Chapter 7.16 - Solved Examples on pH Scale

In the previous section, we saw the details about pH scale. In this section, we will see some solved examples

Solved example 7.52
The concentration of hydrogen ion in a sample of soft drink is 3.8 × 10^-3 M. what
is its pH ?
Solution:
• Given that: [H⁺] of the soft drink = 3.8 × 10^-3 M
• So pH of the soft drink = -log₁₀([H⁺]) = -log₁₀(3.8 × 10^-3) = 2.42
• Since the pH is less than 7, we can write: The soft drink is acidic

Solved example 7.53
Calculate pH of a 1.0 × 10^-8 M solution of HCl.
Solution:
1. Consider 1 L pure water
• In that water, there will be an undisturbed equilibrium according to the equation:
H₂O ⇌ H⁺ + OH^－
   ♦ Number of moles of H⁺ will be 10^-7
   ♦ Number of moles of OH^－ will be 10^-7
2. When HCl is added, that equilibrium will be disturbed
• Given that, 1.0 × 10^-8 moles of HCl is added to that 1 L water
• Since HCl is a strong acid, we assume that, all HCl molecules dissociate into H⁺ and Cl- ions
• So 10^-8 moles of H⁺ are added to the water
   ♦ Then total number of H⁺ ions = (10^-7 + 10^-8) = 1.1 × 10^-7
3. To regain equilibrium, some of the newly added H⁺ will react with the OH^－ ions
So in effect:
   ♦ There will be an increase in the number of H⁺ ions  
   ♦ There will be a decrease in the number of OH^－ ions
   ♦ But K_w will remain constant
4. Let x moles of H⁺ enter into reaction with OH^－
• Then the number of moles of OH^－ entering into reaction will also be x
• So at the new equilibrium,
   ♦ Number of moles of H⁺ ions = (1.1 × 10^-7 -x)
   ♦ Number of moles of OH^－ ions = (10^-7 -x)
5. At the new equilibrium, concentrations will be different. But the equilibrium constant K_w is the same. So we can write:
(1.1 × 10^-7 - x) (10^-7 -x)= 10^-14
⇒ (1.1 × 10^-14) - (10^-7 x)  - (1.1 × 10^-7 x) + x² = 10^-14
⇒ (1.1 × 10^-14) - (2.1 × 10^-7)x + x² = 10^-14
⇒ x² - 2.1 × 10^-7 x + 0.1 × 10^-14 = 0
• Solving this quadratic equation, we get:
x = 2.051 × 10^-7 or 4.875 × 10^-9
6. Calculation of [H⁺]:
• When x = 2.051 × 10^-7, we get: [H⁺] = (1.1 × 10^-7 -x)
= (1.1 × 10^-7 - 2.051 × 10^-7) = -9.512 × 10^-8
• When x = 4.875 × 10^-9, we get: [H⁺] = (1.1 × 10^-7 -x)
= (1.1 × 10^-7 - 4.875 × 10^-9) = 1.051 × 10^-7
• Negative value of concentration is not possible. So we must discard the value (x = 2.051 × 10^-7) obtained by solving the quadratic equation
• The acceptable value is: x = 4.875 × 10^-9
7. Calculation of [OH^－]:
• When x = 4.875 × 10^-9, we get: [OH^－] = (10^-7 -x)
= (10^-7 - 4.875 × 10^-9) = 9.512 × 10^-8
8. Check:
[H⁺][OH^－] = (1.051 × 10^-7)(9.512 × 10^-8) = 1.0 × 10^-14
9. Thus we get: pH = -log₁₀([H⁺]) = -log₁₀(1.051 × 10^-7) = 6.98
10. If required, we can calculate pOH also:
pOH = -log₁₀([OH^-]) = -log₁₀(9.512 × 10^-8) = 7.02
11. Note that: pH + pOH = (6.98 + 7.02) = 14
• The reason can be explained in 4 steps:
(i) We have: [H⁺][OH^－] = 10^-14
(ii) Taking logarithm of both sides, we get:
log₁₀([H⁺]) + log₁₀([OH-]) = log₁₀(10^-14)
⇒ log₁₀([H⁺]) + log₁₀([OH-]) = -14
(iii) Multiplying throughout by '-1', we get:
-log₁₀([H⁺]) + -log₁₀([OH-]) = 14
(iv) But -log₁₀([H⁺]) is pH. Also, -log₁₀([OH-]) is pOH
Thus we get: pH + pOH = 14

