AES and ESS 2004 Inorganic Chemistry Acids and Bases

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AES and ESS 2004Inorganic ChemistryAcids and
Bases

• Acidity is a concept that we have all heard
• about. It is important to us at a biological
  level
• in that we know that if water becomes too
• acidic or alkaline (basic) it can kill off
  aquatic
• life. This can sometimes be the result of acid
  rain or
• industry polluting water. But what exactly are
• these terms acid or base, can we measure them and
• make quantitative predictions.

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Outline

• History
• Arrhenius definition
• Strong and weak acids and bases
• The periodic table and acidity
• Brønsted-Lowry theory
• pH
• pKa
• Lewis Definitions

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History

• Acids and bases have been with us since the dawn
  of history. Acids such as vinegar tasted sour,
  whilst alkalis (bases) have a soapy feel. The
  French Chemist Lavasior proposed that all acids
  contained oxygen. This was shown to be wrong.
  Then it was suggested that all acids contained
  hydrogen but it was not until the Swedish chemist
  Arrenhius came up with a workable definition in
  the late 19th century that a model which could be
  used in calculations became available.

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Arrhenius definition

• An acid is a molecule or polyatomic ion that
  contains hydrogen and reacts with water to
  produce hydronium ions.
• HCl(aq) H2O(l) ? H3O(aq) Cl-(aq)
  . (hydronium
  ion)

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Examples

• Cl-H(aq) H2O(l) ? H3O(aq) Cl-(aq)
• (Hydrochloric acid HCl)
• HOO2S-OH(aq) H2O(l) ? H3O(aq) HSO4-(aq)
• (Sulphuric Acid H2SO4)
• H3CCOO-H(aq) H2O(l) ?H3O(aq) H3CO2-(aq)
• (Acetic Acid H3CCO2H)
• In practice we find that for a hydrogen to react
  with
• water to form a hydronium ion the hydrogen must
  be attached to an electronegative element i.e.
  O, N, F, Cl, Br or I. So if we look at acetic
  acid the maximum number of hydrogen atoms that
  can react is one (How many can react in sulphuric
  acid (H2SO4)?

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Definition of a base

• A base was defined a molecule that produces
  hydroxide ions in water. When an ionic compound
  such as NaOH dissolves in water there is no
  problem in identifying the source of the OH-
  NaOH(aq)
  (H2O)(l) ? Na(aq) OH-(aq)
• For some molecules however the OH- ion has to be
  produced indirectly
  NH3(aq) H2O(l) ? NH4(aq)
  OH-(aq)

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Strong and Weak Acids and Bases

• Since we are talking about how much a molecule is
  dissociated we need to think in terms of
  equilibria.
• Cl-H(aq) H2O(l) H3O(aq)Cl-(aq)
• H3CCO2H(aq)H2O(l) H3O(aq)H3CCO2-(a
  q)

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Table of strong acids and bases

• Strong acids Strong bases
• Hydrochloric acid HCl Group 1 hydroxides
• Hydrobromic acid HBr M(OH)2
• Hydroiodic acid HI M Ca, Sr, Ba
• Nitric acid HONO2 Group 1 Oxides
• Perchloric acid HOClO3 MO
• Chloric acid HOClO2 M Ca, Sr, Ba
• Sulphuric acid HOSO2OH
• (to -SO3OH)

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• Strong acids are those in the previous table and
  are completely dissociated, weak acids (all
  others) are not completely dissociated. Strong
  bases are completely ionized in water (see above)
  Weak bases (not in table) are not completely
  ionized in water.
• The metal oxides of group 1 and MO (M Ca, Sr,
  Ba) are strong bases. This is because the oxide
  ions react completely with water
  
• O2- (aq) H2O(l) 2OH-(aq)
• An example of a weak base is an amine e.g. H3CNH2
  This behaves in the same way as ammonia in water
  
  H3CNH2(aq) H2O(l)
  H3CNH3(aq)OH-(aq)

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Acids and Bases in the context of the periodic
table

• Most solutions of soluble metal oxides give basic
  solutions. But many non-metal oxides react with
  water to give acidic solutions.
• CO2(g) H2O(l) H2CO3 (aq)
  (carbonic acid)
• SO2(g) H2O(l) H2SO3(aq) (sulphurous
  acid)
• P4O10(s) H2O(l) H3PO4(aq) (phosphoric
  acid)
• Many gaseous non-metal oxides SO3, SO2, NO2
  dissolve in atmospheric water so give an acidic
  solution and hence acid rain.
• (remember that in these acids the hydrogen atom
  is attached to the oxygen atoms)

