I. Monoprotic Weak Acid Alpha Fractions HA = Acetic Acid: CH3COOH pk a = logk a = 4.75 K a = 10 4.75 HA H + + A K a = [H+ ][A ] [HA] Two Acid Species: HA and A - Alpha Fractions only depend on ph and pka ph = log[h + ] Robert Corn Chem M3LC
I. Monoprotic Weak Acid Alpha Fractions HA = Acetic Acid: CH3COOH HA H + + A K a = [H+ ][A ] [HA] Two Acid Species: HA and A - Alpha Fractions only depend on ph and pka ph = log[h + ]
Alpha Fraction Plot for Acetic Acid α HA α A Alpha Fractions only depend on ph and pka
Alpha Fraction Plot for Acetic Acid α HA α A α HA = α A = 0.5 @ ph = pka Alpha Fractions only depend on ph and pka
II. Diprotic Weak Acid Alpha Fractions H2A = Oxalic Acid: C2H2O4 pk 1 = 1.25 pk 2 = 4.14 K 1 = [H+ ][HA ] [H 2 A] K 2 = [H+ ][A 2 ] [HA ]
Alpha Fraction Plot for Oxalic Acid α H2 A α HA α A 2 Alpha Fractions only depend on ph, pk1 and pk2
Alpha Fraction Plot for Oxalic Acid α H2 A α HA α A 2 0.5 Alpha Fractions only depend on ph, pk1 and pk2 pk1 pk2
EDTA Metal Ion Complexation Equilibria Ethylene Diamine Tetra-acetic Acid (H4Y) α Y 4 EDTA - the world s best metal ion chelator Y 4- EDTA titrations are in basic buffers Chelate
I. Strong Acid Titration H + + OH H 2 O K t = 1 K w 1 100.0 ml of 0.100 HCl titrated with x ml of 0.100 NaOH
I. Strong Acid Titration H + + OH H 2 O K t = 1 K w 1 100.0 ml of 0.100 HCl titrated with x ml of 0.100 NaOH ph = 7 at the Eq. Pt.
II. Weak Acid Titration HA + OH A + H 2 O 100.0 ml of 0.100 HA titrated with x ml of 0.100 NaOH K t = K a K w 1 Calculate the ph of solution at x = 0.0, 50.0, 100.0 and 150.0 ml Acetic Acid pka = 4.75
II. Weak Acid Titration HA + OH A + H 2 O K t = K a K w 1 100.0 ml of 0.100 HA titrated with x ml of 0.100 NaOH Calculate the ph of solution at x = 0.0, 50.0, 100.0 and 150.0 ml 4. Excess Base 3. Equivalence Pt. 2. Buffer Region 1. Initial Weak Acid
II. Weak Acid Titration HA + OH A + H 2 O K t = K a K w 1 100.0 ml of 0.100 HA titrated with x ml of 0.100 NaOH HA A - OH -
II. Weak Acid Titration HA + OH A + H 2 O K t = K a K w 1 100.0 ml of 0.100 HA titrated with x ml of 0.100 NaOH 1. Initial Weak Acid HA 3. Equivalence Pt. A - 2. Buffer Region 4. Excess Base HA & A - OH -
II. Weak Acid Titration 100.0 ml of 0.100 HA titrated with x ml of 0.100 NaOH Calculate the ph of the solution at points: x = 0.0 ml x = 50.0 ml x = 100.0 ml x = 150.0 ml Initial Weak Acid ph: HA Buffer Region: HA & A - Equivalence Point: A - Excess Base OH - weak acid salt calc. "NaA"
II. Weak Acid Titration HA H + + A pk a = 4.75 K a = [H+ ][A ] [HA] K a = 10 4.75 x = 0.0 ml Initial Weak Acid ph: HA [H + ] = K a C tot HA x = 50.0 ml Buffer Region: HA & A - [H + ] = K a [HA] [A ] = K a x = 100.0 ml x = 150.0 ml Equivalence Point: A - Excess Base OH - [OH ] = K b C tot A [OH ] = m OH V tot C tot A = m HA V tot
I. Monoprotic Weak Acid - Exact Solution Approximate Solution: [H + ] = K a C tot HA
I. Monoprotic Weak Acid - Exact Solution
II. Diprotic Weak Acid - Exact Solution
II. Diprotic Weak Acid - Exact Solution
III. Monoprotic Weak Acid Salt - Exact Solution Approximate Solution: [OH ] = K b C tot A
III. Monoprotic Weak Acid Salt - Exact Solution
IV. Diprotic Ampholyte - Approximate Solution Approximate Solution: ph = pk 1 + pk 2 2
IV. Diprotic Ampholyte - Exact Solution
IV. Diprotic Ampholyte - Exact Solution