1 Buffers & Titrations Chapter 18

2 Buffers A soln that resists change in pH when strong acid or strong base is added A soln that resists change in pH when strong acid or strong base is added Made from weak acid/conjugate-base Made from weak acid/conjugate-base Or Or Made from weak base/conjugate-acid Made from weak base/conjugate-acid Strong acids and strong bases don’t make buffers! Strong acids and strong bases don’t make buffers! So how does it work? So how does it work?

3 How buffers work Acetic acid/acetate buffer system: Acetic acid/acetate buffer system: C 2 H 3 O 2 H (aq) + H 2 O (l)  C 2 H 3 O 2 - (aq) + H 3 O + (l) Add base: Add base: C 2 H 3 O 2 H (aq) + OH - (aq)  C 2 H 3 O 2 - (aq) + H 2 O (l) Add acid: Add acid: H 3 O + (aq) + C 2 H 3 O 2 - (aq)  H 2 O (l) +C 2 H 3 O 2 H (aq)

4 Common Ion Effect: CIE The ionization of an acid or a base is limited by the presence of its conjugate base or acid. The ionization of an acid or a base is limited by the presence of its conjugate base or acid. HAc (aq) + H 2 O (l)  Ac - (aq) + H 3 O + (aq) Acetate ion is added in form of NaAc Acetate ion is added in form of NaAc –Which way will this shift the rxn? –Would you expect a greater or lesser acidity if the CIE was lacking? Let’s look at the next problem Let’s look at the next problem

5 Calculating pH of a buffer soln You have an acetic acid/acetate buffer with a M conc of acetic acid and a M conc of acetate ion. What’s the pH of the buffer? (K a = 1.8 x ) You have an acetic acid/acetate buffer with a M conc of acetic acid and a M conc of acetate ion. What’s the pH of the buffer? (K a = 1.8 x )

6 Solution

7 Let’s work on this Consider mL of a buffer solution that is 1.00M in HAc and 1.00M in NaAc. What is the pH after addition of 25.0 mL of 1.00M NaOH?

8 Solution

9 Henderson-Hasselbalch equation Useful for previous problem Useful for previous problem –Let’s take a look

10 Titrations Used to determine quantity of acid or base in unknown (analyte) Used to determine quantity of acid or base in unknown (analyte)

11 When sample is neutralized (H 3 O + = OH - ) When sample is neutralized (H 3 O + = OH - ) –Equivalence point Determined by use of pH indicator Determined by use of pH indicator

12 Different titration types Strong acid-strong base: equivalence pt = 7.0 (contains neutral salt) Strong acid-strong base: equivalence pt = 7.0 (contains neutral salt)

13 Different titration types Weak acid-strong base: equivalence pt is greater than 7 (basic salt) Weak acid-strong base: equivalence pt is greater than 7 (basic salt) half-equivalence pt (halfway pt) = pK a half-equivalence pt (halfway pt) = pK a

14 Problem We start with 50.0 mL of M HAc mL of M NaOH is then added. What is the pH of the resulting solution?

15 Solution