1. Applications of aqueous equilibria Neutralization Common-Ion effect Buffers Titration curves Solubility and K sp

2. Learning objectives  Determine extent of neutralization  Construct buffer solutions  Derive Henderson-Hasselbalch equation  Apply HH to calculate pH of buffer solutions  Calculate pH titration curves  Write solubility product expressions  Identify factors that affect solubility

3. Neutralization  Acid + Base → Salt + Water  Extent → depends on type of acid and base  What are the major species left when base neutralizes an acid?  Four combinations: Strong-strong Strong-strong Strong-weak Strong-weak Weak-strong Weak-strong Weak-weak Weak-weak

4. Strong-strong  Net ionic equation  K n = 1/K w = 10 14  Neutralization goes to completion  Na +, Cl - and H 2 O

5. Manipulating equilibria expressions  If reaction A can be written as the sum of reactions B + C + D +...+ N (ΣR i )  Then K is product over ΠK i  This approach is used routinely in solution equilibria problems

6. Weak-strong  Acetic acid is not completely ionized  Net ionic equation:  What is K? Write K in terms equilbria we know:  K = 1.8x10 -5 x1.0x10 14 = 1.8x10 9  Goes to completion – attraction of OH - for protons  Na +, Ac -, H 2 O (v. small amount OH - )

7. Strong-weak  Net ionic equation for neutralization of ammonia with HCl  What is K? Write in terms of known equilibria:  Add 1 and 2:  K 1+2 = K 1 K 2 = 1.8x10 -5 x1.0x10 14 = 1.8x10 9  K 1+2 >>1 - Neutralization complete  NH 4 +, Cl -, H 2 O and v. small amount of H 3 O +

8. Weak-weak  Net ionic equation for neutralization of acetic acid by ammonia – neither ionized  Obtain K n from equilibria we know  K 1+2+3 = K 1 K 2 K 3  = 1.8x10 -5 x1.8x10 -5 x1.0x10 14 = 3.2x10 4  NH 4 +, Ac -, H 2 O (v. small amounts of HAc and NH 3 )

9. Common ion effect - buffering  Solutions of weak acid and conjugate bases have important applications for “buffering” pH – resisting change to pH from added acid or base. (weak base and conjugate acid perform the same function)  The pH of operation will depend on the dissociation constants for the particular acid (base).

10. Use weak acid strategy for calculating pH in buffer  Acetic acid and sodium acetate 0.1 M.  Initial species are HAc, Na +, Ac - and H 2 O  Two proton exchange reactions, but one does not alter concentrations

11. The Big Table of concentrations  Determination of final concentrations in terms of initial concentrations  Note difference between acid case and buffer case: [Ac - ] i is 0 with HAc only [Ac - ] i is 0 with HAc only [Ac - ] i is > 0 in the buffer [Ac - ] i is > 0 in the buffer Principal process HAc(aq) + H 2 O = H 3 O + (aq) + Ac - (aq) Initial conc 0.1000.10 Change-xxx Equilibrium conc 0.10 - x x 0.10 + x

12. Solving for x  Put concentrations into expression for K a  0.10 – x ≈ 0.10 ≈ 0.10 + x  x = K a = 1.8 x10 -5 M (life is good)  pH = 4.74  Note: when [HAc] = [Ac - ], pH = pKa  In all buffers pH = pK a when [HB] = [B - ] X << 0.1

13. Common-ion effect – Le Chatelier in action  Without added acetate ion the pH of 0.10 M acetic acid is 2.89.  Addition of Ac - causes [H 3 O + ] to decrease  Consequence of Le Chatelier: increasing [product] equilibrium goes to reactants  pH changes in different buffers

14. Modulating pH in acetic acid by adding acetate ion

15. Do buffers really work?  Compare effect of adding OH - to solution of given pH 1. Buffer solution 2. Strong acid at same initial pH

16. Adding base to buffer  Analytical proof in HAc/Ac - system  Initial pH = 4.74  What happens when 0.01 mol of NaOH is added to I L Principal process HAc(aq) + H 2 O = H 3 O + (aq) + Ac - (aq) Equilibrium conc/M 0.10 - x x 0.10 + x

17. What happens when 0.01 mol of NaOH is added?  Addition of OH - causes conversion of HAc into Ac - Neutralization process HAc(aq) + OH - = H 2 O (aq) + Ac - (aq) Initial conc 0.100.010.10 Change-0.01-0.010.01 Equilibrium conc 0.0900.11

18.  After addition of OH -, recompute [H 3 O + ]  [H 3 O + ] = 1.8x10 -5 x0.09/0.11 = 1.5x10 -5 M  pH = 4.82 – a change of + 0.08 pH units

19. Same deal when adding acid  Almost all the added H + is converted into HAc  In same way as with base, new [HAc] = 0.11 M and new [Ac - ] = 0.09 M  [H 3 O + ] = 1.8x10 -5 x0.11/0.09 = 2.2 x 10 -5 M  pH = 4.66 – a change of -0.08 pH units

20. Adding OH - to strong acid at pH = 4.74  What is change in pH if added to a solution of HCl at pH 4.74?  [H + ] = 1.8 x 10 -5 M  HCl removes 1.8x10 -5 mol of OH -.  Excess OH - = 0.01 - 1.8x10 -5 mol of OH - ≈ 0.01 mol.  New pH = 12 – a change of 7 pH units In buffer at same initial pH, change was only 0.08 units In buffer at same initial pH, change was only 0.08 units

21. Buffer capacity  A measure of the amount of acid or base that a buffer solution can absorb before a significant change in pH occurs.  Or – the amount of acid or base required to yield a given change in pH.  Messin’ with buffers

22. Henderson-Hasselbalch: Send the pain below  Buffer calculation short-cut  Derivation  Assumes that [H + ] << [HA]

23. Applications  Able to predict percent dissociation of an acid from the difference between pH and pK a  Calculate relative quantities of acid and conjugate base required to achieve a given pH (pK a must be reasonably close to pH)

24. The glass half-neutered  Instead of using a source of weak acid and a conjugate base, take a weak acid and neutralize with strong base to make the conjugate base. (HAc and NaOH).  When half the HAc is neutralized [HAc] = [Ac - ] pKa = pH  Useful buffers can be made in pH range of pKa ± 1 pH units