Solved example 7.54
Calculate pH of a 0.002 M solution of HCl.
Solution:
• We will solve this problem first by the long method, then by the short method. After that, we will compare the two methods
Long method:
1. Consider 1 L pure water
• In that water, there will be an undisturbed equilibrium according to the equation:
H₂O ⇌ H⁺ + OH^－
   ♦ Number of moles of H⁺ will be 10^-7
   ♦ Number of moles of OH^－ will be 10^-7
2. When HCl is added, that equilibrium will be disturbed
• Given that, 0.002 moles of HCl is added to that 1 L water
• Since HCl is a strong acid, we assume that, all HCl molecules dissociate into H⁺ and Cl^- ions
• So 0.002 moles of H⁺ are added to the water
   ♦ Then total number of H⁺ ions = (10^-7 + 0.002) = 20001 × 10^-7
3. To regain equilibrium, some of the newly added H⁺ will react with the OH^－ ions
So in effect:
   ♦ There will be an increase in the number of H⁺ ions  
   ♦ There will be a decrease in the number of OH^－ ions
   ♦ But K_w will remain constant
4. Let x moles of H⁺ enter into reaction with OH^－
• Then the number of moles of OH^－ entering into reaction will also be x
• So at the new equilibrium,
   ♦ Number of moles of H⁺ ions = (20001 × 10^-7 -x)
   ♦ Number of moles of OH^－ ions = (10^-7 -x)
5. At the new equilibrium, concentrations will be different. But the equilibrium constant K_w is the same. So we can write:
(20001 × 10^-7 - x) (10^-7 -x)= 10^-14
⇒ (20001 × 10^-14) - (10^-7 x)  - (20001 × 10^-7 x) + x² = 10^-14
⇒ (20001 × 10^-14) - (20002 × 10^-7)x + x² = 10^-14
⇒ x² - 20002 × 10^-7 x + 20000 × 10^-14 = 0
• Solving this quadratic equation, we get:
x = 0.0020001 or 9.99 × 10^-8
6. Calculation of [H⁺]:
• When x = 0.0020001, we get: [H⁺] = (20001 × 10^-7 -x)
= (20001 × 10^-7 - 0.0020001) = -5 × 10^-12
• When x = 9.99 × 10^-8, we get: [H⁺] = (20001 × 10^-7 -x)
= (20001 × 10^-7 - 9.99 × 10^-8) = 0.002
• Negative value of concentration is not possible. So we must discard the value (x = 0.0020001) obtained by solving the quadratic equation
• The acceptable value is: x = 9.99 × 10^-8
7. Calculation of [OH^－]:
• When x = 9.99 × 10^-8, we get: [OH^－] = (10^-7 -x)
= (10^-7 - 9.99 × 10^-8) = 5 × 10^-12
8. Check:
[H⁺][OH^－] = (0.002)(5 × 10^-12) = 1.0 × 10^-14
9. Thus we get: pH = -log₁₀([H⁺]) = -log₁₀(0.002) = 2.70

Short method:
1. In the long method, we saw that [H⁺] = 0.002
2. But it is given that, it is a 0.002 M HCl solution
• The 0.002 moles of HCl will give 0.002 moles of H⁺ ions
3. That means, the 10^-7 moles of H⁺ ions which pre-exist in the water, do not have any effect in deciding the pH
• This is because, 0.002 is very large when compared to 10^-7
4. In the previous example, 10^-8 moles of HCl is added
    ♦ The 10^-8 moles of H⁺ thus derived
    ♦ is comparable to the
    ♦ 10^-7 moles of H⁺ that pre-exist in water
• So we need to take the 10^-7 into account
5. In the short method, we can directly use the 0.002 to find pH
• We get: pH = -log₁₀([H⁺]) = -log₁₀(0.002) = 2.70

