# 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

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