1. Acid-base equilibria & common ions Consider solution containing HF (weak acid) and salt NaF Species in solution: HF, H 2 O, Na +, F - What effect does presence of NaF have on dissociation equilibrium of HF? HF(aq)  H + (aq) + F - (aq) Use Le Chatelier’s principle; extra F - (from NaF) shifts equilibrium to the ….. Because of this, the [H+] will ….. What happens to the pH? This is the Common Ion Effect

2. Equilibrium calculations involving common ions 34.6 g of NH 4 Cl is added to 3.98 L of a 0.0145 M solution of NH 3. K b (NH 3 ) = 1.8 x 10 -5. What is the pH of the original solution before the addition of NH 4 Cl? NH 3 (aq) + H 2 O(l)  NH 4 + (aq) + OH - (aq) So – what’s the pH of the solution after the addition of the NH 4 Cl (assume that the volume stays constant)?

3. Effect of adding common ions to weak acids? What is the general statement that we can make about what happens to the acidity of a weak acid upon addition of a common ion?

4. Buffered solutions Buffer: A solution that resists a change in its pH when either H + or OH - ions are added Very important example; blood (cells can only survive in very narrow pH range, thus constant blood pH is vital.) Buffers can contain weak acids & salts or weak bases & salts (solutions can be buffered at almost any pH) Strong acids / bases cannot be used… But….just how exactly do buffers work?

5. How does buffering work? Buffer contains relatively large concs. of weak acid HA and its conjugate base A -. Add OH - ions into solution, what happens? (OH - is a strong base – it will look for H + ) Weak acid HA is best source of H +. OH - + HA  H 2 O + A - OH - ions cannot accumulate (replaced by A - ions)

6. How does buffering work? If OH - ions converted to A - ions, how does the pH stay so stable? Look at eq. expression for [H + ]; dissociation equilibrium for HA Eq. [H + ] (and thus pH) determined by ratio [HA] / [A - ]. If [HA] and [A - ] are large relative to [OH-] added, change in [HA] / [A - ] will be very small. Thus, [H + ] and pH changes will also be very small. The essence of buffering: [HA] and [A - ] are very large relative to [OH-] added.

7. The essence of buffering

8. Buffer solution pH calculation Buffered solution contains 0.50 M HC 2 H 3 O 2 (K a = 1.8x10 -5 ) and 0.50 M NaC 2 H 3 O 2. Calculate pH of solution. HC 2 H 3 O 2 (aq)  C 2 H 3 O 2 - (aq) + H + (aq)

9. pH changes in buffered solutions Calculate change in pH that occurs when 0.010 mol solid NaOH added to 1.0 L of buffered solution from last question.Compare this with pH change that occurs when 0.010 mol solid NaOH added to 1.0 L of H 2 O. Expected: pH of buffered solution will change very little; pH of water will change by much larger amount.

10. How to handle these problems

11. Adding H +, rather than OH - Exactly the same thinking applied when H + added to buffered solution of weak acid and conjugate base salt. A - (conjugate base) has high affinity for added H +, and will form weak acid HA; H + + A -  HA H + ions cannot accumulate (replaced by HA) Net change of A - to HA, but if [A - ] and [HA] are very large compared to [H + ], little pH change will occur (as expected, in buffered solution).

12. Buffer Capacity Amount of H + /OH - a buffer can absorb without a considerable change in pH Consider two 1.0 L ammonia/ammonium buffer solutions: Buffer 1:1.0 M NH 3, and 1.0 M NH 4 + K a = [NH 3 ][H + ] / [NH 4 + ] [H + ] = K a. [NH 4 + ] / [NH 3 ] = 5.6 x 10 -10 M; pH = ? What happens to the pH if 0.1 moles of NaOH is added? After the reaction with NaOH, calculate the [H + ] and thus the pH….

13. Buffer Capacity Buffer 2:0.01 M NH 3, and 0.01 M NH 4 + K a = [NH 3 ][H + ] / [NH 4 + ] [H + ] = K a. [NH 4 + ] / [NH 3 ] = 5.6 x 10 -10 M; pH = ? What happens to the pH if 0.1 moles of NaOH is added? After the reaction with NaOH, calculate the [H + ] and thus the pH….

14. Buffer Capacity So - what general statement can be made about buffer capacity with respect to the initial concentration of the weak acid and its conjugate base in the buffer? High concentrations: Low concentrations

15. Henderson-Hasselbalch equation Useful equation; allows for the calculation of buffer pH if the concentration of weak acid (HA) and conjugate base (A - )are known. A more convenient method of pH calculation – will give the same answer as with our previous method of pH calculation. Buffered solution contains 0.50 M HC 2 H 3 O 2 (K a = 1.8x10 -5 ) and 0.50 M NaC 2 H 3 O 2. Calculate pH of solution (we did this a few slides ago, but let’s check that H-H equation gives us the same answer). Henderson-Hasselbalch equation

16. Titrations / pH curves Experimental method to determine the concentration of an acid (or base). Think about this: If you were given 25 mL of a unknown concentration of HCl, a solution of 0.100 M NaOH and a burette, how could you determine the concentration of the HCl? What would happen to the pH of the solution of HCl as the NaOH was added to it?

17. Titrations / pH curves

18. Titrations How do you know when all the acid has been neutralized? pH paper pH meter Or…… Indicator can be used Methyl orange; yellow in basic solution and red in acidic solution Many other indicators available

19. Indicators Bromothymol blue (yellow in acidic solution), slight green tint as base is added, and blue form in basic solution

20. Titration Calculations

21. Solubility Equilibria Fluoride – used to combat tooth decay One product of F - reaction at site of teeth is CaF 2 (s) CaF 2 (s)  Ca 2+ (aq) + 2F - (aq) (dissolving in water) Consider equlibrium set up between these species: CaF 2 (s)  Ca 2+ (aq) + 2F - (aq) Equilibrium Expression for this process? K sp = [Ca 2+ ][F - ] 2 (Why is CaF 2 not included in expression?) K sp = Solubility Product Constant (solubility product)

22. Solubility vs K sp K sp : Equilibrium Constant (solubility product) Solubility: Equilibrium Position Copper (I) bromide has a measured aqueous solubility of 2.0 x 10 -4 mol/L at 25 °C. Calculate its K sp value. Equilibrium reaction? Equilibrium Expression? K sp = (equilibrium concentrations) Initial Concentrations?