same

Solved example 7.37
Which of the following reactions will get affected by increasing the pressure?
Also, mention whether change will cause the reaction to go into forward or
backward direction.
(i) COCl₂(g) ⇌ CO(g) + Cl₂(g)
(ii) CH₄(g) + 2S₂(g) ⇌ CS₂(g) + 2H₂S(g)
(iii) CO₂(g) + C(s) ⇌ 2CO(g)
(iv) 2H₂(g) + CO(g) ⇌ CH₃OH(g)
(v) CaCO₃(s) ⇌ CaO(s) + CO₂(g)
(vi) 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g)
Solution:
Part (i):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction
Part (ii):
• Total number of gaseous moles in the reactant side = 3
• Total number of gaseous moles in the product side = 3
◼ When pressure increases, the system will try to decrease the number of gaseous moles. But in this case, the number of gaseous moles is same for both reactants and products
    ♦ So the increase in pressure has no effect in the direction
Part (iii):
• Total number of gaseous moles in the reactant side = 1
• Total number of gaseous moles in the product side = 2
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction
Part (iv):
• Total number of gaseous moles in the reactant side = 3
• Total number of gaseous moles in the product side = 1
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the forward direction
Part (v):
• Total number of gaseous moles in the reactant side = 0
• Total number of gaseous moles in the product side = 1
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction
Part (vi):
• Total number of gaseous moles in the reactant side = 9
• Total number of gaseous moles in the product side = 10
◼ When pressure increases, the system will try to decrease the number of gaseous moles
    ♦ So the reaction will proceed in the backward direction

Solved example 7.38
The equilibrium constant for the following reaction is 1.6 ×10⁵ at 1024K
H₂ (g) + Br₂ (g) ⇌ 2HBr(g)
Find the equilibrium pressure of all gases if 10.0 bar of HBr is introduced into a
sealed container at 1024K.
Solution:
1. At the beginning of the reaction, only HBr is present
• So we have to write the given equation in reverse order:
2HBr(g)  ⇌ H₂ (g) + Br₂ (g)
• The equilibrium constant of this reaction will be: $\mathbf\small{\rm{\frac{1}{1.6 \times 10^5}}}$
2. At equilibrium, let x be the pressure of H₂ and Br₂
• Then the pressure of HBr will be (10.0 - 2x)
3. Thus the equilibrium constant will be given by: $\mathbf\small{\rm{\frac{x^2}{(10.0-2x)^2}}}$
• So we can write: $\mathbf\small{\rm{\frac{x^2}{(10.0-2x)^2}=\frac{1}{1.6 \times 10^5}}}$
4. Taking reciprocals, we get: $\mathbf\small{\rm{\frac{(10.0-2x)^2}{x^2}=\frac{1.6 \times 10^5}{1}}}$
⇒ $\mathbf\small{\rm{\frac{(10.0-2x)}{x}=\sqrt{\frac{16 \times 10^4}{1}}=400}}$
• Thus we get x = 0.0248
5. So we can write:
• At equilibrium,
   ♦ p_H2 = p_Br2 = x = 0.0248 bar
   ♦ p_HBr = (10 - 2x) = (10 - 0.0496) = 9.95 bar

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Link to a few more solved examples are given below:

Solved examples 7.39 to 7.45

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In the next section, we will see the ionic equilibrium in solution


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Chapter 7.9 - Factors Affecting Equilibrium

In the previous section, we saw how equilibrium constant is related to Gibbs energy. In this section, we will see the factors affecting equilibrium

• We know that, equilibrium is attained when:
The concentrations of reactants and products remain constant
◼  Such a constant concentration is achieved when two conditions are satisfied:
Condition 1:
   ♦ The 'number of molecules of reactants' used up during a time interval t in the forward reaction
   ♦ is equal to
   ♦ The ,number of molecules of reactants' produced during the same time interval t in the backward reaction
Condition 2:
   ♦ The 'number of molecules of products' produced during a time interval t in the forward reaction
   ♦ is equal to
   ♦ The 'number of molecules of products' used up during the same time interval t in the backward reaction

