ACIDBASE CHEMISTRY

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ACID-BASE CHEMISTRY
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EXAMPLES OF ACIDS AND BASES PRESENT IN NATURAL
WATERS

• Most important base HCO3-
• Other bases B(OH)4-, PO43-, NH30, AsO43-, SO42-,
  CO32-, etc.
• Most important acid CO2(aq) or H2CO30
• Other acids H4SiO40, NH4, B(OH)30, H2SO40,
  CH3COOH0 (acetic), H2C2O40 (oxalic), etc.

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• Most acid-base reactions in aqueous solutions are
  very fast (almost instantaneous) thermodynamic
  equilibrium is attained and thermodynamic
  principles yield correct answers.
• Acid-base reactions involve proton, but a bare
  proton (H) does not exist in aqueous solution
  it is hydrated, e.g., hydronium ion (H3O) or
  more likely (H9O4).

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BRONSTED DEFINITION

• Acid A substance that can donate a proton to any
  other substance.
• Base A substance that can accept a proton from
  any other substance.

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ACIDS AND BASES ARE ALWAYS PAIRED IN REACTIONS

• H2CO30 H2O ? H3O HCO3-
• NH4 H2O ? H3O NH30
• CH3COOH0 H2O ? H3O CH3COO-
• H2O H2O ? H3O OH-

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SOME DEFINITIONS

• Amphoteric - A substance that can act as either
  an acid or a base, e.g., H2O, HCO3-
• Polyprotic acid or base - An acid or base that
  can donate or accept, respectively, more than one
  proton, e.g., H3PO40, H2CO30, H4EDTA

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SIMPLE METAL IONS ARE ALSO ACIDS

• All metal ions are hydrated in aqueous solution.
  The attached waters can lose protons, and are
  therefore acids. The positive charge of metal ion
  determines the strength of the acid.
• Zn(H2O)62 H2O ? H3O Zn(H2O)5(OH)
• Cu(H2O)42 3H2O ? 3H3O Cu(H2O)(OH)3-

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CONJUGATE ACID-BASE PAIRS

• HCl0, Cl-
• H2CO30, HCO3-
• HSO4-, SO42-
• CH3COOH0, CH3COO-
• Zn(H2O)62, Zn(H2O)5(OH)

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LEWIS DEFINITION

• Acid Any substance that can accept an electron
  pair.
• Base Any substance that can donate an electron
  pair.

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STRENGTH OF AN ACID OR BASE

• Strength The tendency to donate or accept a
  proton, i.e., how readily does the substance
  donate or accept a proton?
• Weak acid has weak proton-donating tendency a
  strong acid has a strong proton-donating
  tendency. Similarly for bases,
• Cannot define strength in absolute sense.
  Strength depends on both the acid and base
  involved in an acid-base reaction.
• Strength measured relative to some reference, in
  our case, the solvent water.

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STRENGTH MEASURED QUANTITATIVELY BY THE
IONIZATION CONSTANT

• HA0 H2O ? H3O A-
• or
• HA0 ? H A-

The larger KA, the stronger the acid the
smaller KA, the weaker the acid
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DEFINITION OF pKA AND pH

• pKA - log KA
• Thus, the larger pKA, the weaker the acid the
  smaller pKA, the stronger the acid.
• Similarly,
• pH - log H
• pOH - log OH-
• pX - log X

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STRENGTH OF A BASE

• A- H2O ? HA0 OH-

pKB - log KB The larger pKB, the weaker the
base the smaller pKB, the stronger the base.
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SELF-IONIZATION OF WATER AND NEUTRAL pH

• H2O ? H OH-
• Neutrality is defined by the condition H
  OH-

At 25oC and 1 bar
Kw H2 log Kw 2 log H -log Kw -2 log
H 14 2 pH pHneutral 7
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CONJUGATE ACIDS-BASES

• H A- ? HA0 1/KA
• H2O ? H OH- Kw
• A- H2O ? HA0 OH- KB
• KB Kw/KA
• The stronger an acid, the weaker the conjugate
  base, and vice versa.

