Chapter 15 Acids and Bases

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Chapter 15Acids and Bases
Chemistry II
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Stomach Acid Heartburn

• the cells that line your stomach produce
  hydrochloric acid
• to kill unwanted bacteria
• to help break down food
• to activate enzymes that break down food
• if the stomach acid backs up into your esophagus,
  it irritates those tissues, resulting in
  heartburn
• acid reflux
• GERD gastroesophageal reflux disease chronic
  leaking of stomach acid into the esophagus

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Curing Heartburn

• mild cases of heartburn can be cured by
  neutralizing the acid in the esophagus
• swallowing saliva which contains bicarbonate ion
• taking antacids that contain hydroxide ions
  and/or carbonate ions

4
Properties of Acids

• sour taste
• react with active metals
• i.e., Al, Zn, Fe, but not Cu, Ag, or Au
• 2 Al 6 HCl ? 2 AlCl3 3 H2
• Corrosive
• react with carbonates, producing CO2
• marble, baking soda, chalk, limestone
• CaCO3 2 HCl ? CaCl2 CO2 H2O
• change color of vegetable dyes
• blue litmus turns red
• react with bases to form ionic salts

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Common Acids
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Structure of Acids

• binary acids have H-atoms attached to a nonmetal
  atom
• e.g. HCl, HF

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Structure of Acids

• oxy acids have H-atoms attached to an O-atom

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Structure of Acids

• carboxylic acids have COOH group
• e.g. Acetic acid CH3COOH

9
Properties of Bases

• also known as alkalis
• taste bitter
• alkaloids plant product that is alkaline
• often poisonous
• solutions feel slippery
• change color of vegetable dyes
• different color than acid
• red litmus turns blue
• react with acids to form ionic salts
• neutralization

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Common Bases
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Structure of Bases

• most ionic bases contain OH- ions
• NaOH, Ca(OH)2
• some contain CO32- ions
• CaCO3 , NaHCO3
• CO32- 2H2O ? HCO3- H2O OH-
• HCO3- H2O OH- ? H2CO3 2OH-

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Indicators

• chemicals which change color depending on the
  acidity/basicity
• many vegetable dyes are indicators
• anthocyanins
• litmus
• from Spanish moss
• red in acid, blue in base
• phenolphthalein
• found in laxatives
• red in base, colorless in acid

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Definitions of Acids and Bases

• What are the main characteristics of molecules
  and ions that exhibit acid and base behavior?
• 3 different definitions
• Arrhenius
• Brønsted-Lowry
• Lewis

14
Arrhenius Theory (1880s)

• acids ionize in water to produce H ions and
  anions
• because molecular acids are not made of ions,
  they cannot dissociate
• they must be pulled apart, or ionized, by the
  water
• HCl(aq) ? H(aq) Cl(aq)
• CH3COOH(aq) ? H(aq) CH3COO(aq)
• bases dissociate in water to produce OH- ions and
  cations
• ionic substances dissociate in water
• NaOH(aq) ? Na(aq) OH(aq)

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Hydronium Ion

• the H ions produced by the acid are so reactive
  they cannot exist in water
• H ions are protons!!
• instead, they react with a water molecule(s) to
  produce complex ions, mainly hydronium ion, H3O
• H H2O ? H3O
• Chemists use H(aq) and H3O(aq) interchangeably

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Arrhenius Acid-Base Reactions

• the H from the acid combines with the OH- from
  the base to make a molecule of H2O
• H OH- ? H2O
• the cation from the base combines with the anion
  from the acid to make a salt
• Na Cl- ? NaCl
• HCl(aq) NaOH(aq) ? NaCl(aq) H2O(l)
• acid base ? salt water

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Problems with Arrhenius Theory

• does not explain why molecular substances, like
  NH3, dissolve in water to form basic solutions
  even though they do not contain OH ions
• does not explain how some ionic compounds, like
  Na2CO3 or Na2O, dissolve in water to form basic
  solutions even though they do not contain OH
  ions
• does not explain why molecular substances, like
  CO2, dissolve in water to form acidic solutions
  even though they do not contain H ions
• does not explain acid-base reactions that take
  place outside aqueous solution

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Brønsted-Lowry Theory (1923)

• in a Brønsted-Lowry Acid-Base reaction, an H is
  transferred
• Acid proton (H) donor
• Base proton (H) acceptor
• base structure must contain an atom with unshared
  pair of e-
• in an acid-base reaction, the acid molecule gives
  an H to the base molecule
• HA B ? A HB

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Brønsted-Lowry Acids

• Brønsted-Lowry acids are H donors
• Anything that has H can potentially be
  Brønsted-Lowry acid
• HCl(aq) is acidic because HCl transfers an H to
  H2O (base or proton acceptor), forming H3O ions

HCl(aq) H2O(l) ? Cl(aq) H3O(aq) acid
base
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Brønsted-Lowry Bases

• Brønsted-Lowry bases are H acceptors
• any material that has atoms with lone pairs can
  potentially be a Brønsted-Lowry base
• NH3(aq) is basic because NH3 accepts an H from
  H2O (acid, proton donor), forming OH(aq)

NH3(aq) H2O(l) ? NH4(aq) OH(aq) base
acid
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Amphoteric Substances

• amphoteric substances can act as either an acid
  or a base
• have both transferable H and atom with lone pair
• water acts as base, accepting H from HCl
• HCl(aq) H2O(l) ? Cl(aq) H3O(aq)
• water acts as acid, donating H to NH3
• NH3(aq) H2O(l) ? NH4(aq) OH(aq)

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Brønsted-Lowry Acid-Base Reactions

• one of the advantages of Brønsted-Lowry theory is
  that it allows reactions to be reversible
• HA B ? A HB
• the original base has an extra H after the
  reaction so it will act as an acid in the
  reverse process
• and the original acid has a lone pair of e-
  after the reaction so it will act as a base in
  the reverse process
• A HB ? HA B

