1. Ch. 15: Applications of Aqueous Equilibria 15.1 Common Ion Effect

2. Common Ion Effect  shift in equilibrium position that happens when an ion already involved in the equilibrium is added to the solution  important with polyprotic acids – Why?

3. Example 1  Ex: A solution of HF (K a =7.2x10 -4 ) & NaF NaF  Na + + F - NaF  Na + + F - HF ⇄ H + + F - HF ⇄ H + + F - major species: HF, Na +, F -, H 2 O major species: HF, Na +, F -, H 2 O common ion : F- from NaF and HF common ion : F- from NaF and HF Consider NaF breaking apart first- Why? Consider NaF breaking apart first- Why? shifts HF dissociation back to the left b/c of F- ions added from NaF shifts HF dissociation back to the left b/c of F- ions added from NaF so [H+] is lower than would be predicted so [H+] is lower than would be predicted pH would be higher than expected pH would be higher than expected

4. Example 2  How would the predicted pH and actual pH compare for: NH 4 Cl and NH 3 ? NH 4 Cl  NH 4 + + Cl - NH 4 Cl  NH 4 + + Cl - NH 3 + H 2 O ⇄ NH 4 + +OH - NH 3 + H 2 O ⇄ NH 4 + +OH - shifts to the left so lowers the [OH-] shifts to the left so lowers the [OH-] increases the [H+] increases the [H+] lowers the pH lowers the pH

5. Example 3  [H+] = 2.7x10 -2 M at equilibrium and % dissociation = 2.7% for a solution of 1.0 M HF.  Find the [H+] and % dissociation of HF in a solution of 1.0 M HF (Ka=7.2x10 -4 ) and 1.0 M NaF. major species: HF, Na +, F-, H 2 O major species: HF, Na +, F-, H 2 O HF ⇄ H + + F - HF ⇄ H + + F -

6. Example 3 HF ⇄ H + + F - HF ⇄ H + + F - I1.001.0 C-x+x+x E1.0-xx1.0+x

7. Ch. 15: Applications of Aqueous Equilibria 15.2 Buffers

8. Buffered Solutions  a solution that resists pH change when H+ or OH- is added  made from: weak acid and its alkaline “salt” weak acid and its alkaline “salt” Ex: HC 2 H 3 O 2 and NaC 2 H 3 O 2 OR weak base and its halogen/nitrate “ salt” weak base and its halogen/nitrate “ salt” Ex: NH 3 and NH 4 Cl Ex: NH 3 and NH 4 Cl

9. pH changes when OH- or H+ are added

10. Example 4  A buffered solution contains 0.5 M HC 2 H 3 O 2 (K a =1.8x10 -5 ) and 0.5 M NaC 2 H 3 O 2. Find pH of the solution major species: HC 2 H 3 O 2, Na +, C 2 H 3 O 2 -, H 2 O major species: HC 2 H 3 O 2, Na +, C 2 H 3 O 2 -, H 2 O HC 2 H 3 O 2 ⇄ H + + C 2 H 3 O 2 - HC 2 H 3 O 2 ⇄ H + + C 2 H 3 O 2 - I0.500.5 C-x+x+x E0.5-xx0.5+x

11. Example 4  What would happen to the pH if we added 0.010 mol of solid NaOH was added to 1.0 L of this buffered solution?

12. Example 4  major species: HC 2 H 3 O 2, Na +, OH-, C 2 H 3 O 2 -, H 2 O  HC 2 H 3 O 2 + OH -  C 2 H 3 O 2 - + H 2 O goes to completion b/c of SB (strong base) goes to completion b/c of SB (strong base) HC 2 H 3 O 2 + OH -  C 2 H 3 O 2 - + H 2 O HC 2 H 3 O 2 + OH -  C 2 H 3 O 2 - + H 2 O B (1.0L)(0.050M) =0.50 mol 0.010 mol (1.0L)(0.050M) =0.50 mol C -x=-0.010 mol +x=+0.010 mol +x=0.010 mol A 0.49 mol 0 0.51 mol

13. Example 4  Now that we know the amount of acetic acid left, we must find out how much H + is created from it. HC 2 H 3 O 2 ⇄ H + + C 2 H 3 O 2 - HC 2 H 3 O 2 ⇄ H + + C 2 H 3 O 2 - I0.4900.51 C-x+x+x E0.49-xx0.51+x

14. How does a buffer maintain the same pH?  if it is a WA (weak Acid) and its salt it contains large amounts of weak acid, HA, and its conjugate base, A -. it contains large amounts of weak acid, HA, and its conjugate base, A -. when OH - is added, the HA donates H + to it so the OH - does not buildup when OH - is added, the HA donates H + to it so the OH - does not buildup when H + is added, A - accepts the H + so H+ does not buildup either when H + is added, A - accepts the H + so H+ does not buildup either as long as the change is small compared to the original conc, pH won’t change much as long as the change is small compared to the original conc, pH won’t change much

15. How does a buffer maintain the same pH?  if it is a WB (weak base) and its salt it contains large amounts of weak base, B, and its conjugate base, BH +. it contains large amounts of weak base, B, and its conjugate base, BH +. when OH - is added, the BH + donates H + to it so the OH - does not buildup when OH - is added, the BH + donates H + to it so the OH - does not buildup when H + is added, B accepts the H + so H + does not buildup either when H + is added, B accepts the H + so H + does not buildup either as long as the change is small compared to the original conc, pH won’t change much as long as the change is small compared to the original conc, pH won’t change much

16. How does it do that?

17. Henderson-Hasselbalch equation  useful for calculating the pH of a solution when the concentrations of acid and base are known and the change (x) is negligible  for a certain acid-base pair, if they have the same ratio, they have the same pH

18. Example 5  Find the pH of 0.75 M HC 3 H 5 O 3 (Ka=1.4x10 -4 ) and 0.25 M NaC 3 H 5 O 3 major species: HC 3 H 5 O 3, Na +, C 3 H 5 O 3 -, H 2 O major species: HC 3 H 5 O 3, Na +, C 3 H 5 O 3 -, H 2 O

19. Neutralizing Buffers Buffer containing HX & X- X- + H+  HX + H 2 O Recalculate [HX] & [X-] & pH HX + OH-  X- + H2O