pH Scale and Acid Dissociation

The pH Scale

  • In aqueous solutions, the concentration of hydronium ions [H3O+][H_3O^+] can range from 1010 M to 10−1510^{-15} M.
  • The pH scale is a logarithmic scale used to compress this range for the intensity of acids in aqueous solutions.
  • The pH of an acid is defined as: pH=−log[H3O+]pH = -log[H_3O^+]
    • Example (a): Pure water at 25°C has [H3O+]=1.0×10−7[H_3O^+] = 1.0 × 10^{-7} M.
      • Therefore, pH=−log[H3O+]=−log[1.0×10−7]=−(−7.00)=7.00pH = -log[H_3O^+] = -log[1.0 × 10^{-7}] = -(-7.00) = 7.00
    • Example (b): An aqueous solution has [H3O+]=0.10[H_3O^+] = 0.10 M (or 1.0×10−11.0 × 10^{-1} M) when an acid is added to it.
      • Therefore, pH=−log[H3O+]=−log[1.0×10−1]=−(−1.00)=1.00pH = -log[H_3O^+] = -log[1.0 × 10^{-1}] = -(-1.00) = 1.00
    • Example (c): An aqueous solution has [OH−]=0.10[OH^-] = 0.10 M (or 1.0×10−11.0 × 10^{-1} M) when a base is added to it.
      • K<em>w=1.0×10−14=[H</em>3O+][OH−]K<em>w = 1.0 × 10^{-14} = [H</em>3O^+][OH^-]
      • [H<em>3O+]=K</em>w[OH−]=1.0×10−140.10=1.0×10−13[H<em>3O^+] = \frac{K</em>w}{[OH^-]} = \frac{1.0 × 10^{-14}}{0.10} = 1.0 × 10^{-13}
      • pH=−log[H3O+]=−log[1.0×10−13]=−(−13.00)=13.00pH = -log[H_3O^+] = -log[1.0 × 10^{-13}] = -(-13.00) = 13.00
  • If pH is given, the concentration of H<em>3O+H<em>3O^+ can be calculated by: [H</em>3O+]=10−pH[H</em>3O^+] = 10^{-pH}
    • This is done on the calculator by using the 10x10^x key.
    • Example (a): Calculate the [H3O+][H_3O^+] for a solution with a pH of 4.80.
      • [H3O+]=10−pH=10−4.80=1.6×10−5[H_3O^+] = 10^{-pH} = 10^{-4.80} = 1.6 × 10^{-5} M
    • Example (b): Calculate the [H3O+][H_3O^+] for a solution with a pH of 8.37.
      • [H3O+]=10−pH=10−8.37=4.3×10−9[H_3O^+] = 10^{-pH} = 10^{-8.37} = 4.3 × 10^{-9} M
  • Overall Properties of pH
    • A high pH signifies a low concentration of H3O+H_3O^+ and vice versa.
    • A change in pH by one unit implies a change in concentration of H3O+H_3O^+ and OH−OH^- by a factor of 10 (one order of magnitude).
    • In terms of pH at 25°C:
    • If pH<7pH < 7, the solution is acidic, and [H3O+]>[OH−][H_3O^+] > [OH^-].
    • If pH=7pH = 7, the solution is neutral, and [H3O+]=[OH−][H_3O^+] = [OH^-].
    • If pH > 7, the solution is basic, and [H_3O^+] < [OH^-].
  • Example Calculation: Calculating pH given [OH−][OH^-].
    • Calculate the pH (at 25°C) of an aqueous solution that has an OH−(aq)OH^-(aq) concentration of 1.2×10−61.2 × 10^{-6} M.
      • K<em>w=[H</em>3O+][OH−]K<em>w = [H</em>3O^+][OH^-]
      • [H<em>3O+]=K</em>w[OH−]=1.00×10−141.2×10−6=8.3×10−9[H<em>3O^+] = \frac{K</em>w}{[OH^-]} = \frac{1.00 × 10^{-14}}{1.2 × 10^{-6}} = 8.3 × 10^{-9} M
      • pH=−log[H3O+]=−log[8.3×10−9]=8.08pH = -log[H_3O^+] = -log[8.3 × 10^{-9}] = 8.08
  • Practice Exercise:
    • Calculate the pH of each solution at 25°C and indicate whether the solution is acidic or basic:
      • (a) [H3O+]=1.8×10−4[H_3O^+] = 1.8 × 10^{-4} M, (b) [OH−]=1.3×10−2[OH^-] = 1.3 × 10^{-2} M.
      • Answer: (a) 3.74, acidic; (b) 12.11, basic.
  • The pH of some grape juice (at 25°C) is 2.85. Calculate [H3O+][H_3O^+] and [OH−][OH^-].
    • pH=−log[H3O+]pH = -log[H_3O^+]
    • [H3O+]=10−pH=10−2.85=1.4×10−3[H_3O^+] = 10^{-pH} = 10^{-2.85} = 1.4 × 10^{-3} M
    • [OH−]=K<em>w[H</em>3O+]=1.00×10−141.4×10−3=7.1×10−12[OH^-] = \frac{K<em>w}{[H</em>3O^+]} = \frac{1.00 × 10^{-14}}{1.4 × 10^{-3}} = 7.1 × 10^{-12}
  • Practice Exercise:
    • The pH of human blood was measured to be 7.41 at 25°C. Calculate [H3O+][H_3O^+] and [OH−][OH^-].
      • Answer: [H3O+]=3.9×10−8[H_3O^+] = 3.9 × 10^{-8} M, [OH−]=2.6×10−7[OH^-] = 2.6 × 10^{-7} M