Comparison between the two methods:
1. When the given solution is dilute, we must use the long method
2. When the given solution is concentrated, we can use the short method

Solved example 7.55
Calculate pH of a 0.00005 M solution of NaOH
Solution:
• We will solve this problem first by the long method, then by the short method. After that, we will compare the two methods
Long method:
1. Consider 1 L pure water
• In that water, there will be an undisturbed equilibrium according to the equation:
H₂O ⇌ H⁺ + OH^－
   ♦ Number of moles of H⁺ will be 10^-7
   ♦ Number of moles of OH^－ will be 10^-7
2. When NaOH is added, that equilibrium will be disturbed
• Given that, 0.00005 moles of NaOH is added to that 1 L water
• Since NaOH is a strong base, we assume that, all NaOH molecules dissociate into OH^- and Na⁺ ions
• So 0.00005 moles of OH^- are added to the water
   ♦ Then total number of OH^- ions = (10^-7 + 0.00005) = 501 × 10^-7
3. To regain equilibrium, some of the newly added OH^- will react with the H⁺ ions
So in effect:
   ♦ There will be an increase in the number of OH^- ions  
   ♦ There will be a decrease in the number of H⁺ ions
   ♦ But K_w will remain constant
4. Let x moles of OH^- enter into reaction with H⁺
• Then the number of moles of H⁺ entering into reaction will also be x
• So at the new equilibrium,
   ♦ Number of moles of OH^- ions = (501 × 10^-7 -x)
   ♦ Number of moles of H⁺ ions = (10^-7 -x)
5. At the new equilibrium, concentrations will be different. But the equilibrium constant K_w is the same. So we can write:
(501 × 10^-7 - x) (10^-7 -x)= 10^-14
⇒ (501 × 10^-14) - (10^-7 x)  - (501 × 10^-7 x) + x² = 10^-14
⇒ (501 × 10^-14) - (502 × 10^-7)x + x² = 10^-14
⇒ x² - 502 × 10^-7 x + 500 × 10^-14 = 0
• Solving this quadratic equation, we get:
x = 5.01 × 10^-5 or 9.98 × 10^-8
6. Calculation of [OH^-]:
• When x = 5.01 × 10^-5, we get: [OH^-] = (501 × 10^-7 -x)
= (501 × 10^-7 - 5.01 × 10^-5) = 0
• When x = 9.98 × 10^-8, we get: [OH^-] = (501 × 10^-7 -x)
= (501 × 10^-7 - 9.98 × 10^-8) = 5.0 × 10^-5
• zero concentration is not possible. So we must discard the value (x = 5.01 × 10^-5) obtained by solving the quadratic equation
• The acceptable value is: x = 9.98 × 10^-8
7. Calculation of [H⁺]:
• When x = 9.98 × 10^-8, we get: [H⁺] = (10^-7 -x)
= (10^-7 - 9.98 × 10^-8) = 1.99 × 10^-10
8. Check:
[H⁺][OH^－] = (1.99 × 10^-10)(5.0 × 10^-5) = 1.0 × 10^-14
9. Thus we get: pH = -log₁₀([H⁺]) = -log₁₀(1.99 × 10^-10) = 9.70

Short method:
1. In the long method, we saw that [H⁺] = 5.0 × 10^-5 = 0.00005
2. But it is given that, it is a 0.00005 M NaOH solution
• The 0.00005 moles of NaOH will give 0.00005 moles of OH^- ions
3. That means, the 10^-7 moles of OH^- ions which pre-exist in the water, do not have any effect in deciding the pH
• This is because, 0.00005 is very large when compared to 10^-7
5. So in the short method, we can directly use the 0.00005 to find pOH
• We get: pOH = -log₁₀([OH^-]) = -log₁₀(0.00005) = 4.30
• Thus pH = (14.0 - pOH) = (14.0 - 4.30) = 9.70