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• The ‘number of molecules in a time interval t’ is simply, the rate of a reaction
• So we can combine the two conditions as:
Equilibrium is achieved when the rate of forward reaction becomes equal to the rate of the backward reaction

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• We have already seen the above facts in the previous sections
• In the present section, we are going to analyze the situations where the equilibrium gets affected
• What do we mean by saying ‘equilibrium gets affected’ ?
The answer can be written in just two steps:
(i) It simply means that, the existing equilibrium suffers a ‘change’
(ii) So a 'new equilibrium' will be reached

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• The manufacturer will try to obtain the ‘new equilibrium’ in such a way that the two conditions are satisfied:
   ♦ The concentrations of the products is maximum
   ♦ The concentrations of the reactants is minimum
• By adjusting the ‘factors which affect equilibrium’ the manufacturer will be able to satisfy the two conditions
• So it is very important that we clearly understand those factors

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• Consider any one factor. It may be concentration, temperature, pressure etc.,
• When we consider any one of those factors, a question immediately arises:
If we bring about a change in that factor, in which direction will the reaction proceed?
• For example:
    ♦ If the temperature is increased, will the reaction proceed in the forward direction, leading to ‘more products?
    ♦ Or, will the reaction proceed in the backward direction, leading to more reactants?

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Le Chatelier’s principle helps us to find answers to such questions
The principle can be stated in 3 steps:
1. Consider a system in equilibrium
• Note down the factors like: concentrations, temperature, pressure etc.,
2. The system is in equilibrium because, those  factors are remaining constant
• If a ‘change’ is brought about in any one of those factors, the equilibrium will change
3. The change will be in such a way that:
The effect of the ‘change’ gets reduced

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• Based on Le Chatelier's principle, we can now analyze various factors
Factor 1: Concentration
This can be written in 2 steps:
1. Concentrations of various reactants and products is an important factor that keeps the system in equilibrium
• This is clear from the expression that we use to find K_c: $\mathbf\small{\rm{K_c=\frac{[C]^c [D]^d}{[A]^a [B]^b}}}$
2. Now we want to know this:
How will the equilibrium be affected, if we change the concentrations of one or more reactants or products ?
• The answer is given by Le Chatelier’s principle. It can be written in two steps:
(i) When a reactant or product is added into the system, we are increasing the concentration
    ♦ In such a situation, the reaction will proceed in that direction which consumes the excess concentration
(ii) When a reactant or product is removed from the system, we are decreasing the concentration
    ♦ In such a situation, the reaction will proceed in that direction which replenishes the lost concentration

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Let us apply this information to a practical situation. It can be written in steps:
1. Consider the reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
• Let it be at equilibrium
2. Suppose that some H₂ is added to the system
• Then the equilibrium is disturbed
3. The system will try to reduce the effect of ‘excess H₂’
• For that, the system will try to use up the excess H₂
4. This can be achieved by moving in the forward direction
• That is., more H₂ will react with I₂ to produce more HI
• When this happens:
    ♦ The concentration of H₂ decreases from the ‘new total value’
    ♦ The concentration of I₂ decreases from the equilibrium value
    ♦ The concentration of HI increase from the equilibrium value
5. Consider the expression for K_c: $\mathbf\small{\rm{K_c=\frac{[HI]^2}{[H_2][I_2]}}}$
• When some H₂ is added in step (2) above, [H₂] increases
    ♦ So K_c decreases
6. In order to bring K_c back, [H₂] must be reduced
• For that, some H₂ combines with I₂ to produce more HI
• Due to this combination, [I₂] decreases and [HI] increases
7. But [H₂] does not decrease all the way back to the original value
• This fact is shown pictorially in fig.7.6 below. It can be explained in 6 steps