pOH and pKw

  • The 'p' in pH signifies the negative log of H3O+H_3O^+. Likewise, the 'p' in pOH signifies the negative log of OH−OH^-.
    • pOH=−log[OH−]pOH = -log[OH^-]
  • Similarly, the 'p' in pKw signifies the negative log of Kw.
    • pK<em>w=−logK</em>w=−log[1.0×10−14]=−(−14.00)=14.00pK<em>w = -log K</em>w = -log [1.0 × 10^{-14}] = -(-14.00) = 14.00
  • At 25°C the pKw of water is 14.00.
  • Also, K<em>w=[H</em>3O+][OH−]=1.0×10−14K<em>w = [H</em>3O^+][OH^-] = 1.0 × 10^{-14}
    • (−logK<em>w)=(−log[H</em>3O+])+(−log[OH−])=(−log[1.0×10−14])(-log K<em>w) = (-log [H</em>3O^+]) + (-log [OH^-]) = (-log [1.0 × 10^{-14}])
    • pKw=pH+pOH=14.00pK_w = pH + pOH = 14.00
  • At 25°C, the sum of pH and pOH is equal to 14.00.

Acids

  • Aqueous solutions of acids have [H_3O^+] > [OH^-].
  • Acids are classified as strong acids or weak acids, depending on whether their reactions with water to give H3O+(aq)H_3O^+(aq) ions go to almost completion or reach an equilibrium somewhat short of completion.
  • Strong acids:
    1. Have K_a >> 1
    2. Dissociate to a large extent (approaches ~100% dissociation).
      • Consequently, we can assume that strong acids react completely with water to produce H3O+(aq)H_3O^+(aq) ions.
    3. Have large [H3O+][H_3O^+] concentrations at equilibrium.
  • Example: HCl(aq) is a strong acid (Ka 107K_a ~10^7 M).
    • We can calculate [H3O+][H_3O^+] and pH in 0.10 M HCl(aq) as follows:
      • HCl(aq)+H<em>2O(l)⇌H</em>3O+(aq)+Cl−(aq)HCl(aq) + H<em>2O(l) \rightleftharpoons H</em>3O^+(aq) + Cl^-(aq)
      • Ka= 107K_a = ~10^7 M
      • (I) Initial concentrations (M): 0.10 - 0 0
      • (E) Equilibrium concentrations (M): ~0 - 0.10 0.10
      • Therefore, [H3O+]=0.10[H_3O^+] = 0.10 M
      • pH=−log(0.10)=−log(1.0×10−1)=1.00pH = -log(0.10) = -log (1.0 × 10^{-1}) = 1.00
      • % dissociation = ~100%
  • Similarly, HClO4(aq) is a strong acid (Ka 107K_a ~10^7 M).
    • We can calculate [H<em>3O+][H<em>3O^+] and pH in 0.010 M HClO</em>4(aq)HClO</em>4(aq) as follows:
