1. SCH 4 U 1

2. What are buffers? Buffers are mixtures of conjugate acid- base pairs that allow a solution to resist changes in pH when acids and/or bases are added. 2

3. When are buffers formed? a) A weak acid is combined with its conjugate base OR b) A weak base is combined with its conjugate acid 3

4. Example 1 Write the equation to represent the acidic buffer made by combining acetic acid with sodium acetate. In solution, acetic acid does the following: (1)CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + In the same solution, sodium acetate does the following: (2)CH 3 COO - + H 2 O  CH 3 COOH + OH - *Note: Why is Na + is ignored? Notice equation (2) is essentially the reverse of equation (1)! Because it is a neutral ion! 4

5. (1)CH 3 COOH + H 2 O  CH 3 COO - + H 3 O + (2)CH 3 COO - + H 2 O  CH 3 COOH + OH - This means that in a buffer the following overall reaction takes place: CH 3 COOH + OH -  CH 3 COO - + H 3 O + OR Weak Acid + OH -  Conjugate Base + H 3 O + This general pattern can be used for any buffers. 5

6. Weak Acid + OH-  Conjugate Base + H 3 O + Using the equation above, we can understand how a buffer works. Consider Le Chatelier’s Principle and determine what will happen in the following situations: A strong acid is added? The equation shifts LEFT to use up the added acid and a new equilibrium is established A strong base is added? The equation shifts RIGHT to use up the added base and a new equilibrium is established 6

7. Buffer Equations The following can be used to determine ion concentrations in buffers: ACID BUFFER[H 3 O + ] = K a x n A n CB Where n A is moles of weak acid and n CB is moles of conj. base BASE BUFFER[OH - ] = K b x n B n CA Where n B is moles of weak base and n CA is moles of conj. acid 7

8. Example 2 A 1.00L sample of an aqueous solution contains 0.200 mol of acetic acid and 0.100 mol of sodium acetate. Calculate: a) The pH of the solution [H 3 O + ] = K a x n A n CB = 1.8 x 10 -5 (0.200 mol/0.100 mol) = 3.6 x 10 -5 mol/L pH = - log[H 3 O + ] pH = 4.444 8 The solution is made up of a weak acid and its conjugate therefore it must be an acidic buffer!

9. Example 2 cont’d b)The pH of the solution after the addition of 1.00 mL of a 12 M HCl solution First create a chemical equation to represent the buffer...... Then we can determine what is happening to offset the addition of the acid! 9 WEAK ACID + CH 3 COOH + OH - CONJ BASE CH 3 COO - + H 3 O + I0.200 mol 0.100 mol0.012 mol C E Note: Any acid added will be used up when equilibrium shifts left! 12 mol/L x 0.001 L + 0.012 mol- 0.012 mol 0.212 mol 0.088 mol0 mol

10. Example 2 cont’d Now find the new pH using our buffer equation... [H 3 O + ] = K a x n A n CB = 1.8 x 10 -5 (0.212 mol/0.088 mol) = 4.3 x 10 -5 mol/L pH = - log[H 3 O + ] pH = 4.367 After the addition of HCl the pH shifted slightly and became more acidic, but the shift is only 0.077 pH units. 10 ↑ very small!

11. Homework Text page 620 #3 – 9 Read section 8.5 if you need clarification 11