Comparison between the two methods:
1. When the given solution is dilute, we must use the long method
2. When the given solution is concentrated, we can use the short method

Solved example 7.56
The pH of a sample of vinegar is 3.76. Calculate the concentration of hydrogen
ion in it.
Solution:
1. We have: pH = -log₁₀([H⁺])
• Substituting the given pH, we get: 3.76 = -log₁₀([H⁺])
2. This is same as: log₁₀([H⁺]) = -3.76
• So [H⁺] will be equal to the antilog of -3.76
3. Thus we get: [H⁺] = antilog (-3.76) = 1.7 × 10^-4 M

Solved example 7.57
Calculate the hydrogen ion concentration in the following biological fluids whose
pH are given below:
(a) Human muscle-fluid, 6.83 (b) Human stomach fluid, 1.2
(c) Human blood, 7.38 (d) Human saliva, 6.4.
Solution:
• We have: pH = -log₁₀([H⁺])
⇒ log₁₀([H⁺]) = -pH
⇒ [H⁺] = antilog (-pH)
Part (a): [H⁺] = antilog (-6.83) = 1.48 × 10^-7 M
Part (b): [H⁺] = antilog (-1.2) = 0.063 M
Part (c): [H⁺] = antilog (-7.38) = 4.17 × 10^-8 M
Part (d): [H⁺] = antilog (-6.4) = 3.98 × 10^-7 M

Solved example 7.58
The pH of milk, black coffee, tomato juice, lemon juice and egg white are 6.8,
5.0, 4.2, 2.2 and 7.8 respectively. Calculate corresponding hydrogen ion
concentration in each
Solution:
• We have: pH = -log₁₀([H⁺])
⇒ log₁₀([H⁺]) = -pH
⇒ [H⁺] = antilog (-pH)
Part (a): [H⁺] = antilog (-6.8) = 1.58 × 10^-7 M
Part (b): [H⁺] = antilog (-5.0) = 1.0 × 10^-5 M
Part (c): [H⁺] = antilog (-4.2) = 6.31 × 10^-5 M
Part (d): [H⁺] = antilog (-2.2) = 0.0063 M
Part (e): [H⁺] = antilog (-7.8) = 1.58 × 10^-8 M 

Solved example 7.59
Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl (b) 0.005 M NaOH (c) 0.002 M HBr (d) 0.002 M KOH
Solution:
Part (a):
• If there is complete dissociation, [H+] will be 0.003 in the 0.003 M HCl solution
• So pH = -log₁₀([H⁺]) = -log₁₀(0.003) = 2.52   
Part (b):
• If there is complete dissociation, [OH-] will be 0.005 in the 0.005 M NaOH solution
• So pH = [14 - pOH] = [14 + log₁₀([OH-]) = [14 + log₁₀(0.005)] = [14 - 2.3] = 11.70   
Part (c):
• If there is complete dissociation, [H+] will be 0.002 in the 0.002 M HCl solution
• So pH = -log₁₀([H⁺]) = -log₁₀(0.002) = 2.70   
Part (d):
• If there is complete dissociation, [OH-] will be 0.002 in the 0.002 M KOH solution
• So pH = [14 - pOH] = [14 + log₁₀([OH-]) = [14 + log₁₀(0.002)] = [14 - 2.7] = 11.30

