3.6: ACIDS AND BASES … Workbook pgs … Buffered Solutions…

BUFFER SOLUTIONS Buffer Causes solutions to be resistant to a change in pH when a strong acid or base is added Mixture of a weak acid and salts of conjugate bases If you add mol of NaOH to 1.0 L of pure water the pH drops from 7 to 2 If you add mol of NaOH to 1.0 L of blood (pH of 7.4) the pH will drop to 7.3 Two requirements for a buffer: An acid capable of reacting with added OH - ions and a base that can consume added H 3 O + ions. The acid and base must not react with each other (acetic acid and acetate ion or ammonia and ammunium ions)

BUFFERS If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH − to make F − and water.

BUFFERS: Similarly, if acid is added, the F − reacts with it to form HF and water.

MATH TIME: What is the pH of a 0.700M solution of acetic acid ? K a = 1.8 x pH = 2.45 What is the pH of a 0.600M solution of sodium acetate? K b = 5.6 x pH = 9.26 What is the pH of an acetic acid/sodium acetate buffer with [CH 3 CO 2 H] = 0.700M and [CH 3 CO 2 - ] = 0.600M? K a = 1.8 x pH = 4.68

BUFFER CALCULATIONS: Consider the equilibrium constant expression for the dissociation of a generic acid, HA: HA + H 2 OH 3 O + + A − [H 3 O + ] [A − ] [HA] K a =

© 2009, Prentice-Hall, Inc. BUFFER CALCULATIONS Rearranging slightly, this becomes [A − ] [HA] K a = [H 3 O + ] Taking the negative log of both side, we get [A − ] [HA] −log K a = −log [H 3 O + ] + − log pKapKa pH acid base

© 2009, Prentice-Hall, Inc. BUFFER CALCULATIONS So pK a = pH − log [base] [acid] Rearranging, this becomes pH = pK a + log [base] [acid] This is the Henderson–Hasselbalch equation.

HENDERSON-HASSELBALCH EQUATION: For when the concentrations of the acid and conjugate base are quite large… … or when you want a short cut to a buffer solution

PRACTICE: What is the pH of a buffer that is 0.12 M in lactic acid, CH 3 CH(OH)COOH, and 0.10 M in sodium lactate? K a for lactic acid is 1.4  10 −4. pH = pK a + log [base] [acid] pH = −log (1.4  10 −4 ) + log (0.10) (0.12) pH = (−0.08) = 3.77

PRACTICE: Benzoic acid (C 6 H 5 CO 2 H, 2.00g) and sodium benzoate (NaC 6 H 5 CO 2, 2.00g) are dissolved in enough water to make 1.00L of solution. Calculate the pH of the solution using the Henderson-Hasselbalch equation. K a = 6.3 x g benzoic acid = mol 2.00g sodium benzoate = mol pH = log( / ) = 4.13

PREPARING BUFFER SOLUTIONS Buffers work on a pH range: The pH range is the range of pH values over which a buffer system works effectively. It is best to choose an acid with a p K a close to the desired pH

PREPARING BUFFER SOLUTIONS: To be useful, a buffer solution must have 2 characteristics: pH control : pH = pK a + log [conjugatebbase] / [acid] Acid is chosen whose pK a is near the intended value of the pH Exact value of pH is then achieved by adjusting the acid-to- conjugate base ratio Buffer Capacity: Buffer should have the ability to keep the pH approximately constant after the addition of reasonable amounts of acid and base.

PREPARING BUFFER SOLUTIONS: You wish to prepare 1.0L of a buffer solution with a pH of A list of possible acids (and their conjugate bases) follows: Which combination should be selected, and what should the ratio of acid to conjugate base be? AcidConjugate Base KaKa pK a HSO 4 - SO x CH 3 CO 2 HCH 3 CO x HCO 3 CO x

PREPARING BUFFER SOLUTIONS: pH = 4.30, so [H 3 O + ] = 5.0 x Acetic acid (K a is the closest value of desire [H 3 O + ]) = 2.8 [CH 3 CO 2 − ] [CH 3 CO 2 H] K a = [H 3 O + ] [CH 3 CO 2 − ] [CH 3 CO 2 H] K a = [H 3 O + ] [CH 3 CO 2 − ] [CH 3 CO 2 H] 5.0 x = 1.8 x 10 -5

PREPARING BUFFER SOLUTIONS: Ratio of conjugate base to acid = 0.36 pH = pK a + log [base] [acid]

PREPARING BUFFER SOLUTIONS: The relative number of moles of acid and conjugate base is important in determining the pH of a buffer solution, not the concentration. Why? Volume cancels out! That means that diluting a buffer solution will not change its pH