Acids

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Acids Bases Brønsted-Lowry Definition

• A. An acid is any substance which can donate a
  proton to any other substance,
• HCl (aq) H2O(l) -----gt H3O(aq) Cl-(aq)
• acid hydronium ion
• B. A base is any substance which can accept a
  proton from any other substance.
• NH3(aq) H2O(l) -----gt NH4(aq) OH-
  (aq)
• base (note says nothing about hydroxides!)
• ---more definitions
• monoprotic capable of donating 1 proton (HCl,
  HNO3, CH3COOH)
• polyprotic capable of donating more than 1
  proton (H2SO4, H3PO4)
• amphiprotic can act as Brønsted acid or base
  (H2O above, HSO4-)
• HSO4- H2SO4 as
  base
• SO4-2 as acid

H
-H
2
Conjugate Acid-Base Pairs

• each reaction weve written has involved the
  transfer of a proton.
• therefore, must have both an acid (H-donor) and
  base (H-acceptor) present.
• HCO3-(aq) H2O(l) ltgt CO3-2(aq)
  H3O(aq)
• bicarbonate carbonate
• bicarbonate and carbonate related to one another
  by the gain or loss of one H
• HCO3-(aq) ltgt CO3-2(aq) H(aq)
• called Conjugate Acid- Base Pair
• Every acid-base reaction involving H transfer
  has two acid-base conjugate pairs.
• HCO3-(aq) H2O(l) ltgt CO3-2(aq)
  H3O(aq)
• acid base conjugate base conjugate
  acid

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Water the pH scale

• H OH-
• H2O ltgt H(aq) OH-(aq) Keq
• H2O
• in all aqueous solutions H2O 55M
• so rewrite Keq H2O Kw H OH-
  1.0 x 10-14
• in pure water H OH- 1.0 x 10-7
• DEFINE Any aqueous solution in which H
  OH- is called a neutral solution
• A neutral solution does not mean there are no H
  or OH- !!
• when H gt OH- acidic solution (H gt
  10-7 M)
• when H lt OH- basic solution (H lt
  10-7 M)
• Sørenson proposed that instead of using
  concentrations, use term called p, where pX
  -log X or pH -log H
• pH lt 7 acidic pH gt 7 basic
  pH 7 neutral

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Strong Acids and Bases

• ionize 100 in solution
• KNOW THESE STRONG ACIDS
• HCl hydrochloric acid
• HBr hydrobromic acid
• HI hydroiodic acid
• HNO3 nitric acid
• H2SO4 sulfuric acid
• HClO4 perchloric acid
• KNOW THESE STRONG BASES
• Hydroxides of Group I metals LiOH, NaOH, KOH
• Hyroxides of Group II metals Mg(OH)2, Ca(OH)2

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Weak Acids Bases

• only partially dissociate in solution
• HCN(aq) ltgt H(aq) CN-(aq)
• NH3(aq) H2O(l) ltgt OH-(aq) NH4(aq)
• notes
• 1. the stronger the acid, the weaker the
  conjugate base
• 2. the stronger the base, the weaker the
  conjugate acid
• 3. all are proton transfer reactions and run from
  the stronger pair towards the weaker pair.
• stronger
• weaker

All species in solution
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Weak Acids Bases in Solution

• weak acids set up the following equilibrium
• H B-
• HB(aq) ltgt H(aq) B-(aq) Ka
  
