Chemistry 100

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Chemistry 100

• Acids and Bases

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The Brønsted Definitions

• Brønsted Acid proton donor
• Brønsted Base proton acceptor
• Conjugate acid - base pair an acid and its
  conjugate base or a base and its conjugate acid

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Example Acid-Base Reactions

• Look at acetic acid dissociating
• CH3COOH(aq) ?CH3COO-(aq) H(aq)
• ?
• Brønsted acid Conjugate base
• Look at NH3(aq) in water
• NH3(aq) H2O(l) ?NH4(aq) OH-(aq)
• ? ?
• Brønsted base conjugate acid

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Representing Protons in Aqueous Solution

• CH3COOH(aq) ? CH3COO-(aq) H(aq)
• CH3COOH(aq) H2O(l) ? CH3COO-(aq) H3O(aq)
• HCl (aq) ? Cl-(aq) H(aq)
• HCl(aq) H2O(l) ? Cl-(aq) H3O(aq)

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Representing Protons

• Both representations of the proton are equivalent
• H5O2 (aq), H7O3 (aq), H9O4 (aq) have been
  observed
• We will use either H(aq) or H3O(aq)

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What is H (aq)?

H3O
H
H5O2
H9O4
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The Hydroxide Bases

• KOH, RbOH, NaOH, are not strictly Brønsted Bases
  since none of these substances accepts a proton
• KOH(aq) ? K(aq) OH-(aq)
• NaOH(aq) ? Na(aq) OH-(aq)
• OH-(aq) H3O(aq) ?2 H2O(l)

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The Autoionization of Water

• Water autoionizes (self-dissociates) to a small
  extent
• 2H2O(l) ? H3O(aq) OH-(aq)
• H2O(l) ? H(aq) OH-(aq)
• These are both equivalent definitions of the
  autoionization reaction. Water is acting as a
  base and an acid in the above reaction ? water is
  amphoteric.

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The Autoionization Equilibrium

• from the preceding chapter

• but we know H2O is constant

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The Defination of Kw

• Keq H2O Kw HOH-
• Ion product constant for water, Kw, is the
  product of the molar concentrations of H and OH-
  ions in pure water at a temperature of 298.15 K
• Kw HOH- 1.0x10-14 at 298.2 K

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The Definition of an Acidic Solution

• We define an acidic solution as one where the
  H in the solution is greater than the the H
  in pure water
• acidic solution ? H gt 1.0 x 10-7 mole/L at
  298.2 K

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The Definition of a Basic Solution

• Basic solutions are those where the H in the
  solution is less than its concentration in pure
  water at 298.2 K.
• Basic solution ? H lt 1.0 x 10-7 mole/L
• An alternative definition of a a basic solution
  is as follows
• Basic solution ? OH- gt 1.0 x 10-7 mole/L

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The Definition of a Neutral Solution

• A neutral solution is defined as one where the
  H in the solution is equal to the hydrogen ion
  concentration in pure water
• Neutral solution ? H OH- 1.0 x 10-7
  mole/L at 298.2 K!

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The Dependence of Kw on Temperature

• In our definitions of an acidic, basic, and a
  neutral solutions, we had explicitly stated the
  temperature as 298.2 K. Why?

Kw is temperature dependent

• How will that affect our definition of an acidic,
  basic, or a neutral solution?

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Neutrality at Body Temperature

• At T 310.15 K (physiological temperature) ? Kw
  2.4 x 10-14
• A neutral solution has H OH- (Kw)½
• At 310.15 K, a neutral solution is one where H
  OH- 1.5 x 10-7 M

(UNLESS OTHERWISE INDICATED, ALL CALCULATIONS
WILL BE AT 298.15 K)
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The pH scale
Sørenson - 1909 pH -log H
Solution Type H / M pH range
neutral solutions H OH- 1.0x10-7 pH 7.00
basic solutions H lt1.0x10-7 pH gt 7.00
acid solutions H gt1.0x10-7 pH lt 7.00
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The Relationship between pH and pOH

• pH -log H
• pOH -log OH-
• From the Kw expression
• Kw HOH- 1.0x10-14 at 298.2 K
• -log (1 x 10-14) -log H -log OH-

14.00 pH pOH
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Acid Strength and Dissociation

• CH3COOH(aq) ? CH3COO-(aq) H(aq)
• HCOOH(aq) ? HCOO-(aq) H(aq)
• both weak acids lt 5 ionized
• Other examples of weak acids ? HF, HNO2, HCN

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Acid Strength

• The strength of an acid is directly dependent on
  its dissociation (? value)
• For an acid

• n the number of groups that donate a proton

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Base Strength and Dissociation

• Strong Bases also 100 ionized in water
• NaOH(aq) Na(aq) OH-(aq)
• Ba(OH)2(aq) Ba2(aq) 2OH-(aq)
• Some bases are weak bases they dont ionize
  completely.
• NH3(aq) H2O(l) ? NH4 (aq) OH-(aq)
• lt 5 ionized in aqueous solution

