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CEE 680 Lecture #16 2/21/2020 Print version Updated: 21 February 2020 Lecture #16 Buffers & Titrations (Benjamin, Chapter 5) (Stumm & Morgan, Chapt.1 3 ) David Reckhow CEE 680 #16 1 Titrations of Acids and Bases Weak acid

  1. CEE 680 Lecture #16 2/21/2020 Print version Updated: 21 February 2020 Lecture #16 Buffers & Titrations (Benjamin, Chapter 5) (Stumm & Morgan, Chapt.1 ‐ 3 ) David Reckhow CEE 680 #16 1 Titrations of Acids and Bases  Weak acid with a strong base NaCl NaOH NaAc HAc O O HCl NaOH 3HC C O H 3HC C O Na H2O 8   [ H ] K a C T 7   4 . 7  3 10 10 ?  10  3 . 85 6 K K pH      [ H ] w w  [ OH ] K C 5 b T  14 10    4 10 9 . 3 10 3   10 7 . 85 3 HAc 0 0.2 0.4 0.6 0.8 1 David Reckhow CEE 680 #16 2 f value 1

  2. CEE 680 Lecture #16 2/21/2020 Defining the Titration Curve  A titration is complete when the equivalents of titrant (t) added equals the equivalents of sample (s) originally present  equ t = equ s  V t N t = V s N s  we can define the extent of a base titration as: V N equ   f B B B V M moles s s s  At any point from the start of the titration, we have a mixed solution of the acid and conjugate base  We must use the ENE in place of the PBE David Reckhow CEE 680 #16 3 Defining the Titration Curve (cont.)  The ENE is:  for this problem (titration of HAc with NaOH):  [Na + ] + [H + ] = [Ac ‐ ] + [OH ‐ ]  and in general, for a base titration:  C B  [Na + ] = [A ‐ ] + [OH ‐ ] ‐ [H + ]  and combining with the definition for f: V N equ C Amount of base added at any point    f B B B B during the titration in equivalents/liter V M moles C s s s T      [ A ] [ OH ] [ H ]  Amount of acid originally present in C moles/liter (which is the same as the T total of acid + conjugate base present    [ OH ] [ H ]    throughout) 1 C T David Reckhow CEE 680 #16 4 2

  3. CEE 680 Lecture #16 2/21/2020 Alpha & f 1  1  0 0.8 curves alpha 0.6 0.4 0.2 0 0 2 4 6 8 10 12 14 11 pH 10 9 8 pH 7 f  1 6 5 4 3 0 0.2 0.4 0.6 0.8 1 f value David Reckhow CEE 680 #16 5 0 OH - H + -1 Log C and f curves HAc Ac - -2 -3 Log C -4  Titration of 10 ‐ 2 M HAc -5 -6  compare to -7 Stumm & 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Morgan’s pH Figure 3.3 0.0 1.0 Mid-point 0.8 pH 4.7 0.2 Starting Point 0.4 0.6 pH 3.35 g f 0.6 0.4 End Point pH 8.35 0.8 0.2 0.0 1.0 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 pH David Reckhow CEE 680 #16 6 3

  4. CEE 680 Lecture #16 2/21/2020 Reverse Titration (acid)  The reverse titration is the addition of a strong acid (e.g., HCl) to the fully titrated acetic acid (e.g., NaAc). This re ‐ forms the original HAc and produces NaCl too.  we can define the extent of an acid titration as: V N equ   A A A g V M moles s s s  As with the forward titration, we have a mixed solution of the acid and conjugate base  We must use the ENE in place of the PBE David Reckhow CEE 680 #16 7 Reverse titration (cont.)  The ENE is:  for this problem (titration of NaAc with HCl):  [Na + ] + [H + ] = [Ac ‐ ] + [OH ‐ ] + [Cl ‐ ] [Cl - ] = [Na + ] - [Ac - ] + [H + ] - [OH - ]  and for an acid titration of a pure base (Na form):  C T  [HA] + [A ‐ ] = [Na + ] C A  [Cl - ] = [HA] + [H + ] - [OH - ]  and combining with the definition for g: V N equ C Amount of acid added at any point    g A A A A during the titration in equivalents/liter V M moles C s s s T     [ HA ] [ H ] [ OH ]  Amount of base originally present in C moles/liter (which is the same as the T total of acid + conjugate base present    [ H ] [ OH ]    throughout) 0 C T David Reckhow CEE 680 #16 8 4

  5. CEE 680 Lecture #16 2/21/2020  For a monoprotic acid/base:  f + g equals 1 throughout a titration      [ OH ] [ H ] [ H ] [ OH ]        f g 1 0 C C T T     1 0  1 David Reckhow CEE 680 #16 9 pH Buffers & Buffer Intensity  Definitions  Buffer: a solution that resists large pH changes when a base or acid is added  commonly a mixture of an acid and its conjugate base  Buffer Intensity: the amount of strong acid or strong base required to cause a small shift in pH  Significance  Natural Waters  wide range  poorly buffered waters are susceptible to acid precipitation David Reckhow CEE 680 #16 10 5