• weak acid conjugate base HB
• weak bases set up the following equilibrium
• BaseH OH-
• Base(aq) H2O(l) ltgt BaseH(aq)
  OH-(aq) Kb
• conjugate acid Base
• Ka and Kb values found in Appendix Tables B.8
  B.9
• Ks measure extent to which the acid/base
  dissociates in water
• (larger value stronger acid/base)
• can use just like other equilibrium expressions
• note not doing reaction here, just putting
  acid/base in water

acid dissociation constant
8
Example

• How many moles of phosphoric acid should be
  placed in 1.0L of solution so that the resulting
  pH 3.00?
• H3PO4 ltgt H H2PO4-
• START x 0 0
• At Equil x-y y y
• by definition pH 3.0, so H 10-3.0
  0.0010 y
• H H2PO4- y y
  (0.0010)2
• Ka 7.1 x 10-3
• H3PO4 x-y
  x - 0.0010
• x 1.14 x 10-3 M
• so, for 1.0L need 1.14 x 10-3 moles

from Table B.8
9
Kas the Quadratic Equation

• can always solve the quadratic equation, although
  many times this is not required
• Why? Kas only known to 2 Sig Figs.
• Therefore, can make mathematical approximation if
  doing so doesnt introduce an error in the 2 Sig
  Figs.
• HB(aq) ltgt H(aq) B-(aq)
• If HBo gt 100Ka , then use approximation
• H B- H2 H2
• Ka ?
• HBe HBo - H HBo
• H2
• Ka if HBo gt 100Ka

equilibrium conc.
initial conc.
The Approximation
10
Example

• 0.020 moles of acetic acid is diluted to 1.00L.
  What is the resulting pH?
• AcOH ltgt AcO- H
• START 0.020 0 0
• At Equil 0.020-y y y
• AcO- H y y
• Ka 1.8 x 10--5
• AcOH 0.020-y
• AcOH gt 100Ka USE APPROXIMATION
• H2 H2
• Ka 1.8 x 10--5
• AcOH 0.020

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Weak Bases Kbs

• can do same sorts of calculations as with Kas
• what amount of ammonia must be placed in 1.00L
  solution so that the pH 8?
• NH3 H2O ltgt NH4 OH-
• START x 0 0
• At Equil x-y y y
• NH4 OH- y y
• Kb 1.8 x 10--5
• NH3 x-y
• pOH 14 - pH 14 - 8 6 OH- 10-6
  y
• (10-6)2
• Kb 1.8 x 10--5 x 1.05 x 10-6 M
• x-10-6
• need 1.05 x 10-6 moles

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Kbs the Approximation

• what is the pH if 0.010 moles of HCO3 - are
  dissolved in 1.00L?
• HCO3- H2O ltgt H2CO3 OH-
  Kb 2.4 x 10-8
• START 0.010 0 0
• At Equil 0.010-y y
  y
• can use approximation if baseo gt 100Kb
  USE APPROXIMATION
• H2CO3 OH- OH- 2
• Kb 2.4 x 10--8
• HCO3 - 0.010
• OH- 1.55 x 10-5
• pOH 4.81
• pH 14 - 4.81 9.19

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Buffers

• a system will maintain a relatively constant pH
  if it already contains a quantity of acid and a
  quantity of base.
• DEFINE A solution that is relatively resistant
  to changes in pH is a BUFFER.
• For this class that will be a solution of a weak
  acid and its conjugate base.
• HCOOH(aq) ltgt HCOO-(aq) H(aq)
• formic acid Na formate
• NOTE With buffers you are NOT starting from
  either side, but from BOTH sides

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Determination of Buffer pH

• HB B- at equilibrium ? concentrations on
  mixing
• HB
• H Ka
• B-
• What is the pH of a formic acid/Naformate buffer
  if 1.0 mole of each are dissolved in 1.0L
  solution?
• HCOOH 1.0M
• H Ka 1.8 x 10-4 1.8 x 10-4
• HCOO- 1.0M
• pH 3.74
• note if HB ???B-, then pH pKa

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Determination of Buffer pH cont.

