Titrations of Acids and Bases

Transcription

1 EE 680 Lecture #16 /6/2018 Updated: 6 March 2018 Print verion Lecture #16 Buffer & itration (Benjamin, hapter 5) (Stumm & Morgan, hapt.1 ) David Reckhow EE 680 #16 1 itration of Acid and Bae NaOH Weak acid with a trong bae Nal HAc NaAc H O O Hl O H NaOH H O Na [ H ] K a [ H ] K w K w 5 [ OH ] Kb HAc David Reckhow EE 680 #16 f value 2? H2O 1

2 EE 680 Lecture #16 /6/2018 Defining the itration urve A titration i complete when the equivalent of titrant (t) added equal the equivalent of ample () originally preent equ t = equ V t N t = V N we can define the extent of a bae titration a: VBN B equb f V M mole At any point from the tart of the titration, we have a mixed olution of the acid and conjugate bae We mut ue the ENE in place of the PBE David Reckhow EE 680 #16 Defining the itration urve (cont.) he ENE i: for thi problem (titration of HAc with NaOH): [Na + ] + [H + ] = [Ac ] + [OH ] and in general, for a bae titration: B [Na + ] = [A ] + [OH ] [H + ] and combining with the definition for f: VBN f V M B equb mole [ A ] [ OH [ OH ] [ H ] 1 B ] [ H ] Amount of bae added at any point during the titration in equivalent/liter Amount of acid originally preent in mole/liter (which i the ame a the total of acid + conjugate bae preent throughout) David Reckhow EE 680 #16 4 2

3 EE 680 Lecture #16 /6/2018 Alpha & f curve alpha f f value David Reckhow EE 680 # Log and f curve-2 itration of 2 M HAc compare to Stumm & Morgan Figure. f Log g H + OH - HAc Ac - Starting Point.5 Mid-point 4.7 End Point David Reckhow EE 680 #16 6

4 EE 680 Lecture #16 /6/2018 Revere itration (acid) he revere titration i the addition of a trong acid (e.g., Hl) to the fully titrated acetic acid (e.g., NaAc). hi re form the original HAc and produce Nal too. we can define the extent of an acid titration a: VAN g V M A with the forward titration, we have a mixed olution of the acid and conjugate bae We mut ue the ENE in place of the PBE A equa mole David Reckhow EE 680 #16 7 Revere titration (cont.) he ENE i: for thi problem (titration of NaAc with Hl): [Na + ] + [H + ] = [Ac ] + [OH ] + [l ] and for an acid titration of a pure bae (Na form): [HA] + [A ] = [Na + ] and combining with the definition for g: VAN g V M A equa mole [ H ] [ OH ] 0 [ HA] [ H ] [ OH ] A [l - ] = [Na + ] - [Ac - ] + [H + ] - [OH - ] A [l - ] = [HA] + [H + ] - [OH - ] Amount of acid added at any point during the titration in equivalent/liter Amount of bae originally preent in mole/liter (which i the ame a the total of acid + conjugate bae preent throughout) David Reckhow EE 680 #16 8 4

5 EE 680 Lecture #16 /6/2018 For a monoprotic acid/bae: f + g equal 1 throughout a titration [ OH ] [ H f g ] [ H ] [ OH 0 ] David Reckhow EE 680 #16 9 Buffer & Buffer Intenity Definition Buffer: a olution that reit large change when a bae or acid i added commonly a mixture of an acid and it conjugate bae Buffer Intenity: the amount of trong acid or trong bae required to caue a mall hift in Significance Natural Water wide range poorly buffered water are uceptible to acid precipitation David Reckhow EE 680 #16 5

6 EE 680 Lecture #16 /6/2018 Engineered Procee certain treatment need large hift (e.g., oftening) other need to reit large hift (e.g., biotreatment) Laboratory buffer needed to calibrate meter ued in experimentation to maintain contant. hi implifie data analyi and interpretation David Reckhow EE 680 #16 11 Making a Buffer Acid & conjugate bae bet to have a reervoir of each o there i reitance to change in both direction A - HA Bae HA A - Acid Mirror quetion Given a deired, what hould the buffer compoition be? Given an acid/conjugate bae mixture, what will the be? David Reckhow EE 680 #

7 EE 680 Lecture #16 /6/2018 Acetic Acid Sytem: 0.4 Alpha & f 0.2 curve 0 11 alpha Bae addition Acid addition f value David Reckhow EE 680 #16 1 Buffer: Acetic Acid & Sodium Acetate Example 1. Lit all pecie preent H +, OH, HAc, Ac, Na + Five total 2. Lit all independent equation equilibria K a = [H + ][Ac ]/[HAc] = 4.77 K w = [H + ][OH ] = 14 ma balance HAc + NaAc = [HAc]+[Ac ] electroneutrality: (poitive charge) = (negative charge) 2 Note: we can t ue the PBE becaue we re adding an acid and it conjugate bae [Na + ] + [H + ] = [OH ] + [Ac ] David Reckhow EE 680 # NaAc = [Na + ] 5 7

8 EE 680 Lecture #16 /6/2018 Simplified HAc/NaAc Example. Ue implified ENE & olve for Ac and HAc 4 [Na + ] + [H + ] = [OH ] + [Ac ] 4+5 [Na + ] [Ac ] NaAc [Ac ] Aume [Na + ]>>[H + ], and [Ac - ]>>[OH - ] NaAc = [Na + ] Plug back in to K a equation and olve for H + 1 K a = [H + ][Ac ]/[HAc] K a = [H + ] NaAc / HAc +4+5 [H + ]= K a HAc / NaAc = pk a + log( NaAc / HAc ) or more generally 2 = pk a + log( A / HA ) HAc + NaAc = [HAc]+[Ac - ] HAc + NaAc = [HAc]+ NaAc HAc = [HAc] K w = [H + ][OH - ] [OH - ] = K w /[H + ] David Reckhow EE 680 #16 15 Henderon Haelbalch Equation laic H H equation Jut a re arrangement of equilibrium equation Alway correct Empirical H H pk [ A ] log [ HA ] Aume buffer alt wamp H + and OH A pka log a HA David Reckhow EE 680 #

9 EE 680 Lecture #16 /6/2018 Simplified HAc/NaAc Example (cont.) Solution #1 NaAc (= A ) = mm HAc (= HA ) = mm A pka log Obervation log HA Solution #2 NaAc (= A ) = 20 mm HAc (= HA ) = 2 mm A pka log log = pk a, when equal amount of acid and conjugate bae are added 2. i independent of (eventually at low thi break down) David Reckhow EE 680 #16 17 HA Exact Solution: Summary Monoprotic Acid: [H + ] + {K a }[H + ] 2 {K w + K a }[H + ] K W K a = 0 Bae: [H + ] + {+K a }[H + ] 2 {K w }[H + ] K W K a = 0 Mixed Acid/Bae (i.e., buffer): [H + ] + { A +K a }[H + ] 2 {K w + K a HA }[H + ] K W K a = 0 David Reckhow EE 680 #

10 EE 680 Lecture #16 /6/2018 o next lecture David Reckhow EE 680 #16 19