      • HClO<em>4(aq)+H</em>2O(l)⇌H<em>3O+(aq)+ClO</em>4−(aq)HClO<em>4(aq) + H</em>2O(l) \rightleftharpoons H<em>3O^+(aq) + ClO</em>4^-(aq)
      • Ka= 107K_a = ~10^7 M
      • (I) Initial concentrations (M): 0.010 - 0 0
      • (E) Equilibrium concentrations (M): ~0 - 0.010 0.010
      • Therefore, [H3O+]=0.010[H_3O^+] = 0.010 M
      • pH=−log(0.010)=−log(1.0×10−2)=2.00pH = -log(0.010) = -log (1.0 × 10^{-2}) = 2.00
      • % dissociation = ~100%
  • HNO3(aq) is a strong acid (Ka 20K_a ~20 M).
    • We can calculate [H<em>3O+][H<em>3O^+] and pH in 1.0 M HNO</em>3(aq)HNO</em>3(aq) as follows:
      • HNO<em>3(aq)+H</em>2O(l)⇌H<em>3O+(aq)+NO</em>3−(aq)HNO<em>3(aq) + H</em>2O(l) \rightleftharpoons H<em>3O^+(aq) + NO</em>3^-(aq)
      • Ka= 20K_a = ~20 M
      • (I) Initial concentrations (M): 1.0 - 0 0
      • (E) Equilibrium concentrations (M): ~0 - 1.0 1.0
      • Therefore, [H3O+]=1.0[H_3O^+] = 1.0 M
      • pH=−log(1.0)=−log(1.0)=0pH = -log(1.0) = -log (1.0) = 0
      • % dissociation = ~100%
  • Practice Exercise:
    • Calculate (a) the pH of a solution that is 0.10 M HNO<em>3(aq)HNO<em>3(aq), (b) the pH of a solution that is 1.0×10−61.0 × 10^{-6} M HCl(aq), and (c) the pH of a solution that is 0.10 M HNO</em>3(aq)HNO</em>3(aq) and 1.0×10−61.0 × 10^{-6} M HCl(aq).
      • Answer: (a) 1.00, (b) 6.00, (c) 1.00

Acid Dissociation Constants:

  • Acid Dissociation Constants in Water at 25 °C:
    • Hydroiodic (HI): Conjugate Base I–, K<em>aK<em>a ~101110^{11}, pK</em>apK</em>a ~–11
    • Hydrobromic (HBr): Conjugate Base Br–, K<em>aK<em>a ~10910^9, pK</em>apK</em>a ~–9
    • Perchloric (HClO<em>4HClO<em>4): Conjugate Base ClO</em>4ClO</em>4–, K<em>aK<em>a ~10710^7, pK</em>apK</em>a ~–7
    • Hydrochloric (HCl): Conjugate Base Cl–, K<em>aK<em>a ~10710^7, pK</em>apK</em>a ~–7
    • Chloric (HClO<em>3HClO<em>3): Conjugate Base ClO</em>3ClO</em>3–, K<em>aK<em>a ~10310^3, pK</em>apK</em>a ~–3
    • Sulfuric (1) (H<em>2SO</em>4H<em>2SO</em>4): Conjugate Base HSO<em>4HSO<em>4–, K</em>aK</em>a ~10210^2, pKapK_a ~–2
    • Nitric (HNO<em>3HNO<em>3): Conjugate Base NO</em>3NO</em>3–, K<em>aK<em>a ~20, pK</em>apK</em>a ~–1.3