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ACTIVITY SCALESINFINITE DILUTION SCALE

• ?A aA/cA
• Based on the fact that solutions approach
  ideality as the total concentration of all ions
  in solutions approaches zero. In other words
• ?A ? 1
• as
• (cA ?ci) ? 0

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ACTIVITY SCALESIONIC MEDIUM SCALE

• Applied to solutions with a dominant
  concentration of a relatively inert electrolyte
  to maintain a constant ionic medium.
• ?A ? 1 as cA ? 0, but ?ci is constant
• If ?ci ? 10 cA, then ?A ? 1
• Seawater is a good example, with approximately
  constant composition of dominantly NaCl

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pH CONVENTIONS

• Infinite dilution scale
• paH - log H - log H - log ?H
• Ionic medium
• pH - log H
• NBS (NIST) scale
• defines pH relative to a series of standard
  buffers

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OPERATIONAL ACIDITY CONSTANTS

• 1) Infinite dilution scale
• 2) Ionic medium scale (concentration quotient or
  conditional constant)
• 3) Mixed constant

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IONIC STRENGTH

• A quantity required to calculate activity
  coefficients.
• Attempts to account for effects of both
  concentration and charge of ion on activity
  coefficients.

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ACTIVITY COEFFICIENT EXPRESSIONS
1) Debye-Hückel Limiting Law Valid at I lt
0.005 M
2) Full Debye-Hückel Equation Valid at I lt
0.1 M
3) Güntelberg Equation Valid at I lt 0.1 M
Useful for mixed electrolytes
4) Davies Equation Valid at I lt 0.5 M
A 0.5 B 0.33 at 25oC and 1 bar
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NUMERICAL EQUILIBRIUM CALCULATIONS

• Monoprotic acid
• What are the pH and the concentrations of all
  aqueous species in a 5 x 10-4 M solution of
  aqueous boric acid (B(OH)3)?
• Steps to solution
• 1) Write down all species likely to be present in
  solution H, OH-, B(OH)30, B(OH)4-.

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• 2) Write the reactions and find the equilibrium
  constants relating concentrations of all species
• H2O ? H OH-

(i)
B(OH)30 H2O ? B(OH)4- H
(ii)
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• 3) Write down all mass balance relationships
• 5 x 10-4 M ?B
• B(OH)4- B(OH)30 (iii)
• 4) Write down a single charge-balance
  (electroneutrality) expressions
• H B(OH)4- OH- (iv)
• 5) Solve n equations in n unknowns.

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EXACT NUMERICAL SOLUTION

• Eliminate OH- in (i) and (iv)
• HOH- Kw
• OH- Kw/H
• H B(OH)4- Kw/H
• H - B(OH)4- Kw/H

(v)
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Solve (iii) for B(OH)30 B(OH)30 ?B -
B(OH)4-
HB(OH)4- KA(?B - B(OH)4-) (vi) Now
solve (v) for B(OH)4- - B(OH)4- Kw/H -
H B(OH)4- H - Kw/H Substitute this
into (vi)
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H(H - Kw/H) KA(?B - H
Kw/H) H2 - Kw KA?B - KAH
KAKw/H H3 - KwH KA?BH - KAH2
KAKw H3 KAH2 - (KA?B Kw)H - KAKw
0 H3 (7x10-10)H2 - (3.6x10-13)H -
(7x10-24) 0 We can solve this by trial and
error, computer or graphical methods. From trial
and error we obtain H 6.1x10-7 M or pH
6.21
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• OH- Kw/H
• OH- 10-14/10-6.21
• OH- 10-7.79 M
• B(OH)4- H - Kw/H
• B(OH)4- 6.1x10-7 - 1.62x10-8
• B(OH)4- 5.94x10-7 M
• B(OH)30 ?B - B(OH)4-
• B(OH)30 5x10-4 - 5.94x10-7 M 4.99x10-4 M

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APPROXIMATE SOLUTION

• Look for terms in additive equations that are
  negligibly small (multiplicative terms, even if
  very small, cannot be neglected.
• Because we are dealing with an acid, we can
  assume that H gtgt OH- so that the mass
  balance becomes
• H B(OH)4-
• and then
• B(OH)30 ?B - H

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(ii)
H2 KA?B-KAH H2 KAH - KA?B
0 This is a quadratic equation of the form ax2
bx c 0 and can be solved using the quadratic
equation
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In our case this becomes
Only the positive root has any physical
meaning. H 5.92 x 10-7 We could have made
this problem even simpler. Because boric as is a
quite weak acid (i.e., very KA value, very little
of it will be ionized, thus B(OH)30 gtgt
B(OH)4- ?B ? B(OH)30 5 x 10-4 M
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H2 3.5 x 10-15 H 5.92 x 10-7 M It is
wise to check your assumptions by back
substituting into original equations. If the
error is ? 5, the approxi- mation is probably
justified because KA values are at least this
uncertain!
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CALULATE THE pH OF A STRONG ACID