Solved example 7.60
Calculate the pH of the following solutions:
(a) 2 g of TlOH dissolved in water to give 2 litre of solution
(b) 0.3 g of Ca(OH)₂ dissolved in water to give 500 mL of solution
(c) 0.3 g of NaOH dissolved in water to give 200 mL of solution
(d) 1 mL of 13.6 M HCl is diluted with water to give 1 litre of solution
Solution:
Part (a):
1. Molar mass of TlOH (Thallium hydroxide) is [204.37+16+1] = 221.37 g
• So 2 g of TlOH will contain ²⁄_221.37 = 0.009 moles of TlOH
2. 2 L of water contains 0.009 moles of TlOH
• So 1 L will contain (^0.009⁄₂) = 0.0045 moles of TlOH
• Thus we get: Initial concentration of TlOH = [TlOH] = 0.0045 M
3. The dissociation of TlOH will be according to the equation:
TlOH ⇌ Tl⁺ + OH^-
• So each mole of TlOH will give one mole of OH^-
• So 0.0045 moles of TlOH will give 0.0045 moles of OH^-
• Thus we get: [OH^-] = 0.0045
4. We have: pH = [14 - pOH] = [14 + log₁₀([OH^-]) = [14 + log₁₀(0.0045)] = [14 - 2.35] = 11.65
5. Note:
• Being a strong base, TlOH will dissociate completely
   ♦ So 0.0045 moles of TlOH will give 0.0045 moles of OH^-
• If it was not a strong base, we would not be able to write this exact number of moles of OH^-
   ♦ We would need the help of ionization constant to find it
Part (b):
1. Molar mass of Ca(OH)₂ is [40+2(16+1)] = 74 g
• So 0.3 g of Ca(OH)₂ will contain ^0.3⁄₇₄ = 0.00405 moles of Ca(OH)₂
2. 500 mL of water contains 0.00405 moles of Ca(OH)₂
• So 1 L will contain (2 × 0.00405) = 0.0081 moles of Ca(OH)2
• Thus we get: Initial concentration of Ca(OH)₂ = [Ca(OH)₂] = 0.0081 M
3. The dissociation of Ca(OH)₂ will be according to the equation:
Ca(OH)₂ ⇌ Ca²⁺ + 2OH^-
• So each mole of Ca(OH)₂ will give two moles of OH^-
• So 0.0081 moles of Ca(OH)₂ will give (2 0.0081) = 0.0162 moles of OH^-
• Thus we get: [OH^-] = 0.0162
4. We have: pH = [14 - pOH] = [14 + log₁₀([OH^-]) = [14 + log₁₀(0.0162)] = [14 - 1.79] = 12.21
5. Note:
• Being a strong base, Ca(OH)₂ will dissociate completely
   ♦ So 0.0081 moles of Ca(OH)2 will give 0.0162 moles of OH^-
• If it was not a strong base, we would not be able to write this exact number of moles of OH^-
   ♦ We would need the help of ionization constant to find it
Part (c):
1. Molar mass of NaOH (Sodium hydroxide) is [23+16+1] = 40 g
• So 0.3 g of NaOH will contain ^0.3⁄₄₀ = 0.0075 moles of NaOH
2. 200 mL of water contains 0.0075 moles of NaOH
• So 1 L will contain (0.0075 × 5) = 0.0375 moles of NaOH
• Thus we get: Initial concentration of NaOH = [NaOH] = 0.0375 M
3. The dissociation of NaOH will be according to the equation:
NaOH ⇌ Na⁺ + OH^-
• So each mole of NaOH will give one mole of OH^-
• So 0.0375 moles of NaOH will give 0.0375 moles of OH^-
• Thus we get: [OH^-] = 0.0375
4. We have: pH = [14 - pOH] = [14 + log₁₀([OH^-]) = [14 + log₁₀(0.0375)] = [14 - 1.43] = 12.57
5. Note:
• Being a strong base, NaOH will dissociate completely
   ♦ So 0.0375 moles of NaOH will give 0.0375 moles of OH^-
• If it was not a strong base, we would not be able to write this exact number of moles of OH^-
   ♦ We would need the help of ionization constant to find it
Part (d):
1. Molar mass of HCl (Hydrochloric acid) is [1+35.45] = 36.45 g
• Given that, concentration of HCl is 13.6 M
• So 1 L solution will contain 13.6 moles of HCl
• So 1 mL will contain ^13.6⁄₁₀₀₀ = 0.0136 moles of HCl
2. This 1 mL is diluted to make 1 L solution
• So the final 1 L solution will contain 0.0136 moles of HCl
• Thus we get: Initial concentration of HCl = [HCl] = 0.0136 M
3. The dissociation of HCl will be according to the equation:
HCl ⇌ H⁺ + Cl^-
• So each mole of HCl will give one mole of H⁺
• So 0.0136 moles of HCl will give 0.0136 moles of H⁺
• Thus we get: [H⁺] = 0.0136
4. We have: pH = log₁₀([H⁺]) = log₁₀(0.0136)] = 1.87]
5. Note:
• Being a strong acid, HCl will dissociate completely
   ♦ So 0.0136 moles of HCl will give 0.0136 moles of H⁺
• If it was not a strong acid, we would not be able to write this exact number of moles of H⁺
   ♦ We would need the help of ionization constant to find it

