Nonmetal oxides react with water to make solutions acidic

1
Nonmetal and Metal Oxides

• Nonmetal oxides react with water to make
  solutions acidic
• CO2(g) H2O(l ) H2CO3(aq)
• H2CO3(aq) H(aq) HCO3-(aq)
• HCO3-(aq) H(aq)
  CO32-(aq)
• 2SO2(g) O2(g) 2SO3(g)
• SO3(g) H2O(l) H2SO4(aq)
• H2SO4(aq) H(aq) HSO4-(aq)
• HSO4-(aq) H(aq)
  SO42-(aq)
• 2NO2(g) H2O(l) HNO3(aq) HNO2(aq)
• HNO3(aq) H(aq) NO3-(aq)
• HNO2 (aq) H (aq) NO2-
  (aq)

2
Nonmetal and Metal Oxides

• Metal oxides react with water to produce basic
  solutions
• CaCO3(s) CaO(s) CO2(g)
• limestone lime
• CaO(s) H2O(l) Ca(OH)2(s)
  Ca(OH)2(s) is soluble in water
• lime slaked lime
• The Hydrogen Ion in water
• H represents a hydrogen atom that has lost its
  electron, and really is a bare proton
• In water, however, H is surrounded by at least
  one, and probably 4, H2O molecules
• Thus hydrogen ion in water is usually represented
  as H3O
• It is likely the formula is H9O4, but this is
  not what chemists usually use to represent the
  hydrated proton

3
Oxidation Numbers

• Oxidation Numbers this is a method of assigning
  a charge to an atom in a molecule or an ionic
  substance
• In molecules, the oxidation number convention
  treats the bonding between an atom and other
  atoms as if it were ionic.
• Oxidation numbers are used to help balance
  oxidation-reduction equations, in naming
  compounds and in some trends in the chemical
  properties of compounds.
• Changes in oxidation number of an atom or ion in
  a chemical equation allow the identification of
  oxidation reduction reactions
• To calculate oxidation numbers of atoms in
  molecules or ions, we use a self-consistent set
  of arbitrary rules.

4
Oxidation Numbers

• The rules
• The oxidation number of an element in its
  elemental form is zero.
  N in N2 has ox. no. 0, Na in Na(s) has ox. no.
  0, P in P4 has ox. no. 0
• The oxidation number of a monatomic ion is its
  charge Cl- has
  ox. no. -1, Ca2 has ox. no. 2
• The sum of the oxidation numbers of all the atoms
  in a neutral compound is zero.
• The sum of the oxidation numbers in a polyatomic
  ion is the charge of the ion. The charge on a
  polyatomic ion is its net oxidation number.
• F always has oxidation number -1
• Group 1A elements (except H) always have
  oxidation number 1
• Group 2A elements always have oxidation number 2
• Group 7A elements have oxidation number -1 in
  binary compounds NaCl, FeCl3
• Oxygen has oxidation number -2 except when bonded
  to F or is peroxide (O22-,-1) or superoxide
  (O2-, )
• Hydrogen has oxidation number 1 except when
  bonded to metals where its -1 NaH

5
Oxidation Numbers

• Oxidation Numbers Examples
• Give the oxidation numbers of each element in the
  following compounds
• SOCl2 Cl, -1 O, -2 S, 4
• S8 S, 0
• ClO4- O, -2 Cl, 7
• NH3 H, 1 N, -3 MnO4- O, -2, Mn, 7
• N2O O, -2 N 1 CrO42-, O, -2, Cr, 6
• NO O, -2 N, 2 Cr2O72- O, -2, Cr, -6
• N2O3, O, -2 N, 3 CrO5 CrO(O2)2 the 1 O, 2
  
• NO2 O, -2 N, 4 the 4 Os in O22-, -1
• N2O5 O, -2 N, 5 the Cr, 6
• HNO3 O, -2 H, 1, N, 5
• HNO2 O, -2 H, 1 N, 3
• HReCl3 Cl, -1 H, -1 Re, 4

6

• MnO4- produces MnO2(s) in neutral or basic
  solution
• CO produces CO2 when reacting with Metal oxides
  such as Fe2O3(s) yielding the metal

7
Oxidation Reduction Reactions

• Recognizing oxidation reduction reactions can be
  accomplished by examining the oxidation numbers
  of each species undergoing chemical change
• This is likely to be tedious!
• A better first step is to try to recognize the
  presence of a substance known to be an oxidizing
  agent or reducing agent
• 2K(s) Br2(l ) 2KBr(s)
• K gives up 1 e- per atom K K is
  oxidized, thus, is the reducing agent
• Br2 gains 2 e- per molecule 2Br- Br2 is
  reduced and is the oxidizing agent
• Zn(s) 2H(aq) Zn2(aq) H2(g)
• Zn is oxidized giving up 2 e- per atom
• H(aq) is reduced, gaining 1 e- per atom
• Cu(s) 2NO3-(aq) 4H(aq) Cu2(aq)
  2NO2(g) 2H2O(l ) -conc. HNO3
• Cu is oxidized and is the reducing agent
• NO3- is reduced and is the oxidizing agent N
  gains 1 e-
• Remember Cu H(aq) No Reaction