• Compute the pH and equilibrium concentrations of
  all species in a 2 x 10-4 M solution of HCl.
• 1) Species H, Cl-, HCl0, OH-
• 2) Mass action laws

3) Mass balance HCl0 Cl- 2 x 10-4 M 4)
Charge balance H Cl- OH-
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• Assumptions HCl is a very strong acid so
• H gtgt OH- and Cl- gtgt HCl0
• Now the only source of H and Cl- are the
  dissociation of HCl, so
• H Cl-
• (this is also apparent from the charge balance)
• Thus, pH - log (2 x 10-4) 3.70, and Cl- 2
  x 10-4 M.
• OH- Kw/H 10-14/2 x 10-4 5 x 10-11 M

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CALCULATE THE pH OF A WEAK MONOPROTIC BASE

• Compute the pH and equilibrium concentrations of
  all species in a 10-4.5 M solution of sodium
  acetate.
• 1) Species H, Na, Ac-, HAc0, OH-
• 2) Mass action laws

3) Mass balances HAc0 Ac- 10-4.5 M
C Na 10-4.5 M C 4) Electroneutrality
Na H Ac- OH- Combine 3) and 4) to
get proton condition HAc0 H OH-
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• We cannot make any approximations relative to the
  concentrations of H and OH- because acetate
  is a weak base and total acetate concentration is
  low. However, because base is weak Ac- gtgt
  HAc0 so
• Ac- ? 10-4.5 M C
• Substitute for OH- in proton condition
• HAc0 H Kw/H
• HAc0 Kw/H - H
• Now substitute into ionization constant expression

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• HC KAKw/H - KAH
• H2C KAKw - KAH2
• H2C H2KA KAKw
• H2(C KA) KAKw
• H2 KAKw/(C KA)
• H (KAKw/(C KA))0.5
• H (10-4.710-14/(10-4.5 10-4.70)
• H 6.2 x 10-8 M
• pH 7.2
• pOH pKw - pH 14.0 - 7.2 6.8
• OH- 1.61 x 10-7

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• Rearranging the proton condition we get
• HAc0 OH- - H 1.6 x 10-7 - 6.2 x 10-8
• 9.8 x 10-8
• Check of assumption
• Ac- C - HAc0 10-4.50 - 9.8 x 10-8
• so Ac- ? 10-4.50
• and the assumption made is valid.

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CALCULATE THE pH OF AN AMPHOLYTE

• Calculate the pH of a 10-3.7 M solution of sodium
  hydrogen phthalate (NaHP).
• 1) Species H2P0, HP-, P2-, H, OH-, Na
• 2) Mass action expressions

COOH
COONa
3) Mass balance expressions PT 10-3.7 M
H2P0 HP- P2-
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• PT 10-3.7 M Na
• 4) Charge balance
• H Na OH- HP- 2P2-
• Now, substitute 3) into 4) to get proton
  condition
• H H2P0 HP- P2- OH- HP-
  2P2-
• H H2P0 OH- P2-
• Because both pK values are less than 7, assume
• OH- ltlt H
• H H2P0 P2-
• H PT 2P2- HP-

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H 2.4 x 10-5 pH 4.62
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CALCULATION OF THE pH OF A POLYPROTIC ACID

• Compute the pH and concentrations of all species
  in equilibrium in a 10-3 M H3PO4 solution.
• 1) Species H, OH-, H3PO40, H2PO4-, HPO4-, PO43-
• 2) Mass action expressions

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• 3) Mass balance
• PT H3PO40 H2PO4- HPO42- PO43-
• 4) Charge balance
• H OH- H2PO4- 2HPO42- 3PO43-
• Assumptions Because phosphoric acid is an acid,
  assume that H gtgt OH-. Also, because KA,2 and
  KA,3 are quite small, then HPO42- and PO43-
  are negligible compared to H3PO40 and H2PO4-.
  The mass balance then becomes
• PT H3PO40 H2PO4-

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• And the charge balance expression becomes
• H H2PO4-
• which can be substituted into the expression for
  KA,1.

