1. Buffers 4/13/2011

2. Definitions Weak Acids: all proton donors that are in equilibrium Conjugate Bases: the ions that are left over after a weak acid loses it H + ions (protons) Buffer: – a solution containing a mix of weak acid and the salt of its conjugate base – A solution that can absorb added acids and bases w/o a large pH change

3. The weak acid rxn HA H + + A - In a pure acid, the [H + ] and [A - ] occur in equal amounts In any solution, addition of H + will decrease A - and vice versa The conjugate base of HA is A - Notice that if [A - ] goes up, the H + will go down. The solution will become more basic.

4. Salt of a conjugate base A “salt” is defined as the product of an acid and a base A salt of a conjugate base is the salt produced when a weak acid is neutralized Examples: Weak acidSalt of Conjugate Base HF  NaF HNO 2  LiNO 2 CH 3 COOH  KCH3CO 3 H 2 SO 3  NaHSO 3 H 2 CO 3  LiHCO 3 To identify the salt: 1.Take any weak acid 2.Remove the H + 3.Replace with any other positive ion Na AH - Weak Acid Conjugate Base Salt of Conjugate Base

5. Non-Buffered Solutions If a strong acid is added to pure water, the pH changes radically Original pH = 7 Imagine 1 mL of 1 Molar HCl (1 x 10 -3 moles) added to 1 Liter of water 1 x 10 -3 HCl [H + ]= 1x10 -3 Cl - (spectator) pH 7 pH 3

6. Buffered Solutions Add roughly equal amounts of weak acid and conjugate base salt An equilibrium is set up  And then altered  HA NaA HA H + A - Na + A - HA  H + + A - H + + A - (The addition of the A - lowers the H + )

7. Buffers in Action The Buffer solution now has both an acid and a base in it. Addition of an acid will add more H +, which will be absorbed (neutralized) by the base. If a base is added, it will react with the H +, shifting the equilibrium to the right, reducing HA and increasing A - HA H + A - Na + A - HA  H + + A - Acid H+H+ Base OH - H2OH2O

8. Buffer Calculations Imagine 1 liter of a buffer with 0.5 moles HA and 0.5 moles NaA Pretend (for now) that Ka = 1 x 10 -7 Ka = [H+] [A-] = (X )[0.5] [HA] [0.5] So…with equal moles of weak acid and conjugate base, H+ = Ka = 1 x 10 -7 And pH = 7 Like before, add 1 mL of 1 molar HCl. (you are adding.001 moles acid) The equilibrium will change… And so will the pH… Ka = [H+] [A-] = (X )[0.499] [HA] [0.501] The H+ has increased and = 1.004 x 10 -7 And pH = 6.998 HA  H + + A - H + + A - Original 0.5 mol 1 x 10 -7 0.5 mol Add Acid + 0.001 mols Increase HA + 0.001 - 0.001 Decrease A- New Values 0.501 mol X 0.499 mol pH 7 pH 6.998

9. Practice calculations What would be the pH of a solution of 0.2 M HF and 0.2 M NaF if the K a = 1.6 x 10 -5 ? What would be the pH if the above solution was diluted by ½ (so each solute is 0.1 M)? The K a for Carbonic acid (H 2 CO 3 ) is 4.5 x 10 -7 (only the first H + is ionzed). If the pH of a solution is 7.4, and the concentration of HCO 3 1- is 0.22 M, what is the concentration of H 2 CO 3 ?

10. Buffers in the real world Buffers are important in the biological world. – (imagine what could happen to a fish tank if the water was not buffered) Your blood is strongly buffered to control the amount of CO 2 in the blood. – CO 2(aq) + H 2 O (l)  H 2 CO 3(aq)  H + (aq) + HCO 3 - (aq) – Your blood measure pH instead of CO 2(aq) – If your pH gets too low, your breathing rate goes up

11. Summary of Buffers They are made of weak acids and conjugate bases, and are in equilibrium with H + Because they contain both base and acid, they can absorb added bases and acids with only a very small change in pH Calculations of acidity can be made if the Ka, acid and base concentrations are known.