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Conjugate Pairs

• In a Brønsted-Lowry Acid-Base reaction, the
  original base becomes an acid in the reverse
  reaction, and the original acid becomes a base in
  the reverse process
• each reactant and the product it becomes is
  called a conjugate pair
• the original base becomes the conjugate acid and
  the original acid becomes the conjugate base

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Brønsted-Lowry Acid-Base Reactions
HA B ? A HB
acid base conjugate conjugate base
acid
HCHO2 H2O ? CHO2 H3O acid
base conjugate conjugate base acid
H2O NH3 ? HO NH4 acid
base conjugate conjugate base acid
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Conjugate (joined) Pairs
In the reaction H2O NH3 ? OH NH4
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Ex 15.1a Identify the Brønsted-Lowry Acids and
Bases and Their Conjugates in the Reaction
H2SO4 H2O ? HSO4 H3O
When the H2SO4 becomes HSO4?, it lost an H ? so
H2SO4 must be the acid and HSO4? its conjugate
base
When the H2O becomes H3O, it accepted an H ? so
H2O must be the base and H3O its conjugate acid
H2SO4 H2O ? HSO4 H3O acid
base conjugate conjugate base
acid
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Ex 15.1b Identify the Brønsted-Lowry Acids and
Bases and Their Conjugates in the Reaction
HCO3 H2O ? H2CO3 HO
When the HCO3? becomes H2CO3, it accepted an H ?
so HCO3? must be the base and H2CO3 its conjugate
acid
When the H2O becomes OH?, it donated an H ? so
H2O must be the acid and OH? its conjugate base
HCO3 H2O ? H2CO3 HO base
acid conjugate conjugate acid
base
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Practice Write the formula for the conjugate
acid of the following bases
H2O NH3 CO32- H2PO4-
H3O
NH4
HCO3-
H3PO4
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Practice Write the formula for the conjugate
base of the following acids
H2O NH3 CO32- H2PO4-
OH-
NH2-
No H, it cannot be an acid
HPO42-
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Strong or Weak

• a strong acid is a strong electrolyte
• practically all the acid molecules ionize, ?
• a strong base is a strong electrolyte
• practically all the base molecules form OH ions,
  either through dissociation or reaction with
  water, ?
• a weak acid is a weak electrolyte
• only a small percentage of the molecules ionize,
  ?
• a weak base is a weak electrolyte
• only a small percentage of the base molecules
  form OH ions, either through dissociation or
  reaction with water, ?

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Strong Acids

• The stronger the acid, the more willing it is to
  donate H
• use water as the standard base
• strong acids donate practically all their Hs
• 100 ionized in water
• strong electrolyte
• H3O strong acid

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Weak Acids

• weak acids donate a small fraction of their Hs
• most of the weak acid molecules do not donate H
  to water
• much less than 1 ionized in water
• H3O ltlt weak acid

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Polyprotic Acids

• often acid molecules have more than one ionizable
  H these are called polyprotic acids
• the ionizable Hs may have different acid
  strengths or be equal
• 1 H monoprotic, 2 H diprotic, 3 H
  triprotic
• HCl monoprotic, H2SO4 diprotic, H3PO4
  triprotic
• polyprotic acids ionize in steps
• each ionizable H removed sequentially
• removing of the first H automatically makes
  removal of the second H harder
• H2SO4 is a stronger acid than HSO4?

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Increasing Basicity
Increasing Acidity
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Strengths of Acids Bases

• commonly, acid or base strength is measured by
  determining the equilibrium constant of a
  substances reaction with water
• HA H2O ? A- H3O
• B H2O ? HB OH-
• the farther the equilibrium position lies to the
  products, the stronger the acid or base
• the position of equilibrium depends on the
  strength of attraction between the base form and
  the H
• stronger attraction means stronger base or weaker
  acid

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General Trends in Acidity

• the stronger an acid is at donating H, the
  weaker the conjugate base is at accepting H
• higher oxidation number stronger oxyacid
• H2SO4 gt H2SO3 HNO3 gt HNO2
• cation stronger acid than neutral molecule
  neutral stronger acid than anion
• H3O gt H2O gt OH- NH4 gt NH3 gt NH2-
• base trend opposite

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Acid Ionization Constant, Ka

• acid strength measured by the size of the
  equilibrium constant when react with H2O
• HA H2O ? A- H3O
• acid ionization constant, Ka
• larger Ka stronger acid

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Autoionization of Water

• Water is actually an extremely weak electrolyte
• therefore there must be a few ions present
• about 1 out of every 10 million water molecules
  form ions through a process called autoionization
• H2O ? H OH
• H2O H2O ? H3O OH
• all aqueous solutions contain both H3O and OH
• the concentration of H3O and OH are equal in
  water
• H3O OH 10-7M _at_ 25C

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Ion Product of Water

• the product of H3O and OH concentrations is
  always the same n
• the number is called the ion product of water -
  symbol, Kw
• Kw H3O OH 1 x 10-14 _at_ 25C
• if you measure one of the concentrations, you can
  calculate the other
• as H3O increases the OH must decrease so
  the product stays constant
• inversely proportional

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Acidic and Basic Solutions

• all aqueous solutions contain both H3O and OH
  ions
• neutral solutions have equal H3O and OH
• H3O OH 1 x 10-7
• acidic solutions have a larger H3O than OH
• H3O gt 1 x 10-7 OH lt 1 x 10-7
• basic solutions have a larger OH than H3O
• H3O lt 1 x 10-7 OH gt 1 x 10-7