KA,1PT-KA,1H H2 H2 KA,1H - KA,1PT
0
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• H 8.986 x 10-4 M
• pH 3.05
• pOH pKw - pH 14 - 3.05 10.95
• OH- 1.122 x 10-11 M
• H3PO40 PT - H2PO4- 10-3 - 8.986 x 10-4
• H3PO40 1.014 x 10-4 M

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PO43- 10-16.15 M 7.079 x 10-17 M A check
of all the assumptions shows that they are
all valid.
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CALCULATION OF pH OF A VOLATILE BASE

• Compute the pH and concentrations of all species
  of a solution exposed to an atmosphere of pNH3
  10-4 atm.
• 1) Species NH30, NH4, OH-, H
• 2) Mass action expressions

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• 3) Charge balance NH4 H OH-
• Assumptions NH3 is a moderate base, so we can
  assume that OH- gtgt H so the charge balance
  becomes
• NH4 OH-
• also
• NH30 pNH3KH 10-4(101.75) 10-2.25 M

OH-2 10-6.75 OH- 10-3.375 M 4.22 x 10-4
M
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• pH pKw - pOH 14 - 3.375 10.625
• so the assumption that OH- gtgt H is valid.
• The concentrations are then
• OH- 4.22 x 10-4 M
• NH4 4.22 x 10-4 M
• H 2.37 x 10-11 M
• NH30 5.62 x 10-3 M

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GRAPHICAL APPROACH TO EQUILIBRIUM CALCULATIONS

• Consider the monoprotic acid HA

CT 10-3 HA0 A- so A- CT -
HA0 KAHA0 HA- KAHA0 H(CT -
HA0) KAHA0 HCT - HHA0 KAHA0
HHA0 HCT
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• CTKA - KAA- HA-
• CTKA A-(H KA)

1) At pH lt pKA, H gtgt KA so H KA ?
H HA0 CT(H/H) CT log HA0 log
CT A- CTKA/H log A- log CT - pKA
pH
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• 2) pH pKA H KA so H KA 2H
• HA0 CTH/(2H) CT/2
• log HA0 log CT - log 2 log CT - 0.301
• A- CT H/(2H) CT/2
• log A- log CT - log 2 log CT - 0.301
• 3) pH gt pKA H ltlt KA so KA H ? KA
• HA0 CTH/KA
• log HA0 log CT pKA - pH
• A- CTKA/KA CT
• log A- CT

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Speciation diagram for HA with pKA 5.5 and CT
10-3
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• To compute the composition of a 10-3 M solution
  of HA, we start with the charge balance
• H A- OH-
• H gtgt OH-
• H ? A-
• To compute composition of 10-3 M NaA solution,
  start with proton condition
• HA0 H OH-
• OH- gtgt H
• HA0 ? OH-

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SPECIATION DIAGRAM FOR A DIPROTIC SYSTEM

• Consider H2S with pK1 7.0, pK2 13.0
• ST 10-3 M H2S0 HS- S2-

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• 1) pH lt pK1 lt pK2 H gt K1 gt K2

log H2S0 log ST
log HS- log (STK1) pH
log S2- log (STK2K1) 2pH
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• 2) pH pK1 lt pK2 H K1 gt K2

log H2S0 log ST - 0.301
log HS- log ST - 0.301
log S2- log (STK2/2) pH
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• 3) pK1 lt pH lt pK2 K1 gt H gt K2

log H2S0 log (STK1) - pH
log HS- log ST
log S2- log (STK2) pH
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• 4) pK1 lt pK2 pH K1 gt H K2

log H2S0 log (STK1/2) -
pH
log HS- log ST - 0.301
log S2- log ST - 0.301
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• 5) pK1 lt pK2 lt pH K1 gt K2 gt H

log H2S0 log (STK1K2) -
2pH
log HS- log (STK2) - pH
log S2- log ST
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Speciation diagram for H2S with CT 10-3 at 25?C
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IONIZATION FRACTIONS

• Monoprotic acid HB
• ?B ?1 ? B/C KA/(KA H)
• ?HB ?0 ? HB/C H/(KA H)
• ?1 ?0 1
• Diprotic acid H2A

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Figure 3.10a-c from Stumm and Morgan
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TITRATION OF ACID OR BASE

• Titration curve plot of pH vs. quantity of base
  added
• At any point on a titration curve,
  electroneutrality must be maintained, e.g.,
  titration of HA with NaOH Na H A-
  OH-
• but Na CB, so CB A- OH- - H
• By combining this equation and a log C vs. pH
  diagram (speciation diagram), we can contruct a
  titration curve.
• Equivalent fraction

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What about dilution during titration?