Solved example 7.61
0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is it's pH?
Solution:
1. Molar mass of KOH (Potassium hydroxide) is [39.1 +16+1] = 56.1 g
• So 0.561 g of KOH will contain ^0.561⁄_56.1 = 0.01 moles of KOH
2. 200 mL of water contains 0.01 moles of KOH
• So 1 L will contain (0.01 × 5) = 0.05 moles of KOH
• Thus we get: Initial concentration of KOH = [KOH] = 0.05 M
3. The dissociation of KOH will be according to the equation:
KOH ⇌ K⁺ + OH^-
• So each mole of KOH will give one mole of OH^-
• So 0.05 moles of KOH will give 0.05 moles of OH^-
• Thus we get: [OH^-] = 0.05
4. We have: pH = [14 - pOH] = [14 + log₁₀([OH^-]) = [14 + log₁₀(0.05)] = [14 - 1.30] = 12.7
5. We have: [H⁺][OH^-] = 10^-14
⇒ [H⁺] 0.05 = 10^-14
⇒ [H⁺] = 2.0 × 10^-13
6. Note:
• Being a strong base, KOH will dissociate completely
   ♦ So 0.05 moles of KOH will give 0.05 moles of OH^-
• If it was not a strong base, we would not be able to write this exact number of moles of OH^-
   ♦ We would need the help of ionization constant to find it

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• In the above three solved examples 7.59, 7.60 and 7.61, the substances were strong acids and strong bases
• So they dissociated completely. As a result, we could know the concentrations of the H⁺ and OH^- ions
• If they were not strong, we would not be able to write the concentrations directly
  ♦ We would need the help of ionization constant to find it
• So our next aim is to study about the ionization constants of weak acids and weak bases
◼  But what about strong acids and strong bases? Wouldn't we need their ionization constants?
• Answer can be written in 3 steps:
(i) The strong acid and strong base will dissociate almost completely
• So [Strong acid]_Equilibrium and [strong base]_Equilibrium will be very small
(ii) Since they come in the denominators, the resulting K will be very close to infinity
(iii) However, we would not need to use those K values because, we assume complete dissociation and so the concentrations of the ions can be directly obtained