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Base Strength

• The strength of a base is also directly dependent
  on its dissociation (? value)
• For a base

• baseo ? the original concentration of base
• m the number of basic groups in the molecule

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Conjugate Acid-Base Strengths

• CH3COOH (aq) ? CH3COO-(aq) H(aq)
• Note that the conjugate base of acetic acid is a
  reasonable proton acceptor
• CH3COO-(aq) H2O (l) ? CH3COOH (aq) OH-(aq)

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Other Examples

• HNO3 (aq) H(aq) NO3-(aq)
• conjugate base (very weak)
• HCOOH (aq) ? HCOO-(aq) H(aq)
• conjugate base is relatively strong
• NH3(aq) H2O(l) ? NH4 (aq) OH-(aq)
• relatively strong
  conjugate acid

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• HCl (aq) ? Cl-(aq) H(aq)
• The conjugate base, Cl- ion, is extremely weak
• Cl-(aq) H2O (l) ? HCl (aq) OH-(aq)

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• S2-(aq) H2O (l) ? HS-(aq) OH- (aq)
• The conjugate acid, the HS- ion, is extremely
  weak
• HS-(aq) ? H (aq) S2- (aq)
• The equilibrium lies very far to the left for
  this reaction

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• The greater the acid strength (large Ka), the
  weaker the conjugate base of that acid
• The weaker the acid (smaller Ka), the stronger
  its conjugate base
• If the base strength is high (Kb is large), its
  conjugate acid is very weak
• The weaker the base (small Kb value), the
  stronger the conjugate acid of the base

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Calculating the pH of Solution of Strong Acids

• For the dissolution of HCl, HI, or any of the
  other seven strong acids in water
• HCl (aq) ? H (aq) Cl- (aq)
• HI (aq) ? H (aq) I- (aq)

• The pH of these solutions is obtained from the
  molarity of the dissolved acid
• pH -log H -logHCl

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Calculating the pH of Solution of Strong Bases

• For the dissolution of NaOH, Ba(OH)2, or any of
  the other strong bases in water
• NaOH (aq) ? Na (aq) OH- (aq)
• Ba(OH)2 (aq) ? Ba2 (aq) 2OH- (aq)

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• The pH of these solutions is obtained by first
  calculating the pOH from the molarity of the
  dissolved base
• pOH -log OH- -logNaOH
• pOH -log OH- -log2 Ba(OH)2
• pH 14.00 - pOH

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The Seven Strong Acids

• chloric acid HClO3
• hydrobromic acid HBr
• hydrochloric acid HCl
• hydroiodic acid HI
• nitric acid HNO3
• perchloric acid HClO4
• sulphuric acid H2SO4
• What about the relative strength of the strong
  acids?

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The Leveling Effect

• H (aq) (or H3O(aq)) is the strongest acid that
  can exist in aqueous solution. Any acid stronger
  than H(aq) reacts with water completely to
  produce H(aq) and the weak conjugate base.

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• HNO3 (aq) is a stronger acid than H(aq) (H3O) \
  reacts with water completely to form H(aq)
• HNO3 (aq) ? H (aq) NO3- (aq)
• Acids weaker than H(aq) have the equilibrium
  lying primarily to the left.
• HNO2(aq) ? H(aq) NO2- (aq)

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• The OH- ion is the strongest base that can exist
  in aqueous solution. Bases stronger than
  OH-(aq) react with water to produce the hydroxide
  ion (OH-).

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• NH2- (the amide ion) is an extremely strong base
  (much stronger than OH-). Therefore,
• NaNH2 (aq) H2O (l) NH3 (aq) NaOH(aq)
• NH2- cannot exist in aqueous solution.
• NH3 is a much weaker base than OH-. Therefore,
  when it reacts with water, the equilibrium
  favours the reactants
• NH3 (aq) H2O (l) ?NH4 (aq) OH-(aq)

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The Leveling Effect Defined

• Any acid that is stronger than H(aq) means that
  we have 100 ionisation of the acid.
• For acids like HCl(aq), HClO4(aq), HNO3(aq), the
  appearance is one of equal acid strength.
• Water is said to have a levelling effect on the
  acid strength

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Equilibria in Aqueous Solutions of Weak Acids/
Weak Bases

• By definition, a weak acid or a weak base does
  not ionize completely in water (? ltlt100).
• How would we calculate the pH of a solution of a
  weak acid or a weak base in water?

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The Ka Value

• To obtain the pH of a weak acid solution, we must
  apply the principles of chemical equilibrium
• Define the acid dissociation constant Ka
• For a general weak acid reaction
• HA (aq) ? H (aq) A- (aq)

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Weak Acid/Bases and pH

• For a solution of hydrofluoric acid in water
• HF (aq) ? H (aq) F- (aq)

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Equilibria of Weak Bases in Water

• To calculate the percentage dissociation of a
  weak base in water (and the pH of the solutions)
  
• CH3NH2 (aq) H2O ? CH3NH3(aq) OH- (aq)
• We approach the problem as in the case of the
  weak acid above, i.e., from the chemical
  equilibrium viewpoint.