• If we initially have v0 mL of acid, the
  concentration of acid at any point becomes

This can be substituted into the previous
expression. Equivalence point the point where
you have added just enough base to just
neutralize the acid, i.e., f CB/C 1, or CB
C. Starting with A- C?1 and CB A- OH-
- H we obtain CB C?1 OH- - H
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• At the equivalence point we have
• H2O A- ? OH- HA0
• and the only HA0 is that due to dissociation of
  A-, so OH- ? HA0
• Now lets titrate the conjugate base of the weak
  acid HA (i.e., KA) with a strong acid (e.g.,
  HCl). Again we must have electroneutrality so
• K H A- OH- Cl-
• C H A- OH- CA
• CA C - A- H - OH-
• CA HA0 H - OH-
• CA C?0 H

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• Define the equivalent fraction of the titrant
  here as

At the endpoint, g 1, so CA C. Note that, g
1 - f. At this equivalence point H A-
because the only A- is due to the reaction HA0 ?
H A- Buffer HA and A- present in near equal
quantities acts as a buffer in that it slows down
the change in pH as an acid or base is added HA0
? H A- HA0 OH- ? A- H2O
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EXAMPLES OF BUFFERS

• Any conjugate acid/base pair can work as a
  buffer, e.g.
• HCO3-/CO32- B(OH)30/B(OH)4- H2PO4-/HPO42-
• Buffers are most effective at pH ? pKA because
  then HA0 ? A-, and acid and base are present
  in approximately equal quantities.

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CALCULATION OF A TITRATION CURVE

• Titrate 0.1 M solution of a weak acid (pKa 5)
  with 0.1 M NaOH.
• 1) f 0 (zero base added or start of titration)
• HA0 ? H A-
• Assume HA0 ? 0.1 M gtgt A-, H ? A-

H2 10-6 pH 3.0
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• 2) f 0.1 (10 titration)
• A-/HA0 10/90
• In the region between endpoints, pH depends only
  on the A-/HA0 ratio, not on absolute
  concentrations.

H 9 x 10-5 pH 4.05 3) f 0.5 (50
titration)
pH 5.0
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• 4) f 0.8 (80 titration)

H 2.5 x 10-6 pH 5.60 5) f 1.0 (endpoint
of titration) Here all original HA has been
converted to A-. Taking dilution into account,
A- ? 0.05 M. The only HA0 present is that due
to the reaction A- H2O ? HA0 OH- also, HA0
OH- and pKB 14.0 - pKA 14.0 - 5.0 9.0
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• pOH 5.15 pH 14.0 -5.15 8.85
• 6) f 1.20 (120 titration)
• The pH is calculated from excess OH-
  concentration alone (OH- is a stronger base than
  A-). Assume we started with 100 mL of 0.1 M HA.
  At f 1.0, we also would have added 100 mL of
  0.1 M NaOH. At f 1.2, we would have added an
  additional 20 mL of 0.1 M NaOH. The total volume
  would then be 220 mL.

pOH 2.04 pH 14.0 - 2.04 11.96
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CURVE FOR TITRATION OF WEAK ACID WITH STONG BASE
pKa 5
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• Buffer capacity number of moles of a strong base
  needed to raise the pH of 1 L of buffer by 1 pH
  unit.

DEMONSTRATION OF BUFFER CAPACITY
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EXAMPLE BUFFER CALCULATION

• In what ratio would o-phthalic acid and sodium
  hydrogen phthalate have to be mixed to get a pH
  of 3.2 (pKA,1 2.92)?

3.2 2.92 log R 0.28 log R R 1.91 We can
make this solution by mixing 0.1 moles of H2P and
0.191 moles of NaHP in 1 L of water.
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Titration of a strong acid with strong base at
various concentrations of acid and base.
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Titration of a strong base with strong acid.
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Titration of weak acids with strong base.
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Buffer regions in titration curves.
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Titration of a weak acid with strong base in
presence of indicators.
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Titration of a weak base with strong acid.
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What can titrations tell us?

• Concentration of acid or base present (analysis)
• pK of acid or base present (thermodynamics)
• Titration of mixtures of acids or bases
• If the pKs are sufficiently different, the
  stronger acid (base) will be titrated first, then
  weaker acid (base). We will be able to see two or
  more breaks in titration curve, i.e., equivalence
  points, if ?pK ? 4 or more. Same rule applies for
  polyprotic acids.

