ph = log[h + ] pk a = logk a = 4.75 K a = HA H + + A K a = [H+ ][A ] [HA] Acid-Base Chemistry: Alpha Fractions, Titrations, Exact Solutions

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1 I. Monoprotic Weak Acid Alpha Fractions HA = Acetic Acid: CH3COOH pk a = logk a = 4.75 K a = 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

2 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 + ]

3 Alpha Fraction Plot for Acetic Acid α HA α A Alpha Fractions only depend on ph and pka

4 Alpha Fraction Plot for Acetic Acid α HA α A α HA = α A = ph = pka Alpha Fractions only depend on ph and pka

5 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 ]

6 Alpha Fraction Plot for Oxalic Acid α H2 A α HA α A 2 Alpha Fractions only depend on ph, pk1 and pk2

7 Alpha Fraction Plot for Oxalic Acid α H2 A α HA α A Alpha Fractions only depend on ph, pk1 and pk2 pk1 pk2

8 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

9 I. Strong Acid Titration H + + OH H 2 O K t = 1 K w ml of HCl titrated with x ml of NaOH

10 I. Strong Acid Titration H + + OH H 2 O K t = 1 K w ml of HCl titrated with x ml of NaOH ph = 7 at the Eq. Pt.

11 II. Weak Acid Titration HA + OH A + H 2 O ml of HA titrated with x ml of NaOH K t = K a K w 1 Calculate the ph of solution at x = 0.0, 50.0, and ml Acetic Acid pka = 4.75

12 II. Weak Acid Titration HA + OH A + H 2 O K t = K a K w ml of HA titrated with x ml of NaOH Calculate the ph of solution at x = 0.0, 50.0, and ml 4. Excess Base 3. Equivalence Pt. 2. Buffer Region 1. Initial Weak Acid

13 II. Weak Acid Titration HA + OH A + H 2 O K t = K a K w ml of HA titrated with x ml of NaOH HA A - OH -

14 II. Weak Acid Titration HA + OH A + H 2 O K t = K a K w ml of HA titrated with x ml of NaOH 1. Initial Weak Acid HA 3. Equivalence Pt. A - 2. Buffer Region 4. Excess Base HA & A - OH -

15 II. Weak Acid Titration ml of HA titrated with x ml of NaOH Calculate the ph of the solution at points: x = 0.0 ml x = 50.0 ml x = ml x = ml Initial Weak Acid ph: HA Buffer Region: HA & A - Equivalence Point: A - Excess Base OH - weak acid salt calc. "NaA"

16 II. Weak Acid Titration HA H + + A pk a = 4.75 K a = [H+ ][A ] [HA] K a = 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 = ml x = 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

17 I. Monoprotic Weak Acid - Exact Solution Approximate Solution: [H + ] = K a C tot HA

18 I. Monoprotic Weak Acid - Exact Solution

19 II. Diprotic Weak Acid - Exact Solution

20 II. Diprotic Weak Acid - Exact Solution

21 III. Monoprotic Weak Acid Salt - Exact Solution Approximate Solution: [OH ] = K b C tot A

22 III. Monoprotic Weak Acid Salt - Exact Solution

23 IV. Diprotic Ampholyte - Approximate Solution Approximate Solution: ph = pk 1 + pk 2 2

24 IV. Diprotic Ampholyte - Exact Solution

25 IV. Diprotic Ampholyte - Exact Solution