Fig.7.6

(i) In the graph
    ♦ Time is plotted along the x-axis
    ♦ Concentration is plotted along the y-axis
(ii) Up to time t₁ (first vertical dashed line), the system is in equilibrium
    ♦ The [H₂] up to t₁ can be denoted as [H₂]_AB
    ♦ This [H₂]_AB is constant. This is indicated by the horizontal nature of line AB
(iii) The excess H₂ is added at t₁
    ♦ Due to the addition of H₂, the [H₂] rises to point C
    ♦ The [H₂] in this situation can be denoted as [H₂]_C
(iv) The reaction between excess H₂ and already existing I₂ begins at t₁ itself
(v) So from t₁ on wards, the decrease in [H₂] begins
    ♦ That is., the concentration begins to decrease from [H₂]_C
        ✰ This is indicated by the yellow curve CD
    ♦ It decreases to a new value
    ♦ This new value is [H₂]_DE
(vi) We see that:
    ♦ [H₂]_DE is lesser than [H₂]_C
    ♦ [H₂]_DE is greater than [H₂]_AB
    ♦ That is., the [H₂] never returns to the original value [H₂]_AB
8. In a similar way, we can analyze the concentrations of I₂ and HI also
• [I₂]_PQ represents the initial concentration of I₂
    ♦ Due to the reaction with H₂, [I₂]_PQ decreases to a new value: [I₂]_RS
• [HI]_UV  represents the initial concentration of HI
    ♦ Due to the increased production, [HI]_UV increases to a new value: [HI]_WX
9. It is clear that:
    ♦ The final value [H₂]_DE is different from the initial value [H₂]_AB
    ♦ The final value [I₂]_RS is different from the initial value [I₂]_PQ
    ♦ The final value [HI]_WX is different from the initial value [HI]_UV
• The final values are all different from their respective initial values
• But yet, the differences occur in such a way as to restore the equilibrium constant back to the original K_c
• That is: $\mathbf\small{\rm{K_c=\frac{[HI]_{UV}^2}{[H_2]_{AB}[I_2]_{PQ}}=\frac{[HI]_{WX}^2}{[H_2]_{DE}[I_2]_{RS}}}}$

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• The above example helps us to obtain a clear idea about how the equilibrium is affected when the concentration of one of the reactants/products is increased
• We can cause a decrease also. That is., we can remove one or more of the reactants/products from the system
• For example, if we remove some HI from the above system at equilibrium, the system will try to replenish the lost concentration by producing more HI
    ♦ That is., the system will move in the forward direction
    ♦ So more H₂ and I₂ will be consumed
• This method is commonly adopted by manufacturers
    ♦ They can thus convert more and more reactants into products
• Let us see two such examples:
◼ In the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the product NH₃ is liquefied and removed
    ♦ This causes the system to convert more and more N₂ and H₂ into NH₃
    ♦ NH₃ (ammonia) is a commercially important substance
◼ In the reaction CaCO₃(s) ⇌ CaO(s) + CO₂(g), the product CO₂ is continuously removed
    ♦ This causes the system to convert more and more CaCO₃ into CaO and CO₂
    ♦ CaO (calcium oxide) is a commercially important substance

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Thus we have seen how 'change in concentration' affects equilibrium. In the next section, we will see the effect of 'change in pressure'


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Chapter 7.3 - Dynamic Equilibrium in Chemical Reactions

In the previous section, we saw the basic features of equilibrium. We also saw an activity which demonstrates the dynamic nature of equilibrium. In this section, we will see the equilibrium in chemical reactions in greater detail

• We have seen some basics about equilibrium in chemical reactions in the first section of this chapter. We saw that:
   ♦ In the forward reaction, the reactants react together to give the products
   ♦ In the backward reaction, the products react together to give the reactants
   ♦ Both forward and backward reactions occur simultaneously
• In the initial stages of the reaction:
   ♦ The rate of forward reaction
   ♦ is greater than
   ♦ The rate of backward reaction
• At equilibrium:
   ♦ The rate of forward reaction
   ♦ is equal to
   ♦ The rate of backward reaction

