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.
- Water autoionizes with an equilibrium constant, Kw:
- 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.
- 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.
NaOH (aq) → Na+(aq) + OH-(aq)
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