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Example 15.2b Calculate the OH? at 25C when
the H3O 1.5 x 10-9 M, and determine if the
solution is acidic, basic, or neutral
H3O 1.5 x 10-9 M OH?
Given Find
Concept Plan Relationships
Solution
Check
The units are correct. The fact that the H3O
lt OH? means the solution is basic
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Complete the TableH vs. OH-
H 100 10-1 10-3 10-5 10-7
10-9 10-11 10-13 10-14
OH-
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Complete the TableH vs. OH-
Acid
Base
H 100 10-1 10-3 10-5 10-7
10-9 10-11 10-13 10-14
OH-10-14 10-13 10-11 10-9 10-7
10-5 10-3 10-1 100
even though it may look like it, neither H nor
OH- will ever be 0
the sizes of the H and OH- are not to scale
because the divisions are powers of 10 rather
than units
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pH

• the acidity/basicity of a solution is often
  expressed as pH
• pH -logH3O, H3O 10-pH
• exponent on 10 with a positive sign
• pHwater -log10-7 7
• need to know the H concentration to find pH
• pH lt 7 is acidic pH gt 7 is basic, pH 7 is
  neutral

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Sig. Figs. Logs

• when you take the log of a number written in
  scientific notation, the digit(s) before the
  decimal point come from the exponent on 10, and
  the digits after the decimal point come from the
  decimal part of the number
• log(2.0 x 106) log(106) log(2.0)
• 6 0.30303 6.30303...
• since the part of the scientific notation number
  that determines the significant figures is the
  decimal part, the sig figs are the digits after
  the decimal point in the log
• log(2.0 x 106) 6.30

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Question

• Complete the following table of pH values

H pH Significant Figures
1 x 10-7 7.0 1
1.0 x 10-7
6.80
4.39
1.78 x 10-11
2
7.00
1.6 x 10-7
2
4.1 x 10-5
2
3
10.750
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pH

• the lower the pH, the more acidic the solution
  the higher the pH, the more basic the solution
• 1 pH unit corresponds to a factor of 10
  difference in acidity
• normal range 0 to 14
• pH can be negative (very acidic) or larger than
  14 (very alkaline)

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pH of Common Substances
Substance pH
1.0 M HCl 0.0
0.1 M HCl 1.0
stomach acid 1.0 to 3.0
lemons 2.2 to 2.4
soft drinks 2.0 to 4.0
plums 2.8 to 3.0
apples 2.9 to 3.3
cherries 3.2 to 4.0
unpolluted rainwater 5.6
human blood 7.3 to 7.4
egg whites 7.6 to 8.0
milk of magnesia (satd Mg(OH)2) 10.5
household ammonia 10.5 to 11.5
1.0 M NaOH 14
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Example 15.3b Calculate the pH at 25C when the
OH? 1.3 x 10-2 M, and determine if the
solution is acidic, basic, or neutral
OH? 1.3 x 10-2 M pH
Given Find
Concept Plan Relationships
Solution
Check
pH is unitless. The fact that the pH gt 7 means
the solution is basic
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pOH

• another way of expressing acidity/basicity of a
  solution is pOH
• pOH -logOH?, OH? 10-pOH
• pOHwater -log10-7 7
• need to know the OH? concentration to find pOH
• pOH lt 7 is basic pOH gt 7 is acidic, pOH 7 is
  neutral

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pH and pOH Complete the Table
pH
H 100 10-1 10-3 10-5 10-7
10-9 10-11 10-13 10-14
OH-10-14 10-13 10-11 10-9 10-7
10-5 10-3 10-1 100
pOH
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pH and pOHComplete the Table
pH 0 1 3 5 7
9 11 13 14
H 100 10-1 10-3 10-5 10-7
10-9 10-11 10-13 10-14
OH-10-14 10-13 10-11 10-9 10-7
10-5 10-3 10-1 100
pOH 14 13 11 9 7
5 3 1 0
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Relationship between pH and pOH

• the sum of the pH and pOH of a solution 14.00
• at 25C
• can use pOH to find pH of a solution

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pK

• a way of expressing the strength of an acid or
  base is pK
• pKa -log(Ka), Ka 10-pKa
• pKb -log(Kb), Kb 10-pKb
• the stronger the acid, the smaller the pKa
• larger Ka smaller pKa
• because it is the log

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Finding the pH of a Strong Acid

• there are two sources of H3O in an aqueous
  solution of a strong acid the acid and the
  water
• for the strong acid, the contribution of the
  water to the total H3O is negligible
• shifts the Kw equilibrium to the left so far that
  H3Owater is too small to be significant
• except in very dilute solutions, generally lt 1 x
  10-4 M
• for a monoprotic strong acid H3O HA
• for polyprotic acids, the other ionizations can
  generally be ignored
• 0.10 M HCl has H3O 0.10 M and pH 1.00

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Finding the pH of a Weak Acid

• there are also two sources of H3O in and aqueous
  solution of a weak acid the acid and the water
• however, finding the H3O is complicated by the
  fact that the acid only undergoes partial
  ionization
• calculating the H3O requires solving an
  equilibrium problem for the reaction that defines
  the acidity of the acid
• HA H2O ? A? H3O

