BUFFERS. Buffers. Buffer Action 5/16/2011

Transcription

1 BUFFERS Acids and Bases Buffers The purpose of a buffer is to maintain a relatively constant ph. Buffers are usually found in most biological systems because drastic changes in ph could cause major problems in the different systems of living organisms. Buffer Action Example: If you add 1 ml of 10 mol/l HCl to neutral saline, you change the ph from 7 to 2 If you add 1 ml of 10 mol/l HCl to blood plasma (contains buffers), you change the ph from 7.4 to 7.2 Shift in ph or ph is only 0.2 1

2 Preparation of Buffers All buffers are mixtures of weak acids and its conjugate base (salt of conjugate base) Or Weak base and salt of conjugate acids. Acidic Buffers An acidic buffer can be prepared by mixing the following compounds Solutions of acetic acid and sodium acetate CH 3 COOH + CH 3 COONa The ph of acidic buffers will always be less than 7 The range of ph of the buffer will depend on the pka value of the acid used Actual values of ph can be determined using Henderson Hasslebach s equation Buffering Action adding OH - ions When OH - ions are added to a buffer it reacts with the acid in the buffer OH - + CH 3 COOH CH 3 COO - + H 2 O The OH - reduces the H 3 O + ion concentration, which causes a shift to the right, forming additional CH 3 COO - ions. For practical purposes each mole of OH - added consumes a mole of CH 3 COOH and produces a mole of CH 3 COO -. 2

3 Buffering Action adding H 3 O + ions When H 3 O + ions are added to a buffer it reacts with the acetate ions in the buffer H 3 O + + CH 3 COO - CH 3 COOH + H 2 O When a strong acid such as HCl is added to the buffer, the hydronium ions react with the CH 3 COO - ions of the salt and form more un-dissociated CH 3 COOH. This effectively does not increase the H 3 O + concentration or lower the ph as otherwise would happen Buffer Capacity As you would expect, there is a limit to the quantity of H + or OH - that a buffer can absorb without undergoing a significant change in ph. If a mole of HCl is added to a litre of buffer solution containing 0.5 moles of sodium acetate/acetic acid buffer the H + completely consumes the buffer and results in a drastic change in ph. Henderson-Hasselbach equation HA = H + + A - Ka = [H + ][A - ] [HA] [H+] = Ka x [HA] now taking log on both sides [A - ] -log[h + ] = -log Ka + (-log[ha]/[a - ]) 3

4 Henderson-Hasselbach equation Factors affecting ph of buffer 1. pk a of the weak acid 2. ratio of the initial molar []'s of the acid and it's salt If we prepare a solution where [anion] = [acid] then the log [anion] = log 1 = 0 [acid] ph = pka of the weak acid ph of a buffer What mostly determines where on a ph scale a buffer can work best is the pk a of the weak acid. By adjusting the ratio of [anion] to [acid], we can cause shifts so that the ph of the buffered solution comes out on one side or the other of this value of ph. Depending on your need you can choose an appropriate weak acid or base and the change in ph is usually ± 1 4

5 Buffer problem 01 A buffer solution is prepared by mixing a 0.11 mol/l CH 3 COONa, sodium acetate, and 0.09 mol/l CH 3 COOH, acetic acid, What is the final ph? Ka for acetic acid is 1.8 x 10-5 Solution problem 01 Calculate the pka of the acid pka = -log 1.8 E-5 pka = 4.74 [acid] = mol/l [anion] = 0.11 mol/l ph = log [0.11] [0.09] = log 1.2 = = 4.82 ph = 4.82 The effectiveness of a buffer Suppose we drop 0.01 moles of strong base into our buffer from the last example. What will be the measured effects? HA + OH - A - + H 2 O start 0.09 M M finish M M 0.08 M 0.0 M 0.12 M 5

6 Effectiveness of buffer contd ph = pk a + log [anion] [acid] = log (0.12) (0.08) = log 1.5 = = 4.92 The change in ph is small compared to what it would have been in pure water! ph = 0.18 Basic Buffers A mixture of ammonia (ammonium hydroxide) and ammonium chloride produces a basic buffer NH 4 OH + NH 4 Cl NH 4 OH is a weak base and NH 4 Cl is the salt of the strong conjugate acid of NH 4 OH ph of this buffer will be greater than 7 Henderson-Hasselbach Basic buffer 6

7 Problem 2 A chemist needs 250 ml of a solution buffered at a ph of How many grams of ammonium chloride have to be added to 250 ml of 0.2 mol/l NH 3 to make such a buffer? (Assumption Volume does not change.) Problem 2 - Solution ph = 9.0 poh = 5.0 [base] = 0.2 pk b of ammonia = 4.74 (from datatable) poh = pk b +log [cation] [base] 5.00 = log [cation] (0.2) 0.26 = log [cation] (0.2) [cation] = X 0.2 = 1.8 * 0.2 = 0.36 mol/l of the NH Problem 2 solution continued But we only need enough for 250 ml so 0.36 mol = x 1000 ml 250 ml x = 0.09 moles of NH 4 +1 ions needed Since the NH 4 +1 comes from NH 4 Cl then we also need 0.09 moles of NH 4 Cl. g = n * Mm = 0.09 moles * 53.5 grams/mole = 4.8 grams of the ammonium salt is required. 7

8 Generalizing Invitro buffers Generalizing Insitu buffers 8