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We can learn some more details with the help of a graph. It can be written in 16 steps:
1. Consider the reaction: A + B ⇌ C + D
2. As this reaction proceeds, we record the following items at regular time intervals:
   ♦ Quantity of reactant A
   ♦ Quantity of reactant B
   ♦ Quantity of product C
   ♦ Quantity of product D
3. We are recording the quantities at regular time intervals
• So, if the time interval is 2 minutes, we take the quantities when the stop watch shows 0, 2, 4, 6 . . . , minutes
4. Now we plot a graph
   ♦ Time is plotted along the x-axis
   ♦ Quantity (or concentration) is plotted along the y-axis
5. The points are plotted according to the following 4 steps:
(i) We will be having a vertical line through the 0 min mark on the x-axis
(This vertical line is the y-axis)
• On that vertical line, we mark two points:
    ♦ Concentration of reactant A at 0 min
    ♦ Concentration of product C at 0 min
(Concentration of C at 0 min will be zero because, no products are formed at 0 min)
(ii) We will be having a vertical line through the 2 min mark on the x-axis
• On that vertical line, we mark two points:
    ♦ Concentration of reactant A at 2 min
    ♦ Concentration of product C at 2 min
(iii) We will be having a vertical line through the 4 min mark on the x-axis
• On that vertical line, we mark two points:
    ♦ Concentration of reactant A at 4 min
    ♦ Concentration of product C at 4 min
- - -
- - -
(iv) In this way, all the readings related to A and C are plotted
    ♦ Draw a smooth red curve through the points related to A
    ♦ Draw a smooth green curve through the points related to C
• The result graph is shown in fig.7.3 below:

Fig.7.3

6. We can plot a graph using the readings related to B and D also. The graph will be similar. There is no need to plot a separate graph for B and D
    ♦ So along the red curve, we write: A or B
    ♦ Also along the green curve, we write: C or D
• The following steps from (7) to (16) give us a detailed analysis of the curves
7. Consider the red curve
• In the initial stages, it is a falling curve. That means, in the initial stages, the concentration of reactant A goes on decreasing
• This is expected. We know that, more and more A is continuously being used up. As time passes by, there will be lesser and lesser A available
8. But in the later stages, the red curve becomes horizontal
• This is not expected. We expect the concentration of A to become lesser and lesser and finally become zero
• That is., we expect the red curve to fall continuously and finally meet the x-axis
• The horizontal nature of the curve indicates that, even if we wait for a very long time, there will be some quantity of A remaining
9. So a question arises:
Why is A not completely used up?
• The answer is that, the products C and D reacts together to give back small quantities of A and B. This is the backward reaction
10. Next, consider the green curve
• In the initial stages, it is a rising curve. That means, in the initial stages, the concentration of product C goes on increasing
• This is expected. We know that, more and more C is continuously being formed. As time passes by, there will be more and more C available
11. But in the later stages, the green curve becomes horizontal
• This is not expected. We want the concentration of C to become larger and larger and reach a maximum
• The horizontal nature of the curve indicates that, even if we wait for a very long time, there will be no further increase in the quantity of C
12. So a question arises:
Why is C not increasing further?
• The answer is that, some of the newly formed C is being used up in the backward reaction
13. Consider the magenta vertical dashed line
• It is drawn just at the beginning of the horizontal portions
• So we have an important point:
The point of intersection of the magenta vertical dashed line and the x-axis
• This point gives us the time at which equilibrium is attained
14. Once equilibrium is attained,
    ♦ The rate of forward reaction
    ♦ will be equal to
    ♦ The rate of backward reaction
(i) That means:
   ♦ The number of reactant molecules used up in unit time
   ♦ will be equal to
   ♦ The number of reactant molecules formed in unit time
(ii) That also means:
   ♦ The number of product molecules formed in unit time
   ♦ will be equal to
   ♦ The number of product molecules used up in unit time
◼  From (i) it is clear that, once equilibrium is attained, the concentration of reactants remain unchanged
◼  From (ii) it is clear that, once equilibrium is attained, the concentration of products remain unchanged
15. Note that, at equilibrium,
    ♦ the concentration of products
    ♦ is greater than
    ♦ the concentration of reactants
• We know this because:
   ♦ The horizontal portion of green curve
   ♦ is at a higher level than
   ♦ The horizontal portion of red curve
◼  So the manufacturer need not worry too much. Even if products are being converted back into reactants, at equilibrium, he has a higher concentration of products
• Also, by making suitable changes to temperature and pressure, he can change the equilibrium in such a way that, more products are obtained
   ♦ We will see those details in later sections
16. It is important to note that, the same equilibrium can be reached by starting the reaction with products C and D, with no presence of A and B
• This is because, as soon as C and D are brought together, they will start to react with each other