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The Kb Value

• Define the base dissociation constant Kb
• For a general weak base reaction with water
• B (aq) H2O (l) ? B (aq) OH- (aq)

• For the above system

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Diprotic/Polyprotic Acids

• Look at the following system.
• H2C2O4 (aq) ?HC2O4- (aq) H (aq) Ka1
• HC2O4- (aq) ?C2O42- (aq) H (aq) Ka2
• For the dissociation of diprotic and polyprotic
  acids, the magnitudes of the dissociation
  constants decrease in the direction
• Ka1 gt Ka2 gt Ka3 etc.

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Example

• For oxalic acid in water,
• Ka1 6.5 x 10-2
• Ka2 6.1 x 10-5
• Since Ka1gtgt Ka2, the H (and the pH) in the
  solution is due primarily to the first
  dissociation ONLY.

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Obtaining the Relationship between Ka and Kb

• We have already seen that there is a relationship
  between the strength of an acid and the ability
  of its conjugate base to hydrolyse.
• HCOOH (aq) ?HCOO- (aq) H (aq)
• Ka (HCOOH) 1.8 x 10-4
• Examine the reverse reaction, the hydrolysis
  (reaction of the substance with water) of HCOO-
• HCOO-(aq) H2O (l) ? HCOOH (aq) OH-(aq)

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Obtaining the Kb of the Conjugate Base

• HCOOH (aq) ? HCOO- (aq) H(aq)
• HCOO- (aq) H2O (l) ?HCOOH (aq) OH- (aq)
• Keq(1) Ka (HCOOH)
• Keq(2) Kb (HCOO-)
• Add the two reactions together
• HCOOH (aq) ? HCOO- (aq) H(aq)
• HCOO- (aq) H2O (l) ?HCOOH (aq) OH- (aq)

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• We are left with the overall reaction
• H2O (l) ?H (aq) OH- (aq)
• Kw H OH-
• From our rules for the equilibria of multiple
  reactions.
• Kw K (1) K (2)

Kw Kb Ka
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Variation of Conjugate base Strength with Ka

• HCOOH (aq) ?HCOO- (aq) H (aq)
• Ka (HCOOH) 1.8 x 10-4
• Kb (HCOO-) 5.6 x 10-11
• CH3COOH (aq) ?CH3COO- (aq) H (aq)
• Ka (CH3COOH) 1.8 x 10-5
• Kb (CH3COO-) 5.6 x 10-10

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Salts of Conjugate Bases

• Look at the dissolution of CH3COONa in water.
• CH3COONa (aq) ? Na (aq) CH3COO- (aq)
• But we know that the acetate ion, CH3COO- (aq),
  will hydrolyze in aqueous solution according to
  the following reaction.
• CH3COO- (aq) H2O (l) ?CH3COOH (aq) OH-(aq)

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Hydrolysis reaction produces OH-
Dissolving a salt of a strong base/ weak acid in
water ? basic solution.
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Salts of Conjugate Acids

• Look at the dissolution of NH4Cl in water
• NH4Cl (aq) ? NH4 (aq) Cl- (aq)
• But we know that the ammonium ion, NH4 (aq),
  will donate a proton aqueous solution
• NH4 (aq) ?NH3 (aq) H(aq)

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Hydrolysis reaction produces H(aq)
Dissolving a salt of a strong acid/ weak base in
water ? acidic solution.
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Both the cation and anion Hydrolyse

• What about salts in which both the cation and
  anion hydrolyze?
• The pH of the solution will depend on the
  magnitude of the Ka and the Kb values of the
  respective acidic and basic ions.

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Salts of Weak Acids/Strong Bases

• How do we prepare a solution of HCOONa?
• Titration of HCOOH with NaOH according to the
  following reaction

HCOOH (aq) NaOH (aq) ? HCOONa (aq) H2O (l)
? ?
? Weak Acid Strong Base
Basic Salt
Dissolution of the salt of a weak acid/strong
base produces a basic solution (pH gt 7.00).
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The Strong Acid/Weak Base Case

• How do we prepare a solution of NH4Cl?
• Titration of HCl with NH3 according to the
  following reaction

HCl (aq) NH3 (aq) ? NH4Cl (aq) ?
? ? Strong Weak
Acidic Acid Base Salt
Dissolution of the salt of a strong acid/weak
base produces a acidic solution (pH lt 7.00).
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The Weak Acid/Weak Base Case

• What would be the pH of a solution of CH3COONH4?
  
• Look at the following reactions
• NH4 (aq) ? NH3 (aq) H (aq)
• K Ka (NH4)
• CH3COO- (aq) H2O (l) ? CH3COOH (aq) OH- (aq)
  
• K Kb (CH3COO-)

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The pH of a solution the salt of a weak acid/weak
base depends on the magnitudes of the equilibrium
constants.