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# ln11s08 - Lecture 11 Solving buffers and neutralization...

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Lecture 11 Solving buffer s and neutralization problems Deriving the Buffer Equation You have had it hammered into your head that there are three simplified equations that can be used to calculate proton or hydroxide concentration in acid-base equilibria: ! the strong acid and base equation ! the weak acid and base equation ! the buffer equation You have seen the strong and weak cases, now it is time to derive the buffer equation. But first, what is a buffer? Very simply, a buffer is the mixture of a weak acid and conjugate base or a weak base and conjugate acid Recall that the word conjugate means that the pair differs by a single proton. For example: acid base Buffer? Example 1 Acetic acid HC 2 H 3 O 2 Sodium acetate NaC 2 H 3 O 2 Yes, weak acid and base differ by one proton Example 2 Ammonium ion NH 4 + ammonia NH 3 Yes, weak acid and base differ by one proton Example 3 Hydrochloric acid HCl Chloride Cl - No, strong acids and their cases can’t be buffers Example 4 Carbonic acid H 2 CO 3 Carbonate ion CO 3 = No, the acid and base differ by two protons H+=Ka(Ca/Cb) OH-=Kb(Cb/Ca)

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Derivation of the buffer equation using the RICE expression We start with the general case of a acid equilibrium HA ! H + + A - with K a = [H + ] [A - ]/[HA] and then construct a RICE expression with the following steps: 1. Start with C a for the amount weak acid HA and C b for the amount of the conjugate weak base 2. We assume 0M is the amount of strong acid, H + , to start 3. Since Q< K the reaction shifts right so we add signs to the change array accordingly 4. Assume that x is the concentration of protons that forms 5. Doing the math, at equilibrium the concentrations are C a -x, x and C b +x respectively Reaction: HA ! H+ + A- ____ ___ Substituting these amounts into the equilibrium expression we get: K a = [x] [C b + x]/[ C a -x] But now we make an important approximation.
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ln11s08 - Lecture 11 Solving buffers and neutralization...

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