Buffers and Titrations South Kingstown igh School缓冲器和滴定南金斯敦高中.pptVIP

Buffers and Titrations South Kingstown igh School缓冲器和滴定南金斯敦高中.ppt

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Buffers and Titrations South Kingstown igh School缓冲器和滴定南金斯敦高中

Buffers A Buffer in Action Definition of a Buffer A buffer solution is one which resists changes in pH when small quantities of an acid or a base are added to it. The most important way that the pH of the blood is kept relatively constant is by buffers dissolved in the blood. H2CO3 + H2O ??HCO3- + OH- HCO3- + H+ ? H2CO3 HCO3- + 3OH ? H2O + H2CO3 When a substance acts as both an acid and a base such as HCO3- does, it is called amphoteric. How do buffer solutions work? A buffer solution contains a weak acid and base, which removes any hydrogen ions or hydroxide ions that are add – keeping the pH constant. (not neutral) The Bicarbonate Buffer: weak acid weak base H2CO3 + H2O ?? HCO3- + OH- Another Buffer solution Ammonia is a weak base, and the position of this equilibrium will be well to the left: Any added hydrogen ions will react with the ammonia to make more of the conjugate acid, maintaining the pH. Another Buffer solutions Any added OH- ions will react with the weak acid to make the conjugate base and again maintain a constant pH. Determination of the pH of a Buffer solution Example: What is the pH of a solution of 0.11 M NaC2H3O2 and 0.090 M HC2H3O2? (ka = 1.8 x 10-5) HC2H3O2 ?? C2H3O2- + H+ Initial 0.09 0.11 0 Change -x +X +x Equil. .09-x 0.11 +X x ka = x (0.11) /.09-x X = 1.47 x 10-5, pH = 4.83 (buffered solution will have a very small dissociation so + or – x can be ignored) Preparing Buffer Solutions The Henderson-Hasselbalch Equation gives the ratio of weak acid to base needed to maintain a constant pH. pH = pka + log [A-]/[HA] (The pH can be approximated: pH = pka +1 Preparing Buffers Example. What is the mole ratio of acetic acid to acetate ion needed to prepare a buffer solution at a pH = 5.00? ka = 1.8 x 10-5 ka = [H+] [A-] = [H+] mol C2H3O2- [HA] mol HC2H3O pH = 5 so, [H+] = 1 x 10-5 mol C2H3O2- = [H+] = 1x 10-5 = .56 mol HC2H3O ka 1.8 x 10-5 Prepari

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