1 A2 Chemistry: F325 Equilibria, Energetics and Elements 5.1.3 Acids, Bases and Buffers. Lesson 6 Buffer Solutions. Learning Outcomes: All Describe what is meant by the term buffer solution (5.1.3 k) Explain what a buffer solution is made from and give an example (5.1.3 l) Most Explain the role of the conjugate base acid pair in an acid buffer solution (5.1.3 m) Some Calculate the ph of a buffer solution from the K a value and the concentrations of the conjugate acid-base pair. (5.1.3 n) Explain the role of carbonic acid hydrogencarbonate as a buffer in the control of blood ph (5.1.3 o) Buffer Solutions. A buffer solution resists change in ph during the addition of acid or alkali. It minimises the change in ph but cannot completely prevent the ph change of the solution. There are two types of buffer solutions; Acidic Buffer Alkaline buffer weak acid and a conjugate-base e.g. ethanoic acid + sodium ethanoate CH 3COOH(aq) + Na(s) CH 3COONa(aq) + H + (aq) weak base and conjugate-acid e.g. ammonia + ammonium chloride NH 3(aq) + HCl(aq) NH 4 +(aq) + Cl - (aq) In this course we will only consider acid buffers solutions which are composed of: A weak acid (HA) A conjugate-base (A - ) An example of an acidic buffer solution (weak acid and a conjugate-base) is ethanoic acid (CH 3COOH) and sodium ethanoate (CH 3COONa). It is essential to have a weak acid for an equilibrium to be present so that ions can be removed and produced, and the dissociation must be small. In the buffer solution of CH 3COOH/CH 3COO - when acid is added the extra H + ions added react with the conjugate base present, A -, to produce undissociated acid. CH 3COOH(aq) CH 3COO - (aq) + H + (aq) Applying Le Chatelier s Principle, the equilibrium moves to the left to minimise the effect of the increase in the H + (aq) concentration. Since most of the H + (aq) added is converted into HA, the ph hardly decreases.
2 The Role of the Weak Acid Conjugate Base in Controlling ph. The weak acid and the conjugate base in a buffer solution are responsible for controlling the ph. In the equilibrium: The weak acid (HA) removes any alkali The conjugate base (A - ) removes the acid When an acid is added to a buffer solution the concentration of H + (aq) ions increases. The conjugate-base will then react with the H + ions shifting the equilibrium to the left to remove the H + ions that were added restoring the equilibrium. Equilibrium shifts to the left to remove the increase in H + ions. When an alkali (OH - ) is added to a buffer solution the concentration of OH - (aq) ions in solution will increase. In order to reduce this increase the H + ions will react with the OH - ions. H + (aq) + OH - (aq) H 2O(l) Reduced the concentration of hydroxide ions is achieved by HA dissociating more, which shifts the equilibrium to the right to increase the amount of H + ions needed which then react with the OH - (aq) ions. Equilibrium shifts to the right to increase the concentration of H + ions. Calculating the ph of a Buffer Solution. The ph of a buffer solution depends on: The acid dissociation constant K a The concentration ratio of the weak and its conjugate base Temperature There are two methods for calculating the ph of a buffer solution. Method 1 Using the K a expression Rearranging this equation gives: K a = [A - ] [H + ] [HA] [H + ] = K a x [HA] [A - ]
3 Once [H + ] has been deduced, ph can be calculated using the ph equation: ph = - log [H + ] Example Calculations. Example 1. Example 2. Example calculations taken from Knock Hardy Buffer Solutions.
4 Method 2 Hendeson-Hasselbalch Relationship This method uses the Hendeson-Hasselbalch relationship: ph = pk a + log [A-] [HA] Where pka = - log [K a ] Use the Hendeson-Hasselbalch relationship to calculate the ph of the two examples above, showing your full working. Buffers are vital to biological systems in order to maintain life. An example of these is in the blood which normally has a ph of 7.40, but if this value drops to 7.35 a condition called acidosis is produced (high acidity of the blood) and a rise to ph 7.45 causes alkalosis, both of which can result in unconsciousness or even a coma. Carbon dioxide is produced during respiration sets up an equilibrium reaction with water to produce carbonic acid. CO 2(g) + H 2O(l) H 2CO 3(aq) Carbonic acid acts as a weak acid and partially dissociates to hydrogen ions and hydrogencarbonate ions H 2CO 3(aq) Weak Acid H + (aq) + HCO 3 -(aq) Conjugate base Equilibrium shifting to the left to use up the added H + ions An increases in the amount of hydrogen ions (H + ) will shift the equilibrium to the left as they are used up by the hydrogencarbonate ions (HCO 3 -) ions.
5 Useful Links: Chem Guide Khan Academy Example calculations http://goo.gl/c6j2s http://goo.gl/jzuyv http://goo.gl/q5qbh Calculate the ph of a buffer containing 8.2g of sodium ethanoate in 100 cm 3 of 0.25M ethanoic acid (K a = 1.77 x 10-5 mol dm -3 ). Questions. 1. What is a buffer solution? 2. What is an acidic buffer solution composed of? 3. What is the Hendeson-Hasselbalch relationship and what is its purpose? 4. What does the ph of a buffer solution depend upon? 5. Explain how a buffer solution resists change in ph using an equilibrium equation.
6 Tasks: 1. Read through pages 150-153 of the A2 textbook. Answer the questions on pages 151 and 153. 2. Complete the questions form the Acids, Bases and Buffer II questions sheet. 3. Complete the Questions on Acids, bases and Buffers 4 worksheet. Summary: In no more than 150 words summarise the key points from this lesson on buffer solutions. Peer review comment: Additional Notes/Questions: Include any questions that you have on this section for your teacher to cover along with any additional notes you wish to make to support your revision.