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Titration of a mixture of weak acids with a
strong base.
pKA 3.88
pKA 4.86
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Titration of a mixture of weak and strong acids
with a strong base.
pKA ? -3
pKA ? 4.86
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BUFFER INTENSITY

• Buffer intensity - A measure of the buffer
  intensity (buffer capacity) would be the inverse
  of the slope of the titration curve at any point.

For a monoprotic system, recalling that CB
C?1 OH- - H
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The maximum buffer intensity occurs at the
inflection point of the titration curve, or where
This condition occurs where ?1 ?0, HA0 A-
or pH pKA. A solution is well buffered at
three points when H is dominant, when OH-
is dominant and where pH pKA.
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Figure 3.10a-c from Stumm and Morgan
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Figure 3.10d from Stumm and Morgan
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BUFFER INTENSITY FOR MIXTURES OF ACIDS
BUFFER INTENSITY FOR POLYPROTIC ACIDS
The above is valid as long as K1/K2 gt 100.
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GRAPHICAL DETERMINATION OF BUFFER INTENSITY

• At pH lt pKA, HA0 A- ? HA0 so
• ? ? 2.3(H OH- A-)
• At pH gt pKA, HA0 A- ? A- so
• ? ? 2.3(H OH- HA0)
• Thus, buffer intensity can be calculated from a
  speciation diagram by summing all concentrations
  represented by a line of slope ?1 and multiplying
  by 2.3025.

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• Base neutralizing capacity BNC the
  equivalent sum of all acids that can be titrated
  with a strong base to an equivalence point. For a
  monoprotic acid HA
• BNC HA0 H - OH-
• BNC is the excess of protons above a reference,
  i.e., pure NaA (f 1), for which the proton
  condition is HA0 H OH-. Addition of
  NaA does not affect BNC!
• Acid neutralizing capacity ANC the
  equivalent sumof all bases that can be titrated
  with a strong acid to an equivalence point. For a
  monoprotic base
• ANC A- OH- - H
• ANC is the deficiency of protons over a
  reference level, i.e., pure HA (f 0) for which
  A- OH- H. Addition of HA does not
  affect ANC!

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EXAMPLE PROBLEM

• Calculate ANC of the following solutions
• a) NH4 NH30 5 x 10-3 M pH 9.3
• b) NH4 NH30 10-2 M pH 9.0
• Which has the greater ANC? pKA 9.3.
• ANC NH30 OH- - H
• ANC C?1 OH- - H
• For a)

ANC 0.5(5 x 10-3 M) 10-4.7 - 10-9.3 ANC ?
2.5 x 10-3 M
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EXAMPLE PROBLEM CONTINUED

• For b)

ANC 0.33(10-2 M) 10-5.0 - 10-9.0 ANC ?
3.3 x 10-3 M Thus, solution b) has the higher
acid neutralizing capacity, even though it has
the lower pH, i.e., is more acidic than a). ANC
should not be confused with pH!
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ANC AND BNC OF MULTIPROTIC ACID/BASE SYSTEMS

• For multiprotic acids we can define various
  reference levels (f 0, 1, 2 ).
• Example Sulfide-containing solution. ANC with
  reference to the equivalence point f 0, g 2,
  i.e., a solution of pure H2S is
• ANCf 0 HS- 2S2- OH- - H
• ANCf 0 ST(?1 2?2) OH- - H
• Example Phosphoric acid system with reference to
  f 2, i.e., a solution of pure Na2HPO4.
• BNCf 2 2H3PO40 H2PO4- H - PO43-
  - OH-
• BNCf 2 PT(2?0 ?1 - ?3) H - OH-

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GENERALIZED EQUATIONS FOR ANC AND BNC

• BNCf n Cn?0 (n-1)?1 (n-2)?2 (n-3)?3
• H - OH-
• ANCf n C-n?0 (1-n)?1 (2-n)?2 (3-n)?3
• - H OH-
• ?x refers to the ionization fraction of the
  species that has lost x protons from the most
  protonated species.

100
MIXED ACID-BASE SYSTEM

• A mixed acid-base system would be, for example, a
  natural water containing bicarbonate, borate and
  ammonium. If we employ H2CO30 as the reference,
  then ANCf 0 represents the equivalent sum of
  all bases stronger than H2CO30 minus the
  equivalent sum of all acids stronger than H2CO30.
• ANCf 0 HCO3- 2CO32- NH30
  B(OH)4-
• OH- - H

101
RELATIONSHIP BETWEEN ANC AND BUFFER INTENSITY

• In carbonate systems, ANCf 0 is called
  alkalinity and BNCf 2 is called acidity.