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Let us see three examples where strong acids react with strong bases
Example 1: 25 mL of 0.1 M HCl reacts with 10 mL of 0.2 M Ca(OH)₂
• Details can be written in 6 steps:
1. Dissociation of HCl
(i) Molarity of the given HCl solution is 0.1
⇒ 1 liter of the solution will contain 0.1 moles of HCl molecules
⇒ 1 mL of the solution will contain 0.1 × 10^-3 moles of HCl molecules
⇒ 25 mL of the solution will contain (25 × 0.1 × 10^-3) = 2.5 × 10^-3 moles of HCl molecules
(ii) HCl dissociates completely into ions as follows:
HCl → H⁺ + Cl^-
• Since there is complete dissociation, the 25 mL sample will contain:
    ♦ 2.5 × 10^-3 moles of H⁺ ions
    ♦ 2.5 × 10^-3 moles of Cl^- ions
2.Dissociation of Ca(OH)₂
(i) Molarity of the given Ca(OH)₂ solution is 0.2
⇒ 1 liter of the solution will contain 0.2 moles of Ca(OH)₂ molecules
⇒ 1 mL of the solution will contain 0.2 × 10^-3 moles of Ca(OH)₂ molecules
⇒ 10 mL of the solution will contain (10 × 0.2 × 10^-3) = 2 × 10^-3 moles of Ca(OH)₂ molecules
(ii) Ca(OH)₂ dissociates completely into ions as follows:
Ca(OH)₂ → Ca²⁺ + 2OH^-
• Since there is complete dissociation, the 10 mL sample will contain:
    ♦ 2 × 10^-3 moles of Ca²⁺ ions
    ♦ (2 × 2 × 10^-3) = 4 × 10^-3 moles of OH^- ions
3. Reaction between ions
• When the two samples are mixed,
(i) Ca²⁺ will react with Cl^- to give CaCl₂ salt
(ii) H⁺ will react with OH^- to give water
• The reaction 3(i) will not affect the pH of the resulting solution. So we will not discuss it here
4. The reaction 3(ii) is:
H⁺ + OH^- → H₂O
• This reaction will affect the pH
• We have:
    ♦ 4 × 10^-3 moles of OH^- ions
    ♦ 2.5 × 10^-3 moles of H⁺ ions
• H⁺ ions are lesser in number. So all of them will be used up
    ♦ (4 × 10^-3 - 2.5 × 10^-3) = 1.5 × 10^-3 moles of OH^- ions will remain
5. Volume of the resulting mixture is (10 + 25) = 35 mL
⇒ 35 mL contain 1.5 × 10^-3 moles of OH^-
⇒ 1 mL will contain [(^1.5⁄₃₅) × 10^-3] moles of OH^-
⇒ 1 liter will contain (^1.5⁄₃₅) = 4.29 × 10^-2 moles of OH^-
⇒ [OH^-] = 4.29 × 10^-2 moles
6. pH = (14 - pOH) = (14 + log₁₀[OH^-])
= (14 + log₁₀(4.29 × 10^-2)) = (14 - 1.37) = 12.6

Example 2: 10 mL of 0.01M H₂SO₄ reacts with 10 mL of 0.01M Ca(OH)2/
• Details can be written in 5 steps:
1. Dissociation of H₂SO₄
(i) Molarity of the given H₂SO₄ solution is 0.01
⇒ 1 liter of the solution will contain 0.01 moles of H₂SO₄ molecules
⇒ 1 mL of the solution will contain 0.01 × 10^-3 moles of H₂SO₄ molecules
⇒ 10 mL of the solution will contain (10 × 0.01 × 10^-3) = 0.1 × 10^-3 moles of H₂SO₄ molecules
(ii) H₂SO₄ dissociates completely into ions as follows:
H₂SO₄ → 2H⁺ + SO₄^2-
• Since there is complete dissociation, the 10 mL sample will contain:
    ♦ 2 × 0.1 × 10^-3 = 0.2 × 10^-3 moles of H⁺ ions
    ♦ 0.1 × 10^-3 moles of SO₄^2- ions
2.Dissociation of Ca(OH)₂
(i) Molarity of the given Ca(OH)₂ solution is 0.01
⇒ 1 liter of the solution will contain 0.01 moles of Ca(OH)₂ molecules
⇒ 1 mL of the solution will contain 0.01 × 10^-3 moles of Ca(OH)₂ molecules
⇒ 10 mL of the solution will contain (10 × 0.01 × 10^-3) = 0.1 × 10^-3 moles of Ca(OH)₂ molecules
(ii) Ca(OH)₂ dissociates completely into ions as follows:
Ca(OH)₂ → Ca²⁺ + 2OH^-
• Since there is complete dissociation, the 10 mL sample will contain:
    ♦ 0.1 × 10^-3 moles of Ca²⁺ ions
    ♦ (2 × 0.1 × 10^-3) = 0.2 × 10^-3 moles of OH^- ions
3. Reaction between ions
• When the two samples are mixed,
(i) Ca²⁺ will react with SO₄^2- to give CaSO₄ salt
(ii) H⁺ will react with OH^- to give water
• The reaction 3(i) will not affect the pH of the resulting solution. So we will not discuss it here
4. The reaction 3(ii) is:
H⁺ + OH^- → H₂O
• This reaction will affect the pH
• We have:
    ♦ 0.2 × 10^-3 moles of OH^- ions
    ♦ 0.2 × 10^-3 moles of H⁺ ions
• H⁺ ions and OH^- ions are equal in number. So all of them will get cancelled
    ♦ The resulting solution will have neither H⁺ nor OH^- ions
    ♦ pH of such a solution is 7