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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution
_at_ 25C
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the H3O from water is 0
HNO2 H2O ? NO2? H3O
HNO2 NO2- H3O
initial
change
equilibrium
HNO2 NO2- H3O
initial 0.200 0 0
change
equilibrium
since no products initially, Qc 0, and the
reaction is proceeding forward
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution
_at_ 25C
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
HNO2 NO2- H3O
initial 0.200 0 0
change
equilibrium
?x
x
x
x
x
0.200 ?x
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution
_at_ 25C
Ka for HNO2 4.6 x 10-4
determine the value of Ka from Table 15.5
since Ka is very small, approximate the HNO2eq HNO2init and solve for x
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.200 x x
0.200 ?x
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution
_at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.200 x x
check if the approximation is valid by seeing if x lt 5 of HNO2init
x 9.6 x 10-3
the approximation is valid
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution
_at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.200-x x x
substitute x into the equilibrium concentration definitions and solve
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.190 0.0096 0.0096
x 9.6 x 10-3
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution
_at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.190 0.0096 0.0096
substitute H3O into the formula for pH and solve
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Ex 15.6 Find the pH of 0.200 M HNO2(aq) solution
_at_ 25C
Ka for HNO2 4.6 x 10-4
HNO2 NO2- H3O
initial 0.200 0 0
change -x x x
equilibrium 0.190 0.0096 0.0096
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
though not exact, the answer is reasonably close
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Practice - What is the pH of a 0.012 M solution
of nicotinic acid, HC6H4NO2? (Ka 1.4 x 10-5 _at_
25C)
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Practice - What is the pH of a 0.012 M solution
of nicotinic acid, HC6H4NO2?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the H3O from water is 0
HC6H4NO2 H2O ? C6H4NO2? H3O
HA A- H3O
initial 0.012 0 0
change
equilibrium
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Practice - What is the pH of a 0.012 M solution
of nicotinic acid, HC6H4NO2?
HC6H4NO2 H2O ? C6H4NO2? H3O
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
HA A- H3O
initial 0.012 0 0
change
equilibrium
?x
x
x
x
x
0.012 ?x
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Practice - What is the pH of a 0.012 M solution
of nicotinic acid, HC6H4NO2? Ka 1.4 x 10-5 _at_
25C
HC6H4NO2 H2O ? C6H4NO2? H3O
determine the value of Ka
since Ka is very small, approximate the HAeq HAinit and solve for x
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012 x x
0.012 ?x
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Practice - What is the pH of a 0.012 M solution
of nicotinic acid, HC6H4NO2? Ka 1.4 x 10-5 _at_
25C
Ka for HC6H4NO2 1.4 x 10-5
check if the approximation is valid by seeing if x lt 5 of HC6H4NO2init
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012 x x
x 4.1 x 10-4
the approximation is valid
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Practice - What is the pH of a 0.012 M solution
of nicotinic acid, HC6H4NO2? Ka 1.4 x 10-5 _at_
25C
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012-x x x
substitute x into the equilibrium concentration definitions and solve
x 4.1 x 10-4
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Practice - What is the pH of a 0.012 M solution
of nicotinic acid, HC6H4NO2? Ka 1.4 x 10-5 _at_
25C
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012 0.00041 0.00041
substitute H3O into the formula for pH and solve
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Practice - What is the pH of a 0.012 M solution
of nicotinic acid, HC6H4NO2? Ka 1.4 x 10-5 _at_
25C
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
HA A2- H3O
initial 0.012 0 0
change -x x x
equilibrium 0.012 0.00041 0.00041
the values match
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Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution _at_ 25C
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the H3O from water is 0
HClO2 H2O ? ClO2? H3O
HClO2 ClO2- H3O
initial 0.100 0 0
change
equilibrium
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Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution _at_ 25C
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
HClO2 ClO2- H3O
initial 0.100 0 0
change -x x x
equilibrium 0.100-x x x
80
Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution _at_ 25C
Ka for HClO2 1.1 x 10-2
determine the value of Ka from Table 15.5
since Ka is very small, approximate the HClO2eq HClO2init and solve for x
HClO2 ClO2- H3O
initial 0.100 0 0
change -x x x
equilibrium 0.100-x x x
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Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution _at_ 25C
Ka for HClO2 1.1 x 10-2
HClO2 ClO2- H3O
initial 0.100 0 0
change -x x x
equilibrium 0.100-x x x
check if the approximation is valid by seeing if x lt 5 of HNO2init
x 3.3 x 10-2
the approximation is invalid
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Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution _at_ 25C
Ka for HClO2 1.1 x 10-2
if the approximation is invalid, solve for x using the quadratic formula
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Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution _at_ 25C
Ka for HClO2 1.1 x 10-2
substitute x into the equilibrium concentration definitions and solve
HClO2 ClO2- H3O
initial 0.100 0 0
change -x x x
equilibrium 0.100-x x x
HClO2 ClO2- H3O
initial 0.100 0 0
change -x x x
equilibrium 0.072 0.028 0.028
x 0.028
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Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution _at_ 25C
Ka for HClO2 1.1 x 10-2
HClO2 ClO2- H3O
initial 0.100 0 0
change -x x x
equilibrium 0.072 0.028 0.028
substitute H3O into the formula for pH and solve
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Ex 15.7 Find the pH of 0.100 M HClO2(aq)
solution _at_ 25C
Ka for HClO2 1.1 x 10-2
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
HClO2 ClO2- H3O
initial 0.100 0 0
change -x x x
equilibrium 0.072 0.028 0.028
the answer matches
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Ex 15.8 - What is the Ka of a weak acid if a
0.100 M solution has a pH of 4.25?
Use the pH to find the equilibrium H3O
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations and H3Oequil
HA H2O ? A? H3O
HA A- H3O
initial
change
equilibrium
HA A- H3O
initial 0.100 0 0
change
equilibrium 5.6E-05
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Ex 15.8 - What is the Ka of a weak acid if a
0.100 M solution has a pH of 4.25?
HA H2O ? A? H3O
fill in the rest of the table using the H3O as a guide
if the difference is insignificant, HAequil HAinitial
substitute into the Ka expression and compute Ka
HA A- H3O
initial 0.100 0 0
change
equilibrium
5.6E-05
5.6E-05
-5.6E-05
0.100 ? 5.6E-05
5.6E-05
5.6E-05
0.100
88
Percent Ionization

• another way to measure the strength of an acid is
  to determine the percentage of acid molecules
  that ionize when dissolved in water this is
  called the percent ionization
• the higher the percent ionization, the stronger
  the acid

