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

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• 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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