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

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 !!!

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

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 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 mL of pure water if the same acid is added? pH = -log[H 3 O + ] = -log(1.8 x ) = pH = 4.74 Before acid added! [H 3 O + ] = K a x = 1.8 x x [CH 3 COOH] [CH 3 COO - ](0.25) = 1.8 x 10 -5

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

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 Change Equilibrium Solving for the quantity ionized: Initial Change -x x +x Equilibrium x x x Assuming: x = & x = pH = -log(1.982 x ) = = 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 x (0.262) (0.238) = x 10 -5

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 = 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 Change Equilibrium Assuming: Again, using x as the quantity of acid dissociated we get: our normal assumptions: x = & x = [H 3 O + ] = 1.8 x x = x pH = -log(1.635 x ) = = 4.79 After base is added!

How a Buffer Works By adding the 1.00mL base to mL of pure water we would get a hydroxide ion concentration of: mL mol OH - [OH - ] = = 3.99 x M OH - This calculates out to give a pH of: The hydrogen ion concentration is: [H 3 O + ] = = = x KwKw [OH - ] 1 x x M pH = -log(2.5 6 x ) = = 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

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