Enduring Understanding 6.C: Acid-Base Reactions

  • Acid-Base reactions are an example of a type of reaction that usually reach equilibrium very quickly, and can have very large or very small equilibrium constants.
  • As has been mentioned before, acids in water form hydronium (H3O+) ions, and bases form hydroxide ions.
  • HBr(aq) + H2O(l) → H3O+(aq) + Br-(aq)
    NaOH (aq) → Na+(aq) + OH-(aq)
  • Water autoionizes with an equilibrium constant, Kw:
  • 2H2O(l) → H3O+(aq) + OH-(aq) Kw = [H3O+] [OH-] = 1 x 10-14
  • Therefore, in pure water at 25 °C, [H3O+] = [OH-] = 1 x 10-7
  • To avoid using very small numbers, negative logarithms are often used:
  • pKw = -log(Kw) = 14.00
  • pH = -log([H3O+])
  • pOH = -log([OH-])
  • In pure water at 25 °C, pH = pOH = 7.00
  • In aqueous solutions, pH + pOH = 14.00
  • Strong acids (e.g. HCl, HBr, H2SO4, HNO3) will completely ionize in water, so the concentration of H3O+ will be approximately equal to the initial concentration of the acid.
  • Example: 0.1M HCl has a [H3O+] of 0.1M.
  • Its pH is -log(0.1) or 1.00
  • Similarly, a strong base (e.g. NaOH) completely ionizes in water. The concentration of OH- will be equal to the initial concentration of the base.
  • Example: 0.1M NaOH has a [OH-] of 0.1M
  • Its pOH is -log(0.1) or 1.00
  • pH is (14.00 - 1.00) or 13.00
  • Weak acids and bases only ionize partially in solution.
  • CH3COOH + H2O → H3O+ + CH3COO-    Ka = 1.8 x 10-5
  • The pKa of acetic acid is -log(1.8 x 10-5) or 4.75
  • Sample Question: Calculate the pH of a 0.1M acetic acid solution.
  • Ka = 1.8 x 10-5 = [H3O+] [CH3COO- ] / [CH3COOH]
  • Since [H3O+] = [CH3COO- ], and [CH3COOH]□0.1
  • 1.8 x 10-5 = [H3O+]2 / 0.1
  • 1.8 x 10-6 = [H3O+]2
  • [H3O+] = 1.3 x 10-3
  • pH = -log([H3O+]) = -log(1.3 x 10-3) = 2.88
  • This corresponds to < 1% ionization of the acetic acid.



Related Links:
Chemistry
Chemistry Quizzes
AP Chemistry Notes
Titration Curves



To link to this Acid-Base Reactions page, copy the following code to your site:


Educational Videos