Buffer Example and Titration Calculations

pH Change to 1M Acetic Acid/1M Acetate Ion Soln. Moles H + Added 0 Moles OH - Added

Buffers Buffer Solutions resist a change in pH Buffers contain relatively large concentrations of either –An acid, HA and its conjugate base A - –A base, B, and its conjugate acid (BH + )

Buffers NH 4 + to react with OH -

Buffers NH 3 to react with H +

Buffers When H + is added, it reacts essentially to completion with the weak base present H + + A -  HA or H + + B  BH +

Buffers When OH - is added, it reacts essentially to completion with the weak acid present OH - + HA  H 2 O + A - OH - + BH +  H 2 O + B

Buffers pH = pK a + log (base/acid) Want pH  pK a  1 pH determined by K a of acid and ratio of acid/conjugate base or K b of base and ratio base/conjugate acid

Buffer Choice Want pH  pK a  1 How do I make a pH 4.0 buffer? Choose a pK a near the desired pH

Buffer Table Formic AcidK a 1.8 X pK a 3.74 Barbituric Acid9.8 X Butanoic Acid1.52 X

Buffer Choice Choose a pK a near the desired pH pH = pK a + log (base/acid) 4.0 = log (base/acid) 0.26 = log (base/acid) = 1.8 = (Na formate / formic acid)

Basic Buffer Choice Ammonia pK b = 4.74 pK a = – 4.74 = 9.26 NH 3 / NH 4 Cl used to buffer around pH 9.26

Buffer Capacity As long as ratio remains virtually constant, the pH will be virtually constant This is true as long as concentrations of buffering materials (HA/A - ) or (B/BH + ) are large compared with H + or OH - added.

25 mL of.2 M HCl titrated with.2 M NaOH Equivalence Point – Where Stoichiometric amounts of acid and base have been added End Point – Where indicator color change occurs Acid / Base Titrations – Strong A & B

VM HCl = 5 X mol -VM NaOH = moles of H + leftover till equivalent point reached At equivalence point, 5 X mol NaCl/.050 L solution pH = 7 Acid / Base Titrations – Strong A & B

Equivalence Point

Acid / Base Titrations – Strong A & B HCl/NaCl buffer

Acid / Base Titrations – Strong A & B Phenolphthalein Bromcresol Green

25 mL of.2 M Acetic Acid (HA) titrated with.2 M NaOH Initial pH calculated as before Acid / Base Titrations – Weak Acid with Strong Base

During titration up to equivalence point VM HA = 5 X mol -VM NaOH = moles of HA leftover VM NaOH = moles OH - added = moles Ac - made Say 10. mL of.2 M NaOH added to 25 mL of.2 M HA Acid / Base Titrations – Weak Acid with Strong Base

HC 2 H 3 O 2  H + + C 2 H 3 O - Acid / Base Titrations – Weak Acid with Strong Base.005 mol-.002 mol mol I.003mol/.035 L mol /.035L C-x+x E.086 M-x .086 x.057 M +x .057

HAc/Ac - buffer Acid / Base Titrations – Weak Acid with Strong Base

At Equivalence Point – all acetic acid converted to Acetate ion At Equivalence Point you have a Sodium Acetate Solution Acid / Base Titrations – Weak Acid with Strong Base

At Equivalence Point you have a Sodium Acetate Solution To determine pH Use K b and C 2 H 3 O H 2 O  HC 2 H 3 O 2 + OH - To determine [OH - ] and [H + ] and pH Acid / Base Titrations – Weak Acid with Strong Base

At Equivalence Point you have a Sodium Acetate Solution pH 7 ? pH = 8.88 Acid / Base Titrations – Weak Acid with Strong Base

Equivalence Pt pH = 8.88

After Equivalence Point VM NaOH – VM HAc(initial) = moles OH - in total volume. From [OH - ] determine [H + ] and pH Acid / Base Titrations – Weak Acid with Strong Base

Phenolphthalein Bromocresol Green

Acid / Base Titrations – Weak Base with Strong Acid

NH3/NH4+ buffer

Acid / Base Titrations – Weak Base with Strong Acid Phenolphthalein Methyl Red