1. Acid-base equilibria & common ions Consider solution containing HF (weak acid) and salt NaF What effect does presence of NaF have on dissociation equilibrium of HF? HF(aq)  H + (aq) + F - (aq) 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. Buffered solutions Buffer: A solution that resists a change in its pH when either H + or OH - ions are added Buffers can contain weak acids & salts or weak bases & salts (solutions can be buffered at almost any pH)

4. 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)

5. 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 The essence of buffering: [HA] and [A - ] are very large relative to [OH-] added.

6. The essence of buffering

7. 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)

8. 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.

9. How to handle these problems

10. 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).

11. Buffer Capacity Amount of H + /OH - a buffer can absorb without a considerable change in pH 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….

12. 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….

13. 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

14. 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?

15. Titrations / pH curves

16. Titration Calculations

17. 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)

18. 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?