1. BUFFERS Mixture of an acid and its conjugate base. Buffer solution  resists change in pH when acids or bases are added or when dilution occurs. Mix: A moles of weak acid + B moles of conjugate base Find: moles of acid remains close to A, and moles of base remains close to B  Very little reaction HA  H + + A - Le Chatelier’s principle

3. How a Buffer Works Consider adding H 3 O + or OH - to water and also to a buffer For 0.01 mol H 3 O + to 1 L water: [H 3 O + ] = 0.01 mol/1.0 L = 0.01 M pH = -log([H 3 O + ]) = 2.0 So, change in pH from pure water:  pH = 7.00 – 2.00 = 5.0 For the H 2 CO 3 - / HCO 3 - system: pH of buffer = 7.38 Addition of 0.01 mol H 3 O + changes pH to 7.46 So change in pH from buffer:  pH = 7.46 – 7.38 = 0.08 !!!

4. How a Buffer Works Consider a buffer made from acetic acid and sodium acetate: CH 3 COOH (aq) + H 2 O (l) CH 3 COO - (aq) + H 3 O + (aq) K a = or [CH 3 COO - ] [H 3 O + ] [CH 3 COOH] [CH 3 COO - ] [H 3 O + ] = K a x

5. How a Buffer Works Let’s consider a buffer made by placing 0.25 mol of acetic acid and 0.25 mol of sodium acetate per liter of solution. What is the pH of the buffer? And what will be the pH of 100.00 mL of the buffer before and after 1.00 mL of concentrated HCl (12.0 M) is added to the buffer? What will be the pH of 300.00 mL of pure water if the same acid is added? pH = -log[H 3 O + ] = -log(1.8 x 10 -5 ) = pH = 4.74 Before acid added! [H 3 O + ] = K a x = 1.8 x 10 -5 x [CH 3 COOH] [CH 3 COO - ](0.25) = 1.8 x 10 -5

6. How a Buffer Works 1.00 mL conc. HCl 1.00 mL x 12.0 mol/L = 0.012 mol H 3 O + Added to 300.00 mL of water : 0.012 mol H 3 O + 301.00 mL soln. = 0.0399 M H 3 O + pH = -log(0.0399 M) pH = 1.40 Without buffer! What is pH if added to pure water?

7. How a Buffer Works After acid is added to buffer: Conc. (M) CH 3 COOH (aq) + H 2 O (aq) CH 3 COO - + H 3 O + Initial 0.250 ---- 0.250 0 Change +0.012 ---- -0.012 0.012 Equilibrium 0.262 ---- 0.238 0.012 Solving for the quantity ionized: Initial 0.262 ---- 0.238 0 Change -x ---- +x +x Equilibrium 0.262 - x ---- 0.238 + x x Assuming: 0.262 - x = 0.262 & 0.238 + x = 0.238 pH = -log(1.982 x 10 -5 ) = 5.000 - 0.297 = 4.70 After the acid is added! Conc. (M) CH 3 COOH (aq) + H 2 O (aq) CH 3 COO - + H 3 O + [CH 3 COOH] [CH 3 COO - ] [H 3 O + ] = K a x =1.8 x 10 -5 x (0.262) (0.238) = 1.982 x 10 -5

8. How a Buffer Works Suppose we add 1.0 mL of a concentrated base instead of an acid. Add 1.0 mL of 12.0 M NaOH to pure water and our buffer, and let’s see what the impact is: 1.00 mL x 12.0 mol OH - /1000mL = 0.012 mol OH - This will reduce the quantity of acid present and force the equilibrium to produce more hydronium ion to replace that neutralized by the addition of the base! Conc. (M) CH 3 COOH (aq) + H 2 O (aq) CH 3 COO - + H 3 O + Initial 0.250 ---- 0.250 0 Change - 0.012 ---- +0.012 +0.012 Equilibrium 0.238 ---- 0.262 +0.012 Assuming: Again, using x as the quantity of acid dissociated we get: our normal assumptions: 0.262 + x = 0.262 & 0.238 - x = 0.238 [H 3 O + ] = 1.8 x 10 -5 x = 1.635 x 10 -5 0.238 0.262 pH = -log(1.635 x 10 -5 ) = 5.000 - 0.214 = 4.79 After base is added!

9. How a Buffer Works By adding the 1.00mL base to 300.00 mL of pure water we would get a hydroxide ion concentration of: 301.00 mL 0.012 mol OH - [OH - ] = = 3.99 x 10 -5 M OH - This calculates out to give a pH of: The hydrogen ion concentration is: [H 3 O + ] = = = 2.506 x 10 -10 KwKw [OH - ] 1 x 10 -14 3.99 x 10 -5 M pH = -log(2.5 6 x 10 -10 ) = 10.000 - 0.408 = 9.59 With 1.0 mL of the base in pure water! In summary: Buffer alone pH = 4.74 Buffer plus 1.0 mL base pH = 4.79 Base alone, pH = 9.59 Buffer plus 1.0 mL acid pH = 4.70 Acid alone, pH = 1.40

10. Problem: Calculate the pH of a solution containing 0.200 M NH 3 and 0.300 M NH 4 Cl given that the acid dissociation constant for NH 4 + is 5.7x10 -10. NH 3 + H 2 O  NH 4 + + OH - acidbase pK a applies to this acid pK a = 9.244 pH = 9.244 + log (0.200) (0.300) pH = 9.07