• I want 1.0L of the above buffer to have a pH
  5.20. If I have a 0.10M formic acid solution, how
  many moles of solid Naformate should be added?
• pH 5.20 H 6.3 x 10-6
• HCOO- Ka HCOOH (1.8 x
  10-4)(0.10M)
• H
  6.3 x 10-6
• HCOO- 2.8M
• 2.8M x 1.0L 2.8 moles

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Effect of Added Strong Acid or Base on Buffers

• if add 0.1 mole H to water, pH will go from 7 to
  1 (106 difference in H).
• HCOOH(aq) ltgt HCOO-(aq) H(aq)
• what will happen if we add a strong acid to a
  HCOOH/HCOO- buffer?
• ---the strong acid will react with the weak base
• HCOO-(aq) H(aq) -----gt HCOOH(aq)
• What happens to a system when its equilibrium is
  perturbed?
• It reestablishes a new equilibrium LeChatliers
  Principle

100 based on this limiting reagent
will change concentrations of both of these
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Quantitative Example

• start with 1.0L buffer which is 1.0M in both acid
  and conjugate base.
• HCOOH 1.0M
• H Ka 1.8 x 10-4 1.8 x 10-4
• HCOO- 1.0M
• pH 3.74
• add 0.10 mole H what is new pH?
• HCOO-(aq) H(aq) -----gt
  HCOOH(aq)
• START 1 mole 0.10 mole 1 mole
• At Equil. 1 - .1 mole 1 .1 mole
• 0.90 mole 1.1 mole
• HCOOH 1.1 M
• H Ka 1.8 x 10-4 2.2 x 10-4
• HCOO- 0.9 M

Calculation works best with moles
Dont forget to convert back to concentration!
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Reactions Between Strong Acids Strong Bases

• in soln, each exists as
• HCl (aq) -----gt H(aq) Cl-(aq)
• .
• NaOH(aq) -----gt Na(aq) OH- (aq)
• mix together
• H(aq) Cl-(aq) Na(aq) OH- (aq) -----gt
  H2O(l) Na(aq) Cl-(aq)
• or
• H(aq) OH- (aq) -----gt H2O(l) neutralizat
  ion
• Cl-(aq) Na(aq) are spectator ions

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Reactions between a Strong Base a Weak Acid

• First step is neutralization
• OH-(aq) HCOOH(aq) ------gt H2O HCOO-(aq)
• goes to completion but note that you are forming
  a weak base.
• What happens when a weak base is in solution?
• Second step is equilibration
• H2O HCOO-(aq) ltgt OH-(aq) HCOOH(aq)
  
• this is an equilibrium it does NOT go back to
  starting point
• if you know the concentration of the weak base
  formed and Kb, you can calculate the pH at the
  equivalence point.

100
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Quantitative Example

• 0.1 mole NaOH and 0.1 mole acetic acid are placed
  in 1.0L of solution. What is the resulting pH?
• OH-(aq) CH3COOH(aq) ------gt H2O
  CH3COO-(aq)
• form 0.1mole of acetate in 1.0L solution 0.1M
• CH3COO-(aq) H2O ltgt CH3COOH(aq)
  OH-(aq)
• START 0.1 0 0
• At Equil. 0.1 - y y y
• CH3COOH OH- y2
• Kb 5.6 x 10-10
• CH3COO- 0.1 - y
• use approximation and solve for y 7.5 x 10-6
  OH- pOH 5.12
• pH 14 - 5.12 8.88

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Reactions between a Strong Acid a Weak Base

• First step is neutralization
• H (aq) NH3(aq) ------gt NH4(aq)
• goes to completion but note that you are forming
  a weak acid.
• What happens when a weak acid is in solution?
• Second step is equilibration
• NH4(aq) ltgt NH3(aq) H (aq)
• this is an equilibrium it does NOT go back to
  starting point
• if you know the concentration of the weak acid
  formed and Ka, you can calculate the pH at the
  equivalence point.

