1. Acid equilibria and alpha plots Chemistry 321, Summer 2014

2. In this lecture Alpha plots show dynamic changes in species concentration during a titration For weak acids, alpha plots mirror the behavior seen on titration plots and yield further information

3. Formal concentration Given: Let C HA = formal concentration of the acid = [HA] + [A – ] = sum of all forms of the acid at equilibrium C HA = [HA] 0 if only the weak acid is added = [A – ] 0 if only the conjugate base is added = [HA] 0 + [A – ] 0 if both are added

4. Alpha is the mole fraction of a species relative to the solution’s formal concentration So, symbolically,For instance, We can rewrite this like: [HA] = α HA C HA or [A – ] = = α A– C HA The constraint is that Σ α i = 1 For a monoprotic acid, α HA + α A– = 1

5. The equation of the alpha plot Start with the definitions of C HA and α HA :

6. The equation of the alpha plot Start with the definitions of C HA and α HA : Invert the expression:

7. The equation of the alpha plot Start with the definitions of C HA and α HA : Invert the expression: Recalling the equilibrium expression: so

8. The equation of the alpha plot (continued) Re-invert the equation: Similarly, we can derive an expression for α A– : Note that both expressions for alpha depend on [H + ] (and, by extension, pH) only! So this can be plotted.

9. The alpha plot Where the two curves cross (each α = 0.5), the x- coordinate is the pK a of the weak acid. pKa

10. Limiting behavior on the graph For acetic acid, K a = 1.8 × 10 –5. Consider an acetic acid solution at pH 8.74 (meaning [H + ] = 1.8 × 10 –9 ). The point is that even at a really high pH, there is still some undissociated acid left (i.e., not zero). When you are dealing with a polyprotic acid or base, then there are pHs at which some species have zero concentration…this is also important when dealing with ligands.

11. Polyprotic acid alpha plots Polyprotic acid dissociation occurs in a stepwise manner; that is, the different H + ions on each molecule dissociate at different pHs, rather than all at once. Note the definitions of the equilibrium constants K a1 and K a2 N.B.: pK a1 = 6.37, and pK a2 = 10.25

12. Setting up the alpha plot equations for a diprotic acid Let C H2A = [H 2 A] + [HA – ] + [A 2– ] ( = formal concentration of H 2 A) Expanding the previous definition of α i : [H 2 A] = α H2A C H2A and [HA – ] = α HA– C H2A and [A 2– ] = α A2- C H2A Note that α H2A + α HA– + α A2- = 1

13. Setting up the alpha plot equations for a diprotic acid Let C H2A = [H 2 A] + [HA – ] + [A 2– ] ( = formal concentration of H 2 A) Expanding the previous definition of α i : [H 2 A] = α H2A C H2A and [HA – ] = α HA– C H2A and [A 2– ] = α A2- C H2A Note that α H2A + α HA– + α A2- = 1

14. Setting up the alpha plot equations for a diprotic acid Inverting the equation yields:

15. Setting up the alpha plot equations for a diprotic acid Inverting the equation yields: Recall the definitions of K a1 and K a2:

16. Setting up the alpha plot equations for a diprotic acid Inverting the equation yields: Recall the definitions of K a1 and K a2: multiply to get:

17. Setting up the alpha plot equations for a diprotic acid Substitute into the original expression:

18. Setting up the alpha plot equations for a diprotic acid Substitute into the original expression: Put it all over a common denominator: Limiting behavior: at low pH (acidic), the [H + ] 2 term dominates the other terms, so α H2A ≈ 1; at high pH (alkaline), [H + ] 2  0, so α H2A ≈ 0.

19. The other α i expressions (α HA– and α A2– ) are derived similarly. Setting up the alpha plot equations for a diprotic acid

20. The alpha plot for carbonic acid What are the pK a s for carbonic acid?

21. The alpha plot for carbonic acid At pH 4, α CO3 2- ≈ 0, according to the graph

22. The alpha plot for carbonic acid not zero, but basically negligible! At pH 4, the only two species that matter are HCO 3 – and H 2 CO 3

23. The titration curve for carbonic acid reflects the stepwise dissociation behavior

24. Challenge problem Consider the stepwise dissociation of phosphoric acid:

25. Challenge problem (continued) Derive the equations to calculate α i for all phosphate- containing species. Draw the alpha plot for all phosphate-containing species; make sure the axes are labeled and the pK a s are sensible. Calculate the mole fraction (α i ) for all phosphate-containing species at blood pH (7.40). Assume all phosphate-containing species have a soluble sodium salt (i.e., if you need PO 4 3–, you will use Na 3 PO 4 ). Which two salts will you use to create pH 7.40 phosphate buffer?