1. Buffer solution دکتر امید رجبی دانشیار گروه شیمی دارویی شیمی عمومی

2. Buffer solutions are solutions that resist change in Hydronium ion and the hydroxide ion concentration (and consequently pH) upon addition of small amounts of acid or base, or upon dilution.solutionsHydroniumhydroxide pHacidbase Buffer solutions are solutions that resist change in Hydronium ion and the hydroxide ion concentration (and consequently pH) upon addition of small amounts of acid or base, or upon dilution.solutionsHydroniumhydroxide pHacidbase

3. When an acid or base is added to water, the pH changes drastically.

4. A buffer solution resists a change in pH when an acid or base is added. A buffer solution resists a change in pH when an acid or base is added. Buffers

5. A buffer solution Contains a combination of acid- base conjugate pairs. Contains a weak acid and a salt of the conjugate base of that acid. Typically has equal concentrations of a weak acid and its salt. May also contain a weak base and a salt with the conjugate acid. A buffer solution Contains a combination of acid- base conjugate pairs. Contains a weak acid and a salt of the conjugate base of that acid. Typically has equal concentrations of a weak acid and its salt. May also contain a weak base and a salt with the conjugate acid. Components of a Buffer

6. The acetic acid/acetate buffer contains acetic acid (CH 3 COOH) and sodium acetate (CH 3 COONa). The salt produces sodium and acetate ions. CH 3 COONa CH 3 COO - + Na + The salt provides a higher concentration of the conjugate base CH 3 COO - than the weak acid. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + small amount Large amount The acetic acid/acetate buffer contains acetic acid (CH 3 COOH) and sodium acetate (CH 3 COONa). The salt produces sodium and acetate ions. CH 3 COONa CH 3 COO - + Na + The salt provides a higher concentration of the conjugate base CH 3 COO - than the weak acid. CH 3 COOH + H 2 O CH 3 COO - + H 3 O + small amount Large amount Buffer Action

7. The function of the weak acid is to neutralize a base. The acetate ion in the product adds to the available acetate. CH 3 COOH + OH - CH 3 COO - + H 2 O The function of the weak acid is to neutralize a base. The acetate ion in the product adds to the available acetate. CH 3 COOH + OH - CH 3 COO - + H 2 O Function of the Weak Acid

8. The function of the acetate ion CH 3 COO - (conjugate base) is to neutralize H 3 O + from acids. The weak acid product adds to the weak acid available. CH 3 COO - + H 3 O + CH 3 COOH + H 2 O The function of the acetate ion CH 3 COO - (conjugate base) is to neutralize H 3 O + from acids. The weak acid product adds to the weak acid available. CH 3 COO - + H 3 O + CH 3 COOH + H 2 O Function of the Conjugate Base

9. The weak acid in a buffer neutralizes base. The conjugate base in the buffer neutralizes acid. The pH of the solution is maintained. The weak acid in a buffer neutralizes base. The conjugate base in the buffer neutralizes acid. The pH of the solution is maintained. Summary of Buffer Action

10. pH of a Buffer The [H 3 O + ] in the K a expression is used to determine the pH of a buffer. Weak acid + H 2 O H 3 O + + Conjugate base K a = [H 3 O + ][conjugate base] [weak acid] [H 3 O + ] = K a x [weak acid] [conjugate base] pH = -log [H 3 O + ] The [H 3 O + ] in the K a expression is used to determine the pH of a buffer. Weak acid + H 2 O H 3 O + + Conjugate base K a = [H 3 O + ][conjugate base] [weak acid] [H 3 O + ] = K a x [weak acid] [conjugate base] pH = -log [H 3 O + ]

11. Calculation of Buffer pH The weak acid H 2 PO 4 - in a blood buffer H 2 PO 4 - /HPO 4 2- has K a = 6.2 x 10 -8. What is the pH of the buffer if it is 0.20 M in both H 2 PO 4 - and HPO 4 2- ? [H 3 O + ] = K a x [H 2 PO 4 - ] [HPO 4 2- ] [H 3 O + ] = 6.2 x 10 -8 x [0.20 M] = 6.2 x 10 -8 [0.20 M] pH = -log [6.2 x 10 -8 ] = 7.21 The weak acid H 2 PO 4 - in a blood buffer H 2 PO 4 - /HPO 4 2- has K a = 6.2 x 10 -8. What is the pH of the buffer if it is 0.20 M in both H 2 PO 4 - and HPO 4 2- ? [H 3 O + ] = K a x [H 2 PO 4 - ] [HPO 4 2- ] [H 3 O + ] = 6.2 x 10 -8 x [0.20 M] = 6.2 x 10 -8 [0.20 M] pH = -log [6.2 x 10 -8 ] = 7.21

12. The Henderson-Hasselbach Equation: A practical application Let's begin our discussion of the Henderson-Hasselbach Equation with a continuation of the dissociation constant derivation. should be a familiar relationship. If we rearrange our equation to provide the [H 3 O + ] on the left side, we get [H 3 O + ] = K a (HA)/(A-) (Notice that the ionized species of the acid is now in the denominator.)

13. Taking the negative log of both sides gives us easier numeric values to work with: which can be rewritten

14. more familiarly Henderson-Hasselbach Equation