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Metals and Metalloids

• But what about the borderline between metals and
  non-metals. What we find is that these oxides
  can behave booth as acids and bases. e.g.
• Al2O3(s) 6HCl(aq) ? 2AlCl3(aq) 3H2O(l) base
  acid
• Al2O3(s) 2NaOH(aq) ? 2NaAl(OH)4)(aq)
  acid base
• We call such a compound amphoteric. The elements
  that show these properties lie at the frontier
  between metals and non-metals

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The frontier zoneA AcidB BaseG
AmphotericThe acid/base character of oxides of
main group elements
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Summary
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Self test

• Will the following compounds produce acidic or
  basic solutions.
• SeO2, CaO, As 2O5, Ga2O3

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A quantitative look at acids and bases

• The theory that is used in assisting us is the
  Brønsted-Lowry theory. According to this theory
• Brønsted acids
• HCl(aq) H2O(l) H3O(aq) Cl-(aq)
• Here the HCl donates a proton to the water
  molecule. Since it is strong acid nearly all the
  protons are donated.
• HOPO32-(aq) H2O(l) HOPO33-(aq)
  H3O(aq)
• Here HOPO32-(aq) is a weak acid and only a few
  of the protons are donated.

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• Brønsted-Lowry Bases
• A Base is a proton acceptor
• CaO(aq) H2O(l) Ca2(aq) 2OH-(aq)
• The small highly charged O2- ion pulls a proton
  from a water molecule
• O2-(aq) H2O(l) 2OH-(aq)
• Since it is a strong base it goes to completion
• NH3(aq) H2O(l) NH4(aq)
  OH-(aq)
• The neutral NH3 molecule has less proton
  pulling power than O2- so far fewer NH3 react so
  that the equilibrium does not go to completion,
  and so this is an example of a weak base.

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The role of the solvent (water)

• We know that it can either donate of accept a
  proton
• O2-(aq) H2O(l) 2OH-(aq)
• HCl(aq) H2O(l) H3O(aq) Cl-(aq)
• And we say it is amphiprotic
• What would happen if we brought 2 water molecules
  together.
• H2O H-OH H3O OH-
  accepts donates
• This reaction occurs it is an equilibrium
  reaction and it is called the

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• 2H2O(l) H3O(aq) OH-(aq)
• The equilibrium lies well over to the left We
  know this because water is such a poor conductor
  of electricity.
• Looking at the equilibrium constant
• Kc H3O OH-
• H2O2
• Because the water is present in such a great
  excess and the reaction lies so far over to the
  left we can ignore the concentration of water
  and define a new constant Kw
• Kc H3O OH-
• This is called the autoprotylysis constant
• Kw 1x10-7-1x10-7 10-14.

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• What this means is that the product of the
  concentration of H3O and OH- is fixed and has a
  value of 10-14. So as H3O goes up OH- goes
  down

Taken from Smith and Jones without permission)
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• Example
• What are the molarities of H3O and OH- in a
  0.005 M Ba(OH)2 (aq) solution
• Ba(OH)2(aq) Ba2(aq) 2OH-(aq)
• So 0.005 moles of Ba(OH)2 produces 0.01 moles of
  OH- ions in solution.
• Since 1 x 10-14 H3OOH-
• H3O 1 x 10-14 1 x 10-14 1 x 10-12
• OH- 0.01
• H3O 1 x 10-12 mol dm-3
• OH- 0.01 mol dm-3

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• Self question
• Estimate the molarities of H3O and OH- in
• 5 x 10-4 M HNO3

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Revision of Logs

• The logarithm (or log) of a number to the base 10
  is the power that the number 10 has to be raised
  to in order to equal that number.
• So 100 102
• Log100 2
• since 0.001 1 x 10-3
• Log 0.001 -3
• So what is log 200
• Log 200 10x
• On a calculator we enter the number and press log
• Log 200 2.301

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• But what if you are given the log of a number (x)
  how do you find that number (x)
• To do this we use 10x. On many calculators this
  can be done by entering 10
• Then xy then entering 2.301
• 102.301 200
• We use log scales when the value of what we are
  interested in changes by several orders of
  magnitude. This is the case in acids and bases
  as for a 0.1 M HCl solution H3O 10-1 but for
  a 0.1M NaOH solution the concentration of H3O is
  10-13 The scale we use to do this is the pH
  scale.