100
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Quantitative Example

• 0.1 mole HCl and 0.1 mole ammonia are placed in
  1.0L of solution. What is the resulting pH?
• H (aq) NH3(aq) ------gt NH4(aq)
• form 0.1mole of ammonium in 1.0L solution 0.1M
• NH4(aq) ltgt NH3(aq) H (aq)
• START 0.1 0 0
• At Equil. 0.1 - y y y
• NH3 H y2
• Ka 5.71 x
  10-10
• NH4 0.1 - y
• use approximation and solve for y 7.55 x 10-6
  H pH 5.12

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Reactions of Weak Acids Weak Bases

• example reaction of methylamine and hypochlorous
  acid
• CH3NH2(aq) HClO(aq) ltgt ClO-(aq)
  CH3NH3(aq)
• these reactions do not go to completion, but set
  up an equilibrium.
• will resulting solution be acidic or basic?
• look at products
• ClO-(aq) H2O(aq) ltgt HClO(aq)
  OH-(aq) Kb 2.9 x 10-7
• CH3NH3(aq) ltgt CH3NH2(aq) H(aq)
  Ka 2.0 x 10-10
• Kb gt Ka therefore, more OH- produced than H
  (pH gt 7)

weak base weak acid
weak base weak acid
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Acid-Base Titrations

• from lab
• adding base from buret until moles of base added
  moles of acid in flask (equivalence point).
• n(OH-) n(H) or MH x VH MOH- x
  VOH-
• how do we determine when we have reached the
  equivalence point?
• INDICATORS weak organic acids or bases which
  have different colors in 2 ionized states.
• HIn ltgt In- H
• e.g. phenolphthalein colorless red
• relative quantities of conjugate acid/base pair
  determines color of the solution
• since the relative quantities depend on H, the
  color will change as pH changes
• LeChateliers Principle, again.

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Choosing Indicators

• it is important to choose an indicator which will
  change color near the pH of the equivalence point
• what will the predicted pH be?
• Strong Acid / Strong Base pH 7
• Strong Acid / Weak Base must calculate
• Strong Base / Weak Acid must calculate
• be careful!
• if indicator changes color before the endpoint,
  the reaction will not be complete.
• if it changes after the endpoint, too much
  reagent will have been added.

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Strong Acid / Strong Base Titration Curve

• --start with 50mL 1.0M HCL pH 0
• --add 10.0mL 1.0M NaOH (0.67 M HCl
  left) pH 0.18
• --add 40.0mL 1.0M NaOH (0.11 M HCl
  left) pH 0.95
• --add 49.0mL 1.0M NaOH (0.010 M HCl
  left) pH 2.0
• --add 49.9mL 1.0M NaOH (0.001 M HCl
  left) pH 3.0
• --add 49.99mL 1.0M NaOH (0.0001 M HCl
  left) pH 4.0
• reach equivalence
  point
• --add 50.01mL 1.0M NaOH (0.0001 M NaOH
  left) pH 10.0
• --add 50.1mL 1.0M NaOH (0.001 M NaOH
  left) pH 11.0

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50.0mL 1.0M HCl 1.0M NaOH
pKa of phenolphthalein
pH
Equivalence point
mL base added
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Strong / Weak Acid / Base Titration Curve

• e.g. titration of 1.0M acetic acid with NaOH
• 1) starting pH (solution is acetic acid)
• C2H3O2H ltgt C2H3O2- H
• Start 1.0M 0 0
• At Equil. 1.0 - y y y
• y2/1.0M Ka 1.8 x 10-5 y
  H 4.24 x 10-3 pH 2.37
• 2) equivalence point (solution is acetate)
• C2H3O2- H2O ltgt C2H3O2H OH-
• Start 1.0M 0 0
• At Equil. 1.0 - y y y
• y2/1.0M Kb 7.1 x 10-10 y
  OH- 2.66 x 10-5
• pOH 4.57 pH 9.43

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Strong / Weak Acid / Base Titration Curve

• 3) Midpoint pH (solution is 5050 acetic acid
  acetate)
• C2H3O2H
• H Ka
• C2H3O2-
• C2H3O2H C2H3O2- , so pH pKa
  4.74
• Note that you have already done all three of
  these calculations.

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50.0mL 1.0M acetic acid 1.0M NaOH

note want indicator which will change color
in this region
starting pH
equivalence point
midpoint pH