Example 3: 10 mL of 0.1M H₂SO₄ reacts with 10 mL of 0.1 M KOH
• Details can be written in 6 steps:
1. Dissociation of H₂SO₄
(i) Molarity of the given H₂SO₄ solution is 0.1
⇒ 1 liter of the solution will contain 0.1 moles of H₂SO₄ molecules
⇒ 1 mL of the solution will contain 0.1 × 10^-3 moles of H₂SO₄ molecules
⇒ 10 mL of the solution will contain (10 × 0.1 × 10^-3) = 1 × 10^-3 moles of H₂SO₄ molecules
(ii) H₂SO₄ dissociates completely into ions as follows:
H₂SO₄ → 2H⁺ + SO₄^2-
• Since there is complete dissociation, the 10 mL sample will contain:
    ♦ 2 × 1 × 10^-3 = 2 × 10^-3 moles of H⁺ ions
    ♦ 1 × 10^-3 moles of SO₄^2- ions
2.Dissociation of KOH
(i) Molarity of the given KOH solution is 0.1
⇒ 1 liter of the solution will contain 0.1 moles of KOH molecules
⇒ 1 mL of the solution will contain 0.1 × 10^-3 moles of KOH molecules
⇒ 10 mL of the solution will contain (10 × 0.1 × 10^-3) = 1 × 10^-3 moles of KOH molecules
(ii) KOH dissociates completely into ions as follows:
KOH → K⁺ + OH^-
• Since there is complete dissociation, the 10 mL sample will contain:
    ♦ 1 × 10^-3 moles of K⁺ ions
    ♦ 1 × 10^-3 moles of OH^- ions
3. Reaction between ions
• When the two samples are mixed,
(i) K⁺ will react with SO₄^2- to give K₂SO₄ salt
(ii) H⁺ will react with OH^- to give water
• The reaction 3(i) will not affect the pH of the resulting solution. So we will not discuss it here
4. The reaction 3(ii) is:
H⁺ + OH^- → H₂O
• This reaction will affect the pH
• We have:
    ♦ 1 × 10^-3 moles of OH^- ions
    ♦ 2 × 10^-3 moles of H⁺ ions
• OH^- ions are lesser in number. So all of them will be used up
    ♦ (2 × 10^-3 - 1 × 10^-3) = 1 × 10^-3 moles of H⁺ ions will remain
5. Volume of the resulting mixture is (10 + 10) = 20 mL
⇒ 20 mL contain 1 × 10^-3 moles of H⁺
⇒ 1 mL will contain [(¹⁄₂₀) × 10^-3] moles of H⁺
⇒ 1 liter will contain (¹⁄₂₀) = 0.05 moles of H⁺
⇒ [H⁺] = 0.05 moles
6. pH = -log₁₀[H⁺]) = -log₁₀(0.05) = 1.3

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


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