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Classification of Reaction types

• Combustion reactions
• For compounds of C and H or C, H and H, complete
  combustion produces CO2 and H2O
• Equations for combustion reactions can be
  balanced by inspection
• Metathesis reactions, also called double
  replacement or exchange reactions involve the
  ions of one compound exchanging places with ions
  of another compound
• The driving force associated with these reactions
  may be
• Formation of a precipitate, an insoluble
  substance
• Formation of a weak electrolyte or nonelectrolyte
• Formation of a gas
• Oxidation reduction reactions involve an electron
  exchange process in which one substance gives up
  electrons and is oxidized and another substance
  gains electrons and is reduced
• The oxidizing agent is reduced as it gains
  electrons
• The reducing agent is oxidized as it loses
  electrons

9
Solutions
Concentration units for stoichiometry
calculations involving solutions Molarity(M) E
xample What is the molarity of a solution made
by dissolving 14.20 g NaCl in enough water to
make 250.0 mL of solution? Step 1 Find the
number of moles of solute, NaCl Step 2 Find the
volume of solution in liters Step 3 Find the
molarity by calculating the moleliter ratio
10
Solutions

• Molarity, volume and moles
• From the definition of molarity the equation can
  be rearranged go calculate moles from molarity
  and volume of solution
• moles molarity x volume
• Example calculate the number of moles NaCl in
  31.5 mL 1.08 M NaCl
• moles NaCl
• From the definition of molarity the equation can
  be rearranged to calculate the volume from
  molarity and moles
• Example calculate the volume of 1.08 M NaCl
  solution that contains 2.00 mol NaCl

11
Solutions

• Preparing solutions by dilution involves
  measuring a known volume of a solution of known
  concentration and adding enough solvent to bring
  the total volume to a new, known volume
• Making use of the defining equation for molarity
• molarity
• The moles of solute in a given volume of solution
  of known concentration can be calculated
• moles molarity x volume or mol M x V
• When a given volume of solution of known molarity
  is placed in another container, the number of
  moles is not changed, even after additional
  solvent is added
• Therefore, Mconc soln x Vconc oln Msolute Mdil
  soln x Vdil soln

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Solutions

• Example If 25.00 mL of 0.09719 M NaCl is added
  to a 100.0 mL volumetric flask and diluted to
  volume, what is the new concentration?

13
Solution Stoichiometry
When solutions of reactants are used,
concentrations and volumes give moles of
reactants To find moles of solute in a solution
of known volume Example how many moles of
NaOH are required to react with 25.0 mL of
0.250 M HCl? NaOH HCl NaCl H2O
14
Solution Stoichiometry
Example how many mL of 0.300 M HCl are required
to neutralize 6.25 x 10-3 mol NaOH? NaOH
HCl NaCl H2O
15
Titrations

• A titration is a method for finding the volume of
  one solution that contains a substance that
  reacts with another substance in a second
  solution.
• If we know
• the number of moles of the second substance
• the balanced equation for the reaction
• the volume of the first solution that exactly
  reacts with the amount of reactant in the
  second solution
• We can calculate the concentration of the first
  solution
• One of the solutions used in a titration is a
  standard solution whose concentration is
  accurately and precisely known.

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Titrations

• The titration procedure involves very carefully
  adding from a buret one of the solutions to the
  second solution.
• The solution in the buret is the titrant.
• The addition is stopped when the volume of
  solution added contains a stoichiometric
  quantity of reactant
• the amount of reactant in the titrant will
  be exactly the amount
  required to consume the substance in the second
  solution.
• The equivalence point is the point in a titration
  when a stoichiometric amount of titrant has
  been added to the second solution
• The titration procedure often uses an indicator.
• An indicator is a substance added to the solution
  titrated that will change color hopefully when
  the equivalence point is reached
• The end point in a titration is the point where
  the indicator changes color.

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Titrations
Example If 45.3 mL of 0.100 M NaOH is required
to exactly neutralize 75.0 mL of H2SO4, what is
the molarity of the H2SO4? H2SO4 2NaOH
Na2SO4 2H2O 2 mol NaOH?1 mol H2SO4
18
Titrations
Example Photo developing laboratories recover
silver from the solutions used to process
film. A 50.0 mL sample from a 5.00 x 103 liter
tank is analyzed by titration with Cl-,
requiring 18.3 mL of 0.100 M Cl- for the
titration. How many grams of Ag are in the
tank? Ag(aq) Cl- AgCl(s) 1 mol
Cl-?1 mol Ag