pH Of Acids And Bases - Practice Questions & MCQ

pH is also referred to as potential or power of hydrogen. Mathematically, it can be represented as follows:

If solution is neutral, then:
K_w = [H₃O⁺][OH^-]
From the ionic product of water, we know:
K_w = 10^-14
[H₃O⁺] = [OH^-] = x (since solution is neutral)
Thus, 10^-14 = K_w = x²
               x = 10^-7
Now, [H₃O⁺] = 10^-7
Thus, pH = - log₁₀(H₃O⁺) = - log₁₀(10^-7) = 7

For Acidic solutions:                                                                                For Basic solutions:
For acidic solutions, we must have [H₃O⁺] > [OH^-]                                       For basic solutions, we must have [H₃O⁺] < [OH^-] 
Thus, [H₃O⁺] > 10^-7                                                                                   Thus, [H₃O⁺] < 10^-7
Thus, [H₃O⁺] for acids can be 10^-6, 10^-5, 10^-4, etc.                                      Thus, [H₃O⁺] for basics can be 10^-8, 10^-9, 10^-10, etc.
Thus, pH of acids can be 6, 5, 4, etc.                                                          Thus, pH of basics can be 8, 9, 10, 11, etc.
Hence, pH of acidic solutions is less than 7                                                  Hence, pH of basic solutions is greater than 7

pH depends upon temperature
We know from ionic product of water that at 63⁰C, the value of K_w = 10^-13.
For neutral solution we know:

Hence, pH depends upon temperature

pH of Strong Acids
Strong acids are those acids which dissociate completely in solutions. For example:

• 2 x 10^-3 M HNO₃
  Since HNO₃ is a strong acid, thus it will dissociate completely into H⁺ and OH^- ions as follows:
  
  
  Thus, pH of HNO₃ is 2.7

• 10^-4 M H₂SO₄
  Since H₂SO₄ is a strong acid, thus it will dissociate completely into H⁺ and OH^- ions as follows:
  
  
  Thus, pH of H₂SO₄ is 3.7

NOTE: If molarity(N) of solution is not given but normality(N) is given, then molarity can be calculated using the following formula:

N = M x n
where, n is the number of moles

pH of Weak Acids
Weak acids are those acids which dissociate partially in solutions. For example:

• 8 M HA (K_a =2 x 10^-8)
  The chemical equation for the dissociation of weak acid HA is as follows:
             
  Initial:   8M            0           0
  Equil:  8 - 8?         8?         8?
  
  The equilibrium constant K_a for the weak acid is given as follows:
  
  Thus, pH of this given acid = 3.4
   
• 0.002N CH₃COOH(? = 0.02)
  The chemical equation for the dissociation of CH₃COOH is as follows:
             
  Initial:           c                              0                   0
  Equil:         c - c?                         c?                c?
  
  The equilibrium constant K_a for the weak acid is given as follows:
  
  Thus, pH of acetic acid = 4.4

Strong bases
Strong bases are those bases that dissociate completely in solution. For example:

• 0.1M NaOH
  Since NaOH is a strong base, thus it will dissociate completely into Na⁺ and OH^- ions. The chemical equation for the dissociation of NaOH is as follows:
  
  Hence, pH of 0.1M NaOH solution is 13.
   
•  0.05M Ba(OH)₂
  Since Ba(OH)₂ is a strong base, thus it will dissociate completely into Ba²⁺ and 2OH^- ions. The chemical equation for the dissociation of Ba(OH)₂ is as follows:
  
  Hence, pH of 0.05M Ba(OH)₂ solution is 13.

• Mixture of Strong Acids:
  
  
  
  
  
  NOTE: Shortcut only for monobasic acids and monoacidic bases:
  
  
   
• Mixture of Strong Bases:
  
  
  Using the shortcut formula for bases as given above, we get:
  
   
• Mixture of Strong Acid and Strong Base:
  
  
  Clearly, moles of (H⁺) = 2 x 10^-2 moles and moles of (OH^-) = 1 x 10^-3 moles. 
  Since moles of (H⁺) is greater than moles of (OH^-), therefore the solution medium will be acidic.
  Now, remaining moles of H⁺ = 2 x 10^-2 - 10^-3 = 19 x 10^-3
  
  Thus, pH of the mixture = 2.2