1. Mullis1 Common Ion Effect and Buffers Ch. 17 in Brown LeMay

2. Mullis2 A buffer is a substance or combination of substances capable of neutralizing limited quantities of either acids or bases.  Why is the pH of some lakes unaffected by acid rain even when they are downwind of big polluters?  The lakes are surrounded by soils which neutralize the acidic precipitation. 1.One way to neutralize acid is to add a base. Limestone (CaCO 3 ) is a weak base. 2.Another way to neutralize either an acid or a base is to add a buffer.

3. Mullis3 How Buffers Work  A buffer has two components:  A weak acid AND a soluble salt of that acid: HA H + + A - and LiA  Li + + A -  Any extra H 3 O + will be neutralized by the A - in the buffer. H 3 O + + A- HA + H 2 O  Any extra OH - that will be neutralized by the acid. HA + OH - A - + H 2 O

4. Mullis4 Buffer Solutions  Resist change in pH  Often made with a weak acid (or base) and its salt  Problems are worked with equilibrium expression and RICE table.  The weak acid keeps addition of a base from changing H + concentration.  The conjugate base ensures that adding an acid will keep original [H + ] close to original- -instead of increasing [H + ] the added base reacts with the conjugate acid.

5. Mullis5 What Mixtures Can Be Buffers?  A chemical buffer consists of a solution of a conjugate acid-base pair with both species present in solution at concentrations to accomplish the neutralization of added species in order to maintain pH of the solution at its original level.  Most buffers are a combination of a weak acid or base and a salt made with its conjugate: Examples are: HCOOH and NaCOOH NH 3 and NH 4 Cl

6. Mullis6 Can a buffer be made with a strong acid or strong base?  Yes, but it is difficult.  Since the presence of both the acid and the base is necessary for the buffer is to be able to deal with both addition and subtraction of hydrogen ions, a buffer could be made using a strong acid and salt in sufficient concentration for its conjugate to produce OH - ions to neutralize added acid.  HI + NaI in water becomes H + + I - + Na + + I -  And--I - + H 2 O  HI + OH -  Although this would be an acidic buffer, small amounts of H + could be added without changing the original pH.

7. Mullis7 Buffers The lake: CaCO 3 + H + Ca 2+ + HCO 3 - + H 2 O The polyatomic ion HCO 3 - acts as a buffer: HCO 3 - + H + 2H 2 O + CO 2 HCO 3 - + OH - H 2 O + CO 3 2- Another buffer is the hydrogen phosphate ion. Possible sources of this ion are Na 2 HPO 4 and NaH 2 PO 4. H 2 PO 4 - HPO 4 2- + H +

8. Mullis8 How Buffers in the Blood Work  Blood must be maintained at pH = 7.35.  This pH is maintained by a buffer system based on H 2 CO 3.  The blood buffer is made up from the dissolved carbon dioxide in the plasma.  CO 2 (g) + H 2 O H 2 CO 3 and H 2 CO 3 HCO 3 - + H 3 O +  When a base is added it reacts with the carbonic acid. OH- + H 3 CO 3 HCO 3 - + H 2 O  When an acid is added it reacts with the bicarbonate ion. H 3 O + + HCO 3 - H 2 CO 3 + H 2 O

9. Mullis9 Buffered solutions contain relatively large portions of weak acid and the corresponding weak base.  Can be HA and A - or B and BH +  When H + added to buffer, it reacts ~ to completion with the weak base: H + + A -  HA or H + + B  BH +  When OH - added to buffer, it reacts ~ to completion with the weak acid: H + + A -  HA or H + + B  BH +  pH in buffered solution is determined by the ratio of the concentrations of weak acid/weak base. pH will be ~ constant as long as: 1. Buffer concentrations are large compared to concentrations of added H + or OH -. 2. The ratio of weak acid/weak base remains constant.

10. Mullis10 Common Ion Effect/Buffer solution: Find pH of 0.50M HC 2 H 3 O 2 and 0.50M NaC 2 H 3 O 2 in solution. K a acetic acid = 1.8 x 10 -5 RHC 2 H 3 O 2 H + +C2H3O2 -C2H3O2 - I0.50 M0 C-x+x E0.50 – xx0.50 + x Major species in solution: HC 2 H 3 O 2, Na +, C 2 H 3 O 2 -, H 2 O Dissociation of HC 2 H 3 O 2 will control pH. 1.8 x 10 -5 = K a = x(.50 + x) = x(.50) =1.8 x 10 -5 (.50 – x) (.50) x = [H + ] = 1.8 x 10 -5 M(< 5% of 0.50 M) pH = -log[1.8 x 10 -5 ] = 4.74

11. Mullis11 Henderson-Hasselbalch Equation H + + A - HA K a = [H + ][A - ][H + ] = [HA](K a ) [HA] [A - ] -log[H + ] = -logK a – log([HA]/[A - ] pH = pK a + log([A - ]/[HA] pH = pK a + log([base]/[acid] ]

12. Mullis12 Practice problems 1. What is the pH of a solution after 0.1 mole of HCl is added to a buffer which contains 0.2 mole of HF and.3 mole of NaF in one liter? 2.Do problem 1 but change the HCl to 0.4 mole. 3.Do problem 1 but change HCl to 0.3 mole. (1) 2.97, (2) 1, (3) 1.7