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Dynamic equilibrium in chemical reactions

• Consider the reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
• After some time, the reaction will attain equilibrium
• Even after attaining equilibrium, both forward and backward reactions will be still going on
• Haber proved this using isotopes. His experiment can be explained in 15 steps:
1. Take H₂ and N₂ in a container
• Maintain a high temperature and pressure inside the container. This is because, the forward reaction is favored under high temperature and pressure
2. Measure the quantities of N₂, H₂ and NH₃ at regular intervals
3. Plot the Concentration vs Time graph. It will be as shown in fig.7.4(a) below:

Fig.7.4

 ◼  We see that:
• The red curve is high above in the graph
   ♦ That means, we are taking a greater quantity of H₂
   ♦ This is because, we need 3 mol H₂ for every one mol N₂
◼   Also we see these:
• Initially the concentration of H₂ decreases
   ♦ This is because, H₂ is being used up in the forward reaction
• Initially the concentration of N₂ also decreases
   ♦ This is because, N₂ is also being used up in the forward reaction
• Initially the concentration of NH₃ increases
   ♦ This is because, NH₃ is being formed in the forward reaction
4. But after some time, all the three graphs become horizontal
• This indicates equilibrium
5. Next, we want N₂ to react with D₂ to form ND₃
   ♦ That is., we want the reaction: N₂(g) + 3D₂(g) ⇌ 2ND₃(g)
   ♦ [D₂ (Deuterium) is the isotope of H₂]
• For that, take D₂ and N₂ in a second container
• The quantities must be same as those taken in the first container
   ♦ Quantity of D₂ must be same as that of H₂ in the first container
   ♦ Quantity of N₂ must be same as that of N₂ in the first container
• Maintain a high temperature and pressure inside the container. This is because, the forward reaction is favored under high temperature and pressure
6. Measure the quantities of N₂, H₂ and NH₃ at regular intervals
7. Plot the Concentration vs Time graph. It will be as shown in fig.7.4(b) above
◼  We see that:
• The graph is fig.b is similar to the graph in fig.a
• That means, at equilibrium,
   ♦ Quantity of D₂ remaining will be same as that of H₂ in the first container
   ♦ Quantity of N₂ remaining will be same as that of N₂ in the first container
   ♦ Quantity of ND₃ formed will be same as that of NH₃ in the first container
8. The position of the vertical magenta dashed line is also same in both the graphs (a) and (b)
• That means, the 'time at which equilibrium is attained' is same for both cases
9. Now we have two mixtures
• In the first container, we have a mixture of N₂, H₂ and NH₃
    ♦ This mixture is at a temperature T and pressure P
    ♦ This mixture is at equilibrium
• In the second container, we have a mixture of N₂, D₂ and ND₃
    ♦ This mixture is also at the same temperature T and pressure P
    ♦ This mixture is also at equilibrium
10. The two mixtures are mixed together in a third container
    ♦ The temperature and pressure in this third container are the same T and P
◼ We would expect the following composition in the third container:
• Quantity of H₂:
    ♦ Same as that in container 1
• Quantity of N₂:
    ♦ Double of that in container 1
• Quantity of D₂:
    ♦ Same as that in container 2
• Quantity of NH₃:
    ♦ Same as that in container 1
• Quantity of ND₃:
    ♦ Same as that in container 2
11. If we analyze the composition of the mixture in the third container, by ordinary methods, we will see that, the results mentioned in (10) are true
12. But if we analyze using a mass spectrometer, we will see that:
• The mixture in the third container has the following three compounds also:
    ♦ NH₂D, NHD₂ and HD
13. We would not expect to find these three compounds because:
• In the first container, there was no D₂
    ♦ So in the first container, there will not be any NH₂D, NHD₂ or HD
• In the second container, there was no H₂
    ♦ So in the second container, there will not be any NH₂D, NHD₂ or HD
14. So clearly, the three compounds are formed after the inter mixing in the third container
• But in the third container, every thing is in equilibrium. There is no change in composition. Then how did the new compounds form?
• The answer can be written in 3 steps:
(i) Even at equilibrium, the forward and backward reactions are taking place simultaneously
(ii) So in the new forward reaction, N₂ reacts with both H₂ and D₂ to give
NH₂D and NHD₂
(iii) Also, in the new backward reaction, both NH₂D and NHD₂ get decomposed
15. If the equilibrium was static, there would have been no mixing of isotopes in this way
◼ So the formation of NH₂D, NHD₂ and HD is clear evidence for the dynamic nature of equilibrium