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Figure 15.8 Very large ranges of numbers are
difficult to represent graphically. However,
their logarithms span a much smaller range and
can be represented easily. Note how the numbers
shown here range over 10 orders of magnitude
(from 10?4 to 106), but their logarithms range
over 10 units (from ?4 to 6). Negative values of
logarithms correspond to numbers between 0 and 1
positive values correspond to numbers greater
than 1.(Taken from Atkins and Jones Molecules
Matter and change without permission)
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The pH Scale

• pH is defined as
• pH - log H3O
• Thus for a 0.1m HCl solution
• pH -log 0.01 1
• For a 0.1 m NaOH solution
• pH -log 1 x 10-13 13
• So we can say

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Typical pH values of common aqueous solutions.
The brown regions indicate the pH for liquids
regarded as corrosive.(Taken from Atkins and
Jones Molecules matter and change without
permission)

• We can generalise pH to pX
• Such that
• pX -logX
• So pOH - log OH-
• and
• pKw - logKw

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Acidity Constants

• Since an acid reacts with water to give hydronium
  ions we have an equilibrium and in the case of a
  weak acid
• H3CCO2H(aq)H2O(l) H3O(aq)
• 
  H3CCO2-(aq)
• Kc H3CCO2-H3O
• H3CCO2HH2O
• Since the solution is dilute and the solvent is
  almost pure (and un dissociated) water the
  concentration of the water does not change to
  any significant extent so we can define a new
  constant Ka

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• Ka H3CCO2-H3O
• H3CCO2H
• For acetic acid Ka at 25 C is 1.77 x 10-5 so in
  H3CCO2H very few of the molecules have
  undergone dissociation.

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Kb

• We can also do the same for a base
• NH3(aq) H2O(l) NH4(aq) OH-(aq)
• Kc NH4OH-
• NH3H2O
• Kb NH4OH-
• NH3

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pKa and pKb

• These acidity and basicisity constant can be
  used in a pX scale where
• pKa - log Ka and pKb - -log Kb
• So the lower the value of pKa the stronger the
  acid
• So the lower the value of pKb the stronger the
  base
• Acids pKa Bases pKb
• Sulphurous H2SO3 1.81 ammonia NH3 4.75
• Acetic H3CCO2H 4.75 nicotine C10H14N2
  5.98
• Phenol C6H5OH 9.89 urea CO(NH2)2 13.90

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Calculation of pH from pKa

• Calculate the pH of a 0.1 M solution of acetic
  acid given that pKa 4.752
• pKa -log Ka
• Ka 10 4.752 1.770 x 10-5
• So Ka H3CCO2-H3O 1.770 x 10-5
• H3CCO2H
• We need to find H3O
• If the amount of H3CCO2H that has dissociated is
  x then
• x moles of H3CCO2H and x moles of H3O have been
  formed and there are (0.1-x) moles of H3CCO2H
  left.

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• 1.77 x 10 -5 xx/0.1 x
• Now in reality Ka values are only accurate to
  about 5 so if we stay within 5 we can make
  approximations. The approximation we use is that
  is most weak acids are less than 5 dissociated.
  This means that that the concentration of the
  acid does not change a great deal and instead of
  using 0.1-x in the equation above we can use 0.1.
  So the equation becomes (You can always make
  this approximation with a weak acid).
• 1.77 x 10-5 xx/0.1
• 1.77 x 10-6 x2
• x 1.33 x 10-3 M
• pH -logH3O - log 1.33 x 10-3 2.876
• (we ought to check that the acid is less than5
  dissociated)
• (1.33x10-3/0.1)x 100 1.33 so our
  approximation is valid

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• Self test question
• What is the pH of a 0.3 M solution of benzoic
  acid given that the pKa value for benzoic acid is
  4.19
• (answer is 2.356 )

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Lewis acids and bases

• There is one further definition of acids and
  bases that is useful and that is due to Lewis
  (The same Lewis as of Lewis structures)
• Lewis definition
• An acid is electron pair acceptor.
• A base is an electron pair donor.
• The advantage of this is that it get away from
  hydrogen and solvents. Two examples will show
  this

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Examples

• Cl- H(g) NH3(g) ? (NH4)(Cl -) (s)
• Here the nitrogen of the ammonia molecule donates
  a pair of electrons to the hydrogen and the
  electron pair that was in the H-Cl bond ends up
  on the Cl, so a Cl- ion is produced and the
  final product (NH4)(Cl -) is formed.
• Cl3B(l) NH3(l) ? Cl3B?NH3(s)
• Cl3B has an empty p orbital on it that can accept
  a lone pair from the nitrogen of the ammonia
  molecule. To form a dative bond