• since ionized acidequil H3Oequil

89
Ex 15.9 - What is the percent ionization of a 2.5
M HNO2 solution?
Write the reaction for the acid with water
Construct an ICE table for the reaction
Enter the Initial Concentrations
Define the Change in Concentration in terms of x
Sum the columns to define the Equilibrium Concentrations
HNO2 H2O ? NO2? H3O
HNO2 NO2- H3O
initial
change
equilibrium
HNO2 NO2- H3O
initial 2.5 0 0
change
equilibrium
?x
x
x
x
x
2.5 ? x
90
Ex 15.9 - What is the percent ionization of a 2.5
M HNO2 solution?
Ka for HNO2 4.6 x 10-4
determine the value of Ka from Table 15.5
since Ka is very small, approximate the HNO2eq HNO2init and solve for x
HNO2 NO2- H3O
initial 2.5 0 0
change -x x x
equilibrium 2.5-x 2.5 x x
91
Ex 15.9 - What is the percent ionization of a 2.5
M HNO2 solution?
HNO2 H2O ? NO2? H3O
substitute x into the Equilibrium Concentration definitions and solve
HNO2 NO2- H3O
initial 2.5 0 0
change -x x x
equilibrium 2.5 0.034 0.034
x 3.4 x 10-2
2.5 ? x
x
x
92
Ex 15.9 - What is the percent ionization of a 2.5
M HNO2 solution?
HNO2 H2O ? NO2? H3O
Apply the Definition and Compute the Percent Ionization
HNO2 NO2- H3O
initial 2.5 0 0
change -x x x
equilibrium 2.5 0.034 0.034
since the percent ionization is lt 5, the x is
small approximation is valid
93
Relationship Between H3Oequilibrium
HAinitial

• increasing the initial concentration of acid
  results in increased H3O concentration at
  equilibrium
• increasing the initial concentration of acid
  results in decreased percent ionization
• this means that the increase in H3O
  concentration is slower than the increase in acid
  concentration

94
Why doesnt the increase in H3O keep up with the
increase in HA?

• the reaction for ionization of a weak acid is
• HA(aq) H2O(l) ? A-(aq) H3O(aq)
• according to Le Châteliers Principle, if we
  reduce the concentrations of all the (aq)
  components, the equilibrium should shift to the
  right to increase the total number of dissolved
  particles
• we can reduce the (aq) concentrations by using a
  more dilute initial acid concentration
• the result will be a larger H3O in the dilute
  solution compared to the initial acid
  concentration
• this will result in a larger percent ionization

95
Finding the pH of Mixtures of Acids

• generally, you can ignore the contribution of the
  weaker acid to the H3Oequil
• for a mixture of a strong acid with a weak acid,
  the complete ionization of the strong acid
  provides more than enough H3O to shift the
  weak acid equilibrium to the left so far that the
  weak acids added H3O is negligible
• for mixtures of weak acids, generally only need
  to consider the stronger for the same reasons
• as long as one is significantly stronger than the
  other, and their concentrations are similar

96
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Write the reactions for the acids with water and determine their Kas
If the Kas are sufficiently different, use the strongest acid to construct an ICE table for the reaction
Enter the initial concentrations assuming the H3O from water is 0
HF H2O ? F? H3O Ka 3.5 x 10-4
HClO H2O ? ClO? H3O Ka 2.9 x 10-8
H2O H2O ? OH? H3O Kw 1.0 x 10-14
HF F- H3O
initial 0.150 0 0
change
equilibrium
97
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
HF F- H3O
initial 0.150 0 0
change
equilibrium
?x
x
x
x
x
0.150 ?x
98
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF 3.5 x 10-4
determine the value of Ka for HF
since Ka is very small, approximate the HFeq HFinit and solve for x
HF F- H3O
initial 0.150 0 0
change -x x x
equilibrium 0.150 x x
0.150 ?x
99
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF 3.5 x 10-4
HF F- H3O
initial 0.150 0 0
change -x x x
equilibrium 0.150 x x
check if the approximation is valid by seeing if x lt 5 of HFinit
x 7.2 x 10-3
the approximation is valid
100
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF 3.5 x 10-4
HF F- H3O
initial 0.150 0 0
change -x x x
equilibrium 0.150-x x x
HF F- H3O
initial 0.150 0 0
change -x x x
equilibrium 0.143 0.0072 0.0072
substitute x into the equilibrium concentration definitions and solve
x 7.2 x 10-3
101
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF 3.5 x 10-4
HF F- H3O
initial 0.150 0 0
change -x x x
equilibrium 0.143 0.0072 0.0072
substitute H3O into the formula for pH and solve
102
Ex 15.10 Find the pH of a mixture of 0.150 M
HF(aq) solution and 0.100 M HClO2(aq)
Ka for HF 3.5 x 10-4
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
HF F- H3O
initial 0.150 0 0
change -x x x
equilibrium 0.143 0.0072 0.0072
though not exact, the answer is reasonably close
103
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104
Strong Bases

• the stronger the base, the more willing it is to
  accept H
• use water as the standard acid
• for strong bases, practically all molecules are
  dissociated into OH or accept Hs
• strong electrolyte
• multi-OH strong bases completely dissociated

105
Example 15.11b Calculate the pH at 25C of a
0.0015 M Sr(OH)2 solution and determine if the
solution is acidic, basic, or neutral
Sr(OH)2 1.5 x 10-3 M pH
Given Find
Concept Plan Relationships
OH?2Sr(OH)2
Solution
OH? 2(0.0015) 0.0030 M
Check
pH is unitless. The fact that the pH gt 7 means
the solution is basic
106
Example 15.11b (alternative) Calculate the pH
at 25C of a 0.0015 M Sr(OH)2 solution and
determine if the solution is acidic, basic, or
neutral
Sr(OH)2 1.5 x 10-3 M pH
Given Find
Concept Plan Relationships
OH?2Sr(OH)2
Solution
OH? 2(0.0015) 0.0030 M
Check
pH is unitless. The fact that the pH gt 7 means
the solution is basic
107
Practice - Calculate the pH of a 0.0010 M Ba(OH)2
solution and determine if it is acidic, basic,
or neutral
108
Practice - Calculate the pH of a 0.0010 M Ba(OH)2
solution and determine if it is acidic, basic,
or neutral
Ba(OH)2 Ba2 2 OH- therefore OH- 2 x
0.0010 0.0020 2.0 x 10-3 M
Kw H3OOH?
pH -log H3O -log (5.0 x 10-12) pH 11.30
pH gt 7 therefore basic
109
Weak Bases