  6. CEE 680 Lecture #16 2/21/2020  Engineered Processes  certain treatments need large pH shifts (e.g., softening)  others need to resist large shifts (e.g., biotreatment)  Laboratory  buffers needed to calibrate pH meters  used in experimentation to maintain constant pH. This simplifies data analysis and interpretation David Reckhow CEE 680 #16 11 Making a Buffer  Acid & conjugate base  best to have a reservoir of each so there is resistance to change in both directions A -  Mirror questions  Given a desired pH, Base what should the buffer composition be? HA A -  Given an acid/conjugate base mixture, what will Acid the pH be? HA David Reckhow CEE 680 #16 12 6

  7. CEE 680 Lecture #16 2/21/2020 1  1 Acetic Acid  0 0.8 System: alpha 0.6 0.4 Alpha & f 0.2 curves 0 0 2 4 6 8 10 12 14 11 pH 10 9 Base addition 8 pH 7 Acid addition 6 5 4 3 0 0.2 0.4 0.6 0.8 1 f value David Reckhow CEE 680 #16 13 Buffers: Acetic Acid & Sodium Acetate Example  1. List all species present Five total  H + , OH ‐ , HAc, Ac ‐ , Na +  2. List all independent equations  equilibria 1  K a = [H + ][Ac ‐ ]/[HAc] = 10 ‐ 4.77  K w = [H + ][OH ‐ ] = 10 ‐ 14 2  mass balances 5 C NaAc = [Na + ] 3  C HAc + C NaAc = [HAc]+[Ac ‐ ]  electroneutrality:  (positive charges) =  (negative charges)  Note: we can’t use the PBE because we’re adding an acid and its conjugate base 4  [Na + ] + [H + ] = [OH ‐ ] + [Ac ‐ ] David Reckhow CEE 680 #16 14 7

  8. CEE 680 Lecture #16 2/21/2020 Simplified HAc/NaAc Example  3. Use simplified ENE & solve for Ac ‐ and HAc 4  [Na + ] + [H + ] = [OH ‐ ] + [Ac ‐ ]  [Na + ]  [Ac ‐ ] Assumes [Na + ]>>[H + ], and [Ac - ]>>[OH - ]  C NaAc  [Ac ‐ ] 4+5 5 C NaAc = [Na + ]  4. Plug back in to K a equation and solve for H + 3 C HAc + C NaAc = [HAc]+[Ac - ] 1  K a = [H + ][Ac ‐ ]/[HAc] C HAc + C NaAc = [HAc]+C NaAc 3+4+5  K a = [H + ] C NaAc / C HAc 1+3+4+5 C HAc = [HAc]  [H + ]= K a C HAc /C NaAc  pH = pK a + log(C NaAc /C HAc ) K w = [H + ][OH - ]  or more generally 2 [OH - ] = K w /[H + ]  pH = pK a + log(C A /C HA ) David Reckhow CEE 680 #16 15 Henderson ‐ Hasselbalch Equation  Classic H ‐ H equation  Just a re ‐ arrangement of equilibrium equation  Always correct  [ A ]   pH pK log HA a [ ]  Empirical H ‐ H  Assumes buffer salts swamp H + and OH ‐ C   pH pK log A a C HA Lawrence Henderson was a biochemist, born 3 Jun 1878 in Lynn MA, established the fatigue lab at Harvard David Reckhow CEE 680 #16 16 8

  9. CEE 680 Lecture #16 2/21/2020 Simplified HAc/NaAc Example (cont.)  Solution #1  Solution #2  C NaAc (= C A ) = 10 mM  C NaAc (= C A ) = 20 mM  C HAc (= C HA ) = 10 mM  C HAc (= C HA ) = 2 mM C C     pH pK log A pH pK log A a a C C HA HA 10 20   4 . 7 log   4 . 7 log 10 2  4 . 7  5 . 7 Observations 1. pH = pK a , when equal amounts of acid and conjugate base are added 2. pH is independent of C T (eventually at low C T this breaks down) David Reckhow CEE 680 #16 17 Exact Solutions: Summary  Monoprotic  Acids:  [H + ] 3 + {K a }[H + ] 2 ‐ {K w + K a C}[H + ] ‐ K W K a = 0  Bases:  [H + ] 3 + {C+K a }[H + ] 2 ‐ {K w }[H + ] ‐ K W K a = 0  Mixed Acid/Bases (i.e., buffers):  [H + ] 3 + {C A +K a }[H + ] 2 ‐ {K w + K a C HA }[H + ] ‐ K W K a = 0 David Reckhow CEE 680 #16 18 9

  10. CEE 680 Lecture #16 2/21/2020  To next lecture David Reckhow CEE 680 #16 19 10

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