Common Ion Effect and Buffers Ch 17 in

Common Ion Effect and Buffers Ch. 17 in Brown Le. May Mullis 1

A buffer is a substance or combination of substances capable of neutralizing limited quantities of either acids or bases. o o Why is the p. H 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. 2. One way to neutralize acid is to add a base. Limestone (Ca. CO 3) is a weak base. Another way to neutralize either an acid or a base is to add a buffer. Mullis 2

How Buffers Work o o A buffer has two components: A weak acid AND a soluble salt of that acid: HA <-> H+ + Aand Li. A Li+ + AAny 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 Mullis 3

Buffer Solutions o o o Resist change in p. H 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. Mullis 4

What Mixtures Can Be Buffers? o o o 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 p. H 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: n HCOOH and Na. COOH n NH 3 and NH 4 Cl But---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 + Na. I in water becomes H+ + I- + Na+ + IAnd--I- + H 2 O HI + OHAlthough this would be an acidic buffer, small amounts of H + could be added without changing the original p. H. Mullis 5

Buffers The lake: Ca. CO 3 + H+ Ca 2+ + HCO 3 - + H 2 O The polyatomic ion HCO 3 - acts as a buffer: HCO 3 - + H+ 2 H 2 O + CO 2 HCO 3 - + OHH 2 O + CO 32 Another buffer is the hydrogen phosphate ion. Possible sources of this ion are Na 2 HPO 4 and Na. H 2 PO 4. H+ + H 2 PO 4 2 H 2 PO 4 - + H 2 O Mullis 6

How Buffers in the Blood Work o o o Blood must be maintained at p. H = 7. 35. This p. H 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 When a base is added it reacts with the carbonic acid. OH- + H 3 CO 3 o HCO 3 - + H 3 O+ 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 Mullis 7

Buffered solutions contain relatively large portions of weak acid and the corresponding weak base. Ø Ø Can be HA and Aor 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+ p. H in buffered solution is determined by the ratio of the concentrations of weak acid/weak base. p. H 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. Mullis 8

Common Ion Effect/Buffer solution: Find p. H of 0. 50 M HC 2 H 3 O 2 and 0. 50 M Na. C 2 H 3 O 2 in solution. Ka acetic acid = 1. 8 x 10 -5 R HC 2 H 3 O 2 I 0. 50 M C -x E 0. 50 – x H+ + C 2 H 3 O 2 - 0 0. 50 M +x +x x 0. 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 p. H. 1. 8 x 10 -5 = Ka = 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) p. H = -log[1. 8 x 10 -5] = 4. 74 Mullis 9

Henderson-Hasselbalch Equation H+ + A HA Ka = [H+][A-] [H+] = [HA](Ka) [HA] [A-] -log[H+] = -log. Ka – log([HA]/[A-] p. H = p. Ka + log([base]/[acid] Mullis ] 10

Practice problems 1. What is the p. H 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 Na. F 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 Mullis 11