• in weak bases, only a small fraction of molecules
  accept Hs
• weak electrolyte
• most of the weak base molecules do not take H
  from water
• much less than 1 ionization in water
• HO ltlt weak base
• finding the pH of a weak base solution is similar
  to finding the pH of a weak acid

110
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111
Structure of Amines
112
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Write the reaction for the base with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the OH? from water is 0
NH3 H2O ? NH4 OH?
NH3 NH4 OH?
initial
change
equilibrium
NH3 NH4 OH?
initial 0.100 0 0
change
equilibrium
since no products initially, Qc 0, and the
reaction is proceeding forward
113
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
NH3 NH4 OH?
initial 0.100 0 0
change
equilibrium
?x
x
x
x
x
0.100 ?x
114
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 1.76 x 10-5
determine the value of Kb from Table 15.8
since Kb is very small, approximate the NH3eq NH3init and solve for x
NH3 NH4 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 x x
0.100 ?x
115
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 1.76 x 10-5
NH3 NH4 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 x x
check if the approximation is valid by seeing if x lt 5 of NH3init
x 1.33 x 10-3
the approximation is valid
116
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 1.76 x 10-5
substitute x into the equilibrium concentration definitions and solve
NH3 NH4 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 ?x x x
NH3 NH4 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.099 1.33E-3 1.33E-3
x 1.33 x 10-3
117
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 1.76 x 10-5
use the OH- to find the H3O using Kw
substitute H3O into the formula for pH and solve
NH3 NH4 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.099 1.33E-3 1.33E-3
118
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 1.76 x 10-5
use the OH- to find the pOH
use pOH to find pH
NH3 NH4 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.099 1.33E-3 1.33E-3
119
Ex 15.12 Find the pH of 0.100 M NH3(aq) solution
Kb for NH3 1.76 x 10-5
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
NH3 NH4 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.099 1.33E-3 1.33E-3
though not exact, the answer is reasonably close
120
Practice Find the pH of a 0.0015 M morphine
solution, Kb 1.6 x 10-6
121
Practice Find the pH of a 0.0015 M morphine
solution
Write the reaction for the base with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the OH? from water is 0
B H2O ? BH OH?
B BH OH?
initial 0.0015 0 0
change
equilibrium
since no products initially, Qc 0, and the
reaction is proceeding forward
122
Practice Find the pH of a 0.0015 M morphine
solution
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
B BH OH?
initial 0.0015 0 0
change
equilibrium
?x
x
x
x
x
0.0015 ?x
123
Practice Find the pH of a 0.0015 M morphine
solution
Kb for Morphine 1.6 x 10-6
determine the value of Kb
since Kb is very small, approximate the Beq Binit and solve for x
B BH OH?
initial 0.0015 0 0
change -x x x
equilibrium 0.0015 x x
0.0015 ?x
124
Practice Find the pH of a 0.0015 M morphine
solution
Kb for Morphine 1.6 x 10-6
check if the approximation is valid by seeing if x lt 5 of Binit
B BH OH?
initial 0.0015 0 0
change -x x x
equilibrium 0.0015 x x
x 4.9 x 10-5
the approximation is valid
125
Practice Find the pH of a 0.0015 M morphine
solution
Kb for Morphine 1.6 x 10-6
substitute x into the equilibrium concentration definitions and solve
B BH OH?
initial 0.0015 0 0
change -x x x
equilibrium 0.0015 ?x x x
B BH OH?
initial 0.0015 0 0
change -x x x
equilibrium 0.0015 4.9E-5 4.9E-5
x 4.9 x 10-5
126
Practice Find the pH of a 0.0015 M morphine
solution
Kb for Morphine 1.6 x 10-6
use the OH- to find the H3O using Kw
substitute H3O into the formula for pH and solve
B BH OH?
initial 0.0015 0 0
change -x x x
equilibrium 0.0015 4.9E-5 4.9E-5
127
Practice Find the pH of a 0.0015 M morphine
solution
Kb for Morphine 1.6 x 10-6
use the OH- to find the pOH
use pOH to find pH
B BH OH?
initial 0.0015 0 0
change -x x x
equilibrium 0.0015 4.9E-5 4.9E-5
128
Practice Find the pH of a 0.0015 M morphine
solution
Kb for Morphine 1.6 x 10-6
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
B BH OH?
initial 0.0015 0 0
change -x x x
equilibrium 0.0015 4.9E-5 4.9E-5
the answer matches the given Kb
129
Acid-Base Properties of Salts

• salts are water soluble ionic compounds
• salts that contain the cation of a strong base
  and an anion that is the conjugate base of a weak
  acid are basic
• NaHCO3 solutions are basic
• Na is the cation of the strong base NaOH
• HCO3- is the conjugate base of the weak acid
  H2CO3
• salts that contain cations that are the conjugate
  acid of a weak base and an anion of a strong acid
  are acidic
• NH4Cl solutions are acidic
• NH4 is the conjugate acid of the weak base NH3
• Cl- is the anion of the strong acid HCl

130
Acid-Base Properties of SaltsAnions as Weak Bases

• every anion can be thought of as the conjugate
  base of an acid
• therefore, every anion can potentially be a base
• A-(aq) H2O(l) ? HA(aq) OH-(aq)
• the stronger the acid is, the weaker the
  conjugate base is
• an anion that is the conjugate base of a strong
  acid is pH neutral
• Cl-(aq) H2O(l) ? HCl(aq) OH-(aq)
• since HCl is a strong acid, this equilibrium lies
  practically completely to the left
• an anion that is the conjugate base of a weak
  acid is basic
• F-(aq) H2O(l) ? HF(aq) OH-(aq)
• since HF is a weak acid, the position of this
  equilibrium favors the right

131
Ex 15.13 - Use the Table to Determine if the
Given Anion Is Basic or Neutral

• NO3-
• the conjugate base of a strong acid, therefore
  neutral
• NO2-
• the conjugate base of a weak acid, therefore
  basic

132
Relationship between Ka of an Acid and Kb of Its
Conjugate Base

• many reference books only give tables of Ka
  values because Kb values can be found from them

when you add equations, you multiply the Ks
133
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq)
solution
Na is the cation of a strong base pH neutral. The CHO2- is the anion of a weak acid pH basic
Write the reaction for the anion with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the OH? from water is 0
CHO2- H2O ? HCHO2 OH?
CHO2- HCHO2 OH?
initial 0.100 0 0
change
equilibrium
134
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq)
solution
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
Calculate the value of Kb from the value of Ka of the weak acid from Table 15.5
substitute into the equilibrium constant expression
CHO2- HCHO2 OH?
initial 0.100 0 0
change
equilibrium
x
x
?x
0.100 ?x
x
x
135
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq)
solution
Kb for CHO2- 5.6 x 10-11
since Kb is very small, approximate the CHO2-eq CHO2-init and solve for x
CHO2- HCHO2 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 x x
0.100 ?x
136
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq)
solution
Kb for CHO2- 5.6 x 10-11
CHO2- HCHO2 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 x x
check if the approximation is valid by seeing if x lt 5 of CHO2-init
x 2.4 x 10-6
the approximation is valid
137
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq)
solution
Kb for CHO2- 5.6 x 10-11
CHO2- HCHO2 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 -x x x
CHO2- HCHO2 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 2.4E-6 2.4E-6
substitute x into the equilibrium concentration definitions and solve
x 2.4 x 10-6
138
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq)
solution
Kb for CHO2- 5.6 x 10-11
use the OH- to find the H3O using Kw
substitute H3O into the formula for pH and solve
CHO2- HCHO2 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 2.4E-6 2.4E-6
139
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq)
solution
Kb for CHO2- 5.6 x 10-11
CHO2- HCHO2 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 2.4E-6 2.4E-6
use the OH- to find the pOH
use pOH to find pH
140
Ex 15.14 Find the pH of 0.100 M NaCHO2(aq)
solution
Kb for CHO2- 5.6 x 10-11
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Kb to the given Kb
CHO2- HCHO2 OH?
initial 0.100 0 0
change -x x x
equilibrium 0.100 2.4E-6 2.4E-6
though not exact, the answer is reasonably close
141
Polyatomic Cations as Weak Acids

• some cations can be thought of as the conjugate
  acid of a base
• others are the counterions of a strong base
• therefore, some cation can potentially be an acid
• MH(aq) H2O(l) ? MOH(aq) H3O(aq)
• the stronger the base is, the weaker the
  conjugate acid is
• a cation that is the counterion of a strong base
  is pH neutral
• a cation that is the conjugate acid of a weak
  base is acidic
• NH4(aq) H2O(l) ? NH3(aq) H3O(aq)
• since NH3 is a weak base, the position of this
  equilibrium favors the right

142
Metal Cations as Weak Acids

• cations of small, highly charged metals are
  weakly acidic
• alkali metal cations and alkali earth metal
  cations pH neutral
• cations are hydrated
• Al(H2O)63(aq) H2O(l) ? Al(H2O)5(OH)2 (aq)
  H3O(aq)

143
Ex 15.15 - Determine if the Given Cation Is
Acidic or Neutral

• C5N5NH2
• the conjugate acid of a weak base, therefore
  acidic
• Ca2
• the counterion of a strong base, therefore
  neutral
• Cr3
• a highly charged metal ion, therefore acidic

144
Classifying Salt Solutions asAcidic, Basic, or
Neutral

• if the salt cation is the counterion of a strong
  base and the anion is the conjugate base of a
  strong acid, it will form a neutral solution
• NaCl Ca(NO3)2 KBr
• if the salt cation is the counterion of a strong
  base and the anion is the conjugate base of a
  weak acid, it will form a basic solution
• NaF Ca(C2H3O2)2 KNO2

145
Classifying Salt Solutions asAcidic, Basic, or
Neutral

• if the salt cation is the conjugate acid of a
  weak base and the anion is the conjugate base of
  a strong acid, it will form an acidic solution
• NH4Cl
• if the salt cation is a highly charged metal ion
  and the anion is the conjugate base of a strong
  acid, it will form an acidic solution
• Al(NO3)3

146
Classifying Salt Solutions asAcidic, Basic, or
Neutral

• if the salt cation is the conjugate acid of a
  weak base and the anion is the conjugate base of
  a weak acid, the pH of the solution depends on
  the relative strengths of the acid and base
• Ion with higher K value dominates
• NH4F NH4, F-
• NH4 is conjugate acid of weak base (NH3)
  acidic
• F- is conjugate base of weak acid basic
• Ka of NH4 (5.68 x 10-10) is larger than Kb of
  the F- (2.9 x 10-11) therefore the solution will
  be acidic

147
Ex 15.16 - Determine whether a solution of the
following salts is acidic, basic, or neutral

• SrCl2
• Sr2 is the counterion of a strong base, pH
  neutral
• Cl- is the conjugate base of a strong acid, pH
  neutral
• solution will be pH neutral
• AlBr3
• Al3 is a small, highly charged metal ion, weak
  acid
• Cl- is the conjugate base of a strong acid, pH
  neutral
• solution will be acidic
• CH3NH3NO3
• CH3NH3 is the conjugate acid of a weak base,
  acidic
• NO3- is the conjugate base of a strong acid, pH
  neutral
• solution will be acidic

148
Ex 15.16 - Determine whether a solution of the
following salts is acidic, basic, or neutral

• NaCHO2
• Na is the counterion of a strong base, pH
  neutral
• CHO2- is the conjugate base of a weak acid,
  basic
• solution will be basic
• NH4F
• NH4 is the conjugate acid of a weak base,
  acidic
• F- is the conjugate base of a weak acid, basic
• Ka(NH4) gt Kb(F-) solution will be acidic

149
Polyprotic Acids

• since polyprotic acids ionize in steps, each H
  has a separate Ka
• Ka1 gt Ka2 gt Ka3
• generally, the difference in Ka values is great
  enough so that the second ionization does not
  happen to a large enough extent to affect the pH
• most pH problems just do first ionization
• except H2SO4 ? use H2SO4 as the H3O for the
  second ionization
• A2- Ka2 as long as the second ionization is
  negligible

150
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151
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution _at_ 25C
H2SO4 H2O ? HSO4? H3O
Write the reactions for the acid with water
Construct an ICE table for the reaction
Enter the initial concentrations assuming the HSO4- and H3O is H2SO4
HSO4? H2O ? SO42? H3O
HSO4 ? SO42 ? H3O
initial 0.0100 0 0.0100
change
equilibrium
152
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution _at_ 25C
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
HSO4 ? SO42 ? H3O
initial 0.0100 0 0.0100
change -x x x
equilibrium 0.0100 -x x 0.0100 -x
153
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution _at_ 25C
Ka for HSO4- 0.012
expand and solve for x using the quadratic formula
154
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution _at_ 25C
Ka for HSO4- 0.012
substitute x into the equilibrium concentration definitions and solve
HSO4 ? SO42 ? H3O
initial 0.0100 0 0.0100
change -x x x
equilibrium 0.0100 -x x 0.0100 -x
HSO4 ? SO42 ? H3O
initial 0.0100 0 0.0100
change -x x x
equilibrium 0.0055 0.0045 0.0145
x 0.0045
155
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution _at_ 25C
Ka for HSO4- 0.012
substitute H3O into the formula for pH and solve
HSO4 ? SO42 ? H3O
initial 0.0100 0 0.0100
change -x x x
equilibrium 0.0055 0.0045 0.0145
156
Ex 15.18 Find the pH of 0.0100 M H2SO4(aq)
solution _at_ 25C
Ka for HSO4- 0.012
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
HSO4 ? SO42 ? H3O
initial 0.0100 0 0.0100
change -x x x
equilibrium 0.0055 0.0045 0.0145
the answer matches
157
Strengths of Binary Acids

• the more d H-X d- polarized the bond, the more
  acidic the bond
• the stronger the H-X bond, the weaker the acid
• binary acid strength increases to the right
  across a period
• H-C lt H-N lt H-O lt H-F
• binary acid strength increases down the column
• H-F lt H-Cl lt H-Br lt H-I

158
Strengths of Oxyacids, H-O-Y

• the more electronegative the Y atom, the stronger
  the acid
• helps weakens the H-O bond
• the more oxygens attached to Y, the stronger the
  acid
• further weakens and polarizes the H-O bond

159
Lewis Acid - Base Theory

• electron sharing
• electron donor Lewis Base nucleophile
• must have a lone pair of electrons
• electron acceptor Lewis Acid electrophile
• electron deficient
• when Lewis Base gives electrons from lone pair to
  Lewis Acid, a covalent bond forms between the
  molecules
• Nucleophile Electrophile ? NucleophileElectrop
  hile
• product called an adduct
• other acid-base reactions also Lewis

160
Example - Lewis Acid-Base ReactionsLabel the
Nucleophile and Electrophile
OH-1 ?
161
Practice - Lewis Acid-Base ReactionsLabel the
Nucleophile and Electrophile

• BF3 HF ?
• CaO SO3 ?
• KI I2 ?

162
Practice - Lewis Acid-Base ReactionsLabel the
Nucleophile and Electrophile

• BF3 HF ? HBF4-
• CaO SO3 ? Ca2SO4-2
• KI I2 ? KI3

163
What Is Acid Rain?

• natural rain water has a pH of 5.6
• naturally slightly acidic due mainly to CO2
• rain water with a pH lower than 5.6 is called
  acid rain
• acid rain is linked to damage in ecosystems and
  structures

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What Causes Acid Rain?

• many natural and pollutant gases dissolved in the
  air are nonmetal oxides
• CO2, SO2, NO2
• nonmetal oxides are acidic
• CO2 H2O ? H2CO3
• 2 SO2 O2 2 H2O ? 2 H2SO4
• processes that produce nonmetal oxide gases as
  waste increase the acidity of the rain
• natural volcanoes and some bacterial action
• man-made combustion of fuel
• weather patterns may cause rain to be acidic in
  regions other than where the nonmetal oxide is
  produced

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pH of Rain in Different Regions
166
Sources of SO2 from Utilities
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Damage from Acid Rain

• acids react with metals, and materials that
  contain carbonates
• acid rain damages bridges, cars, and other
  metallic structures
• acid rain damages buildings and other structures
  made of limestone or cement
• acidifying lakes affecting aquatic life
• dissolving and leaching more minerals from soil
• making it difficult for trees

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Acid Rain Legislation

• 1990 Clean Air Act attacks acid rain
• force utilities to reduce SO2
• result is acid rain in northeast stabilized and
  beginning to be reduced

169
Damage from Acid Rain