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• Let us analyze another reaction: H₂(g) + I₂(g) ⇌ 2HI(g)
• It is the formation of Hydrogen iodide from H₂ and I₂
• We can analyze this reaction with the help of the graph in fig.7.5(a) below:

Fig.7.5

• The analysis can be written in 9 steps:
1. In the initial stages, concentration of H₂ and I₂ (red curve) decreases
• This is because, they are being use up in the forward reaction
2. Also in the initial stages, concentration of HI (green curve) increases
• This is because, it is being formed in the forward reaction
3. We would expect the red curve to fall continuously and reach the x-axis
• But in reality, it stops falling and becomes horizontal
• This is because, H₂ and I₂ gets newly formed due to the backward reaction
4. We would expect the green curve to rise continuously and reach a maximum
• But in reality, it stops rising and becomes horizontal
• This is because, HI begins to get decomposed due to the backward reaction
5. As time passes by,
    ♦ the rate of forward reaction
    ♦ becomes equal to
    ♦ the rate of backward reaction
(i) That means:
    ♦ Number of molecules of H₂ and I₂ used up in forward reaction
    ♦ becomes equal to
    ♦ Number of molecules of H₂ and I₂ newly formed in backward reaction
(ii) That also means:
    ♦ Number of molecules of HI being decomposed in backward reaction
    ♦ becomes equal to
    ♦ Number of molecules of HI newly formed in forward reaction
◼  From (i) it is clear that, once equilibrium is attained, the concentration of reactants remain unchanged
◼  From (ii) it is clear that, once equilibrium is attained, the concentration of products remain unchanged
• This condition is indicated by the horizontal nature of the red and green curves
6. We know that, one mol H₂ reacts with one mol I₂ to give two mol HI
• So, if we take n mol H₂ and n mol I₂, we will get 2n mol HI
7. Suppose that, the curves in fig.7.5(a) is obtained when n mol each of H₂ and I₂ are mixed
• Theoretically, we must receive 2n mol HI
• But the reaction will not proceed up to the formation of 2n mol HI
• This is because, some of the newly formed HI will be decomposed back to H₂ and I₂
8. Suppose that, initially we take 2n mol HI
• Then this HI will decompose to give n mol H₂ and n mol I₂
• This is shown graphically in fig. 7.5(b) above
    ♦ For better comparison, time is plotted from right to left
• Theoretically, we must receive n mol H₂ and n mol I₂
• But the decomposition will not proceed to that extent
• This is because, some of the newly formed H₂ and I₂ will combine to give back HI
• So in this decomposition reaction also, an equilibrium will be attained
• This equilibrium is indicated by the horizontal portions in fig.b
9. Let us note an interesting point. It can be written in 3 steps:
(i) Number of moles:
    ♦ In step (7), we took n mol each of H₂ and I₂
    ♦ In step (8), we took 2n mol HI
• So the number of moles is the same in two cases
(ii) If number of moles is same, the number of atoms will also be the same. Then the curves will coincide
• That is:
    ♦ The green curve in fig.b
    ♦ will be at the same level as
    ♦ The green curve in fig.a
• Also:
    ♦ The red curve in fig.b
    ♦ will be at the same level as
    ♦ The red curve in fig.a
(iii) Since the two curves in each graph coincide, we can combine them. This is shown in fig.c
• By making suitable changes to temperature and pressure, the manufacturer can force the magenta dashed line to be more towards the right
• Then at equilibrium, there will be a greater concentration of HI
• In later sections, we will see how this can be achieved

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• In the next section, we